The Lax equation and weak regularity of asymptotic estimate Lie groups

Abstract

We investigate the Lax equation in the context of infinite-dimensional Lie algebras. Explicit solutions are discussed in the sequentially complete asymptotic estimate context, and an integral expansion (sums of iterated Riemann integrals over nested commutators with correction term) is derived for the situation that the Lie algebra is inherited by an infinite-dimensional Lie group in Milnor’s sense. In the context of Banach Lie groups (and Lie groups with suitable regularity properties), we generalize the Baker–Campbell–Dynkin–Hausdorff formula to the product integral (with additional nilpotency assumption in the non-Banach case). We combine this formula with the results obtained for the Lax equation to derive an explicit representation of the product integral in terms of the exponential map. An important ingredient in the non-Banach case is an integral transformation that we introduce. This transformation maps continuous Lie algebra-valued curves to smooth ones and leaves the product integral invariant. This transformation is also used to prove a regularity statement in the asymptotic estimate context.

1 Introduction

We investigate regularity properties of Lie groups in Milnor’s sense [3, 19] and extend the Baker–Campbell–Dynkin–Hausdorff formula for the exponential map to the product integral. The product integral generalizes the Riemann integral (for curves in Hausdorff locally convex vector spaces) to Lie groups, so that Lie algebra-valued curves are “integrated” to Lie group elements. The results are based on a deeper analysis of the Lax equation in the context of asymptotic estimate and sequentially complete Lie algebras. A further ingredient is an integral transformation that we introduce. This transformation maps continuous Lie algebra-valued curves to smooth ones and leaves the product integral invariant.

Explicitly, let G be a Lie group in Milnor’s sense¹ [3, 19] that is modeled over the Hausdorff locally convex vector space E. We denote the Lie algebra of G by $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$, the adjoint action by $\text {Ad} :G\times \mathfrak {g}\rightarrow \mathfrak {g}$, and set $\text {Ad} _g:=\text {Ad} (g,\cdot )$ for each $g\in G$. The left and the right translation by some $g\in G$ is denoted by $\text {L} _g$ and $\text {R} _g$, respectively. For $a<b$, we set

$$\begin{aligned} C^1_*([a,b],G):=\{\mu \in C^{1}([a,b],G)\,|\,\mu (a)=e\}. \end{aligned}$$

The evolution map is the inverse of the right logarithmic derivative²             ($a<b$)

and is denoted by

The product integral is defined by

and for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, we set

We equip $C^k([0,1],\mathfrak {g})$ for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ with the $C^k$-topology, as well as $\mathfrak {D}\cap C^k([0,1],\mathfrak {g})$ with the corresponding subspace topology³:

• We say that G is $C^k$-semiregular if $C^k([0,1],\mathfrak {g})\subseteq \mathfrak {D}$ holds.

• We say that G is weakly $C^k$-regular if G is $C^k$-semiregular and ${\text {evol} }_k$ is differentiable.

• We say that G is $C^k$-regular if G is $C^k$-semiregular and ${\text {evol} }_k$ is smooth.

We briefly want to report on the continuity problem (under which circumstances is ${\text {evol} }_k$ continuous w.r.t. the $C^k$-topology):

• In [8], this problem had been solved for the case $k=0$. Specifically, it was shown that $C^0$-continuity of ${\text {evol} }_0$ on its domain (also if $C^0$-semiregularity is not assumed) is equivalent to local $\mu$-convexity of G [4, 8]. Roughly speaking, local $\mu$-convexity is a generalized triangle inequality for the Lie group multiplication.

• In [9], the continuity problem was completely solved in the asymptotic estimate context that we also consider in this paper (simply put, Theorem 1 in [9] states that all continuity notions are equivalent in the asymptotic estimate context).

A Lie group is said to be asymptotic estimate if its Lie algebra is asymptotic estimate. More generally, we have the following definitions for an infinite-dimensional Lie algebra $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$⁴

• A subset $\text {M} \subseteq \mathfrak {q}$ is said to be an $\text {AE}$-set if for each $\textsf {v} \in \textsf {Sem}(\mathfrak {q})$, there exists $\textsf {v} \le \textsf {w} \in \textsf {Sem}(\textsf {q} )$ with

  $$\begin{aligned} \textsf {v} (\varvec{[}X_1,\varvec{[}X_2,\varvec{[}\dots ,\varvec{[}X_n,Y\varvec{]}{\dots }\varvec{]}\varvec{]}])\le \textsf {w} (X_1)\cdot {\dots }\cdot \textsf {w} (X_n)\cdot \textsf {w} (Y) \end{aligned}$$

  for all $X_1,\dots ,X_n,Y\in \text {M}$, and $n\ge 1$.

• The Lie algebra $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ is said to be asymptotic estimate if $\mathfrak {q}$ is an $\text {AE}$-set.

For instance, abelian Lie groups are asymptotic estimate, and the same is true for Lie groups with nilpotent Lie algebras. Also Banach Lie groups are asymptotic estimate, because their Lie bracket is submultiplicative. Notably, the class of asymptotic estimate Lie algebras has good permanence properties, as it is closed under passage to subalgebras, Hausdorff quotient Lie algebras, as well as closed under arbitrary Cartesian products (hence, e.g., under projective limits).

As already indicated above, Theorem 1 in [9] states that if G is asymptotic estimate, then G is locally $\mu$-convex if and only if ${\text {evol} }_k$ is $C^k$-continuous for some (and then each) $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$. Based on the semiregularity results obtained in [8], this statement was used in [9] to prove that $C^\infty$-regularity implies $C^k$-regularity for each $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, whereby for $k=0$ additional completeness assumptions (on the Lie group not the Lie algebra) are required. Complementary to that, we prove the following theorem in the weakly regular context (cf. Theorem 2):

Theorem I

Assume that $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is asymptotic estimate and sequentially complete. If G is weakly $C^\infty$-regular, then G is weakly $C^k$-regular for each $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$.

This theorem is obtained by a comprehensive analysis of the Lax equation in the sequentially complete asymptotic estimate context, as well as by application of an integral transformation that we introduce. As a by-product of these investigations, we obtain a generalization of the Baker–Campbell–Dynkin–Hausdorff formula (BCDH formula) in the Banach case (Proposition 4) as well as in the nilpotent weakly $C^k$-regular case (Theorem 1). In particular, this yields an explicit formula for the product integral in terms of the exponential map that involves iterated Riemann integrals over nested commutators (see Point (a)). More explicitly:

Given an infinite-dimensional Lie algebra $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ (not necessarily the Lie algebra of a Lie group), we denote its completion by $\widehat{\mathfrak {q}}$ and set

$$\begin{aligned} \text {ad} _{Z}:\mathfrak {q}\rightarrow \mathfrak {q},\qquad Y\mapsto \varvec{[}Z, Y \varvec{]} \end{aligned}$$

for each $Z\in \mathfrak {q}$. For $\psi \in C^0([a,b],\mathfrak {q})$ and $X\in \mathfrak {q}$, we define the maps ($\ell \ge 1$)

$$\begin{aligned} \textstyle \varvec{\lambda }_{0,\psi }[X]:[a,b]\rightarrow \widehat{\mathfrak {q}},\qquad&t\mapsto X\\ \textstyle \varvec{\lambda }_{\ell ,\psi }[X]:[a,b]\rightarrow \widehat{\mathfrak {q}},\qquad&\textstyle t\mapsto \int _a^{t}\text {d} s_1\int _a^{s_1}\text {d} s_2 \, {\dots } \int _a^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(X), \end{aligned}$$

and set

$$\begin{aligned} \qquad \varvec{\lambda }_{\ell ,\psi }[t]:\mathfrak {q}\rightarrow \widehat{\mathfrak {q}},\qquad X\mapsto \varvec{\lambda }_{\ell ,\psi }[X](t)\qquad \qquad \forall \, t\in [a,b],\, \ell \in \mathbb {N}. \end{aligned}$$

(1)

We have the following results:

• (a) Let G be a Banach Lie group such that the norm on $\mathfrak {g}$ fulfills $\Vert [X,Y]\Vert \le \Vert X\Vert \cdot \Vert Y\Vert$ for all $X,Y\in \mathfrak {g}$. Then, there exists $\mathfrak {r}>0$ such that             (Corollary 9 in Sect. 4.1)

  

  (2)

  holds for all $a<b$ and $\phi \in C^0([a,b],\mathfrak {g})$ with $\int _a^b \Vert \phi (s)\Vert \, \text {d} s< \mathfrak {r}$.

  The same formula is proven in the nilpotent weakly $C^k$-regular context (see Corollary 10 in Sect. 4.2.1), whereby the occurring sums are finite. For instance, if G is abelian and weakly $C^k$-regular for some $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, then (2) specializes to the well-known formula [18]

  

  Note that (2) provides an explicit formula for holonomies in principal bundles, as such holonomies are locally given by product integrals of curves that are pairings of (smooth) connections with derivatives of (smooth) curves in the base manifold of the principal bundle.

• (b) The maps (1) are the elementary building blocks of solutions to the Lax equation

  $$\begin{aligned} {\dot{\alpha }}=\varvec{[}\psi , \alpha \varvec{]}\qquad \quad \text {with initial condition}\qquad \quad \alpha (a)=X, \end{aligned}$$

  (3)

  where $X\in \mathfrak {q}$, $\psi \in C^0([a,b],\mathfrak {q})$ ($a<b$) are fixed parameters, and $\alpha \in C^1([a,b],\mathfrak {q})$ is the solution.

• Assume that $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})=(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is the Lie algebra of a Lie group G:

  • For $\psi \in \mathfrak {D}_{[a,b]}$, the unique solution to (3) is given by (cf. Corollary 7).

  • For $\psi \in \mathfrak {D}_{[a,b]}$ and $\ell \ge 1$, define

    

    A straightforward induction shows (cf. Lemma 13)

    

    In particular, if $\text {M}$ is an $\text {AE}$-set with $\text {im} [\psi ]\subseteq \text {M}$, then we have (cf. Corollary 8)

    

    (4)

    Notably, for $\psi :[0,1]\ni t\mapsto Y\in \mathfrak {g}$ constant, (4) reproduces the well-known formula

    

• Assume that $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ is sequentially complete and asymptotic estimate:

  Mimicking (4), for $\psi \in C^0([a,b],\mathfrak {q})$ and $X\in \mathfrak {q}$, we define

  $$\begin{aligned} \textstyle {\textbf {Ad}} _\psi [X]:[a,b]\rightarrow \mathfrak {q},\qquad t\mapsto \sum _{\ell =0}^\infty \varvec{\lambda }_{\ell ,\psi }[X](t). \end{aligned}$$

  (5)

  • It is straightforward to see that $\alpha :={\textbf {Ad}} _\psi [X]$ solves the Lax equation (3)—and is of class $C^{k+1}$ if $\psi$ is of class $C^k$ for $k\in \mathbb {N}\cup \{\infty \}$.

  • It is ad hoc not clear that this solution is unique. To prove uniqueness, one basically has to show that for each $t\in [a,b]$ one has

    $$\begin{aligned} \textsf {Aut} (\mathfrak {q})\ni {\textbf {Ad}} _\psi [t]:\mathfrak {q}\rightarrow \mathfrak {q},\qquad X\mapsto {\textbf {Ad}} _\psi [X](t). \end{aligned}$$

    Indeed, it is not hard to see that ${\textbf {Ad}} _\psi [t]$ admits a left inverse; but, to prove the existence of a right inverse some effort is necessary.

In this paper, we use the solutions (5) to prove Theorem I. We combine them with an integral transformation that we introduce, and that is also used to prove Formula (2) in the nilpotent weakly $C^k$-regular situation. It naturally arises in the context of semiregular Lie groups:

• Let G be $C^k$-semiregular for $k\in \mathbb {N}\cup \{\text {lip} , \infty \}$. Then, the integral transformation

  

  (6)

  is defined for all $a<b$ (cf. Proposition 1) and has the invariance property

  

• Let $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ be sequentially complete and asymptotic estimate. Motivated by (6), we consider the integral transformation (defined by Lemma 24)

  $$\begin{aligned} \textstyle \mathfrak {T}:C^0([a,b],\mathfrak {q}) \rightarrow&~ C^\infty ([0,1],\mathfrak {q})\nonumber \\ \phi \mapsto&~\textstyle \big [ [0,1]\ni t \mapsto \int _a^b {\textbf {Ad}} _{t\cdot \phi |_{[s,b]}}[b](\phi (s)) \,\text {d} s\,\big ], \end{aligned}$$

  (7)

  for which we have (Corollary 19)

  

  (8)

  If G is weakly $C^\infty$-semiregular and $\phi \in C^0([0,1],\mathfrak {q})$, then

  

  is of class $C^1$, whereby (8) implies which establishes Theorem I.

This paper is organized as follows. In Sect. 2, we fix the notations and provide some basic facts and definitions concerning locally convex vector spaces, Lie groups, Lie algebras, and mappings. We furthermore discuss some elementary properties of power series that we need in the main text. Section 3 contains some preliminary results. We discuss the integral transformation (6), and provide an integral expansion for the adjoint action. We furthermore prove a differentiation result for the product integral in the weakly regular context. In Sect. 4, we generalize the Baker–Campbell–Dynkin–Hausdorff formula to the product integral and give some applications to it. In particular, we prove formula (2). In Sect. 5, we discuss the Lax equation for asymptotic estimate and sequentially complete Lie algebras. We also investigate the elementary properties of the integral transformation (7) that we finally use to prove Theorem I.

2 Preliminaries

In this section, we fix the notations and provide definitions and elementary facts concerning Lie groups, Lie algebras, and locally convex vector spaces that we shall need in the main text.

2.1 Conventions

Given sets X, Y, the set of all mappings $X\rightarrow Y$ is denoted by $\textsf {Maps} (X,Y)\equiv Y^X$. Let Z be a topological space, and $S\subseteq Z$ a subset:

• We denote the closure of S in Z by $\textsf {clos} _{Z}(S)$, or simply by $\textsf {clos} (S)$ if it is clear from the context which topological space Z is meant.

• We denote the interior of S in Z by $\textsf {int} _{Z}(S)$, or simply by $\textsf {int} (S)$ if it is clear from the context which topological space Z is meant.

The class of Hausdorff locally convex vector spaces is denoted by $\textsf {hlcVect}$. Given $F\in \textsf {hlcVect}$, the system of continuous seminorms on F is denoted by $\textsf {Sem}(F)$. We define

$$\begin{aligned} \text {B} _{\textsf {q} ,\varepsilon }:=\{X\in E\,|\, \textsf {q} (X)<\varepsilon \}\qquad \quad \text {and}\qquad \quad \overline{\text {B} }_{\textsf {q} ,\varepsilon }:=\{X\in E\,|\, \textsf {q} (X)\le \varepsilon \} \end{aligned}$$

for each $\textsf {q} \in \textsf {Sem}(F)$ and $\varepsilon >0$. The completion of F is denoted by $\widehat{F}\in \textsf {hlcVect}$. For each $\textsf {q} \in \textsf {Sem}(F)$, we let $\widehat{\textsf {q} }:\widehat{F}\rightarrow [0,\infty )$ denote the continuous extension of $\textsf {q}$ to $\widehat{F}$. Given $F_1,\dots ,F_n\in \textsf {hlcVect}$, we have $F_1\times {\dots }\times F_n\in \textsf {hlcVect}$ via the Tychonoff topology generated by the seminorms

$$\begin{aligned} \textsf {max} [\textsf {q} _1,\dots ,\textsf {q} _n]:F_1\times {\dots }\times F_n\ni (X_1,\dots ,X_n)\mapsto \max \{\textsf {q} _k(X_k)\,|\, k=1,\dots ,n\}\in [0,\infty ) \end{aligned}$$

with $\textsf {q} _k\in \textsf {Sem}(F_k)$ for $k=1,\dots ,n$. If $\Phi :F_1\times {\dots }\times F_n\rightarrow F$ is a continuous n-multilinear map, then $\widehat{\Phi }:\widehat{F}_1\times {\dots }\times \widehat{F}_n\rightarrow \widehat{F}$ denotes its continuous n-multilinear extension.

In this paper, manifolds and Lie groups are always understood to be in the sense of [3], i.e., smooth, Hausdorff, and modeled over a Hausdorff locally convex vector space.⁵ If $f:M\rightarrow N$ is a $C^1$-map between the manifolds M and N, then $\text {d} f:TM \rightarrow TN$ denotes the corresponding tangent map between their tangent manifolds, and we write $\text {d} _xf\equiv \text {d} f(x,\cdot ):T_xM\rightarrow T_{f(x)}N$ for each $x\in M$. An interval is a non-empty, non-singleton connected subset $D\subseteq \mathbb {R}$. A curve is a continuous map $\gamma :D\rightarrow M$ for an interval and a manifold M. If $D\equiv I$ is open, then $\gamma$ is said to be of class $C^k$ for $k\in \mathbb {N}\cup \{\infty \}$ if it is of class $C^k$ when considered as a map between the manifolds I and M. If D is an arbitrary interval, then $\gamma$ is said to be of class $C^k$ for $k\in \mathbb {N}\cup \{\infty \}$ (we write $\gamma \in C^k(D,M)$) if $\gamma =\gamma '|_D$ holds for a $C^k$-curve $\gamma ':I\rightarrow M$ that is defined on an open interval I containing D. If $\gamma :D\rightarrow M$ is of class $C^1$, then we denote the corresponding tangent vector at $\gamma (t)\in M$ by ${\dot{\gamma }}(t)\in T_{\gamma (t)}M$. The above conventions also hold if $M\equiv F\in \textsf {hlcVect}$ is a Hausdorff locally convex vector space.

Throughout this paper, G denotes an infinite-dimensional Lie group in the sense of [3] that is modeled over $E\in \textsf {hlcVect}$. We fix a chart $\Xi :G\supseteq \mathcal {U}\rightarrow \mathcal {V}\subseteq E$, with $\mathcal {V}$ convex, $e\in \mathcal {U}$, and $\Xi (e)=0$. We denote the Lie algebra of G by $(\mathfrak {g}, \varvec{[} \cdot , \cdot \varvec{]})$, and set⁶

$$\begin{aligned} \textsf {p} (X):=(\textsf {p} \circ \text {d} _e\Xi )(X)\qquad \quad \forall \,\textsf {p} \in \textsf {Sem}(E),\, X\in \mathfrak {g}. \end{aligned}$$

We let $\textsf {m} :G\times G\rightarrow G$ denote the Lie group multiplication, $\text {R} _g:=\textsf {m} (\cdot , g)$ the right translation by $g\in G$, $\textsf {inv} :G\ni g\mapsto g^{-1}\in G$ the inversion, and $\text {Ad} :G\times \mathfrak {g}\rightarrow \mathfrak {g}$ the adjoint action. We have

$$\begin{aligned} \text {Ad} (g,X)\equiv \text {Ad} _g(X):=\text {d} _e\text {Conj} _g(X)\qquad \text {for}\qquad \text {Conj} _g:G\ni h\mapsto g\cdot h\cdot g^{-1}\in G \end{aligned}$$

for each $g\in G$ and $X\in \mathfrak {g}$, as well as

$$\begin{aligned} \,\,\varvec{[}X, Y \varvec{]}:=\text {d} _e\text {Ad} (Y)(X)\qquad \forall \, X\in \mathfrak {g}\qquad \qquad \text {for}\qquad \quad \text {Ad} (Y):G\ni g\mapsto \text {Ad} _g(Y)\in \mathfrak {g}. \end{aligned}$$

(9)

We recall the Jacobi identity (cf. also Example 5)

$$\begin{aligned} \textstyle \varvec{[}Z, \varvec{[}X, Y \varvec{]} \varvec{]}= \varvec{[}\varvec{[}Z, X \varvec{]}, Y \varvec{]}+ \varvec{[}X, \varvec{[}Z, Y \varvec{]} \varvec{]}\qquad \quad \,\forall \, X,Y,Z\in \mathfrak {g}\end{aligned}$$

(10)

as well as the product rule

$$\begin{aligned} \text {d} _{(g,h)}\textsf {m} (v,w)= \text {d} _g\text {R} _h(v) + \text {d} _h\text {L} _g(w)\qquad \quad \forall \, g,h\in G,\,\, v\in T_gG,\,\, w\in T_h G. \end{aligned}$$

(11)

2.2 Locally convex vector spaces

In this section, we collect some elementary facts and definitions concerning locally convex vector spaces that we shall need in the main text.

2.2.1 Sets of curves

Let $F\in \textsf {hlcVect}$ be given. For each $X\in F$, we let $\mathcal {C}_X:\mathbb {R}\ni t\mapsto X \in F$ denote the constant curve X. Let $a<b$ be given. We set $C^\text {c} ([a,b],F):=\{\mathcal {C}_X|_{[a,b]}\,|\, X\in F\}$; and let $C^\text {lip} ([a,b],F)$ denote the set of all Lipschitz curves on [a, b], i.e., all curves $\gamma :[a,b]\rightarrow F$ with

$$\begin{aligned} \textstyle \mathfrak {L}(\textsf {q} ,\gamma ):= \sup \Big \{\frac{\textsf {q} (\gamma (t)-\gamma (t'))}{|t-t'|}\,\Big |\, t,t'\in [a,b],\, t\ne t'\Big \}\in [0,\infty )\qquad \quad \forall \, \textsf {q} \in \textsf {Sem}(F). \end{aligned}$$

We define $\text {c} +1:=\infty$, $\infty +1:=\infty$, $\text {lip} +1:=1$; as well as

for $\textsf {q} \in \textsf {Sem}(F)$, $k\in \mathbb {N}\cup \{\infty ,\text {c} \}$, $0\le \text {s} \preceq k$. Here, the notation $s\preceq k$ means

• $\text {s} = \text {lip}$             for  $k= \text {lip}$,

• $\mathbb {N}\ni \text {s} \le k$         for  $k\in \mathbb {N}$,

• $\text {s} \in \mathbb {N}$                 for  $k=\infty$,

• $\text {s} =0$                  for  $k= \text {c}$.

Given $k\in \mathbb {N}\cup \{\text {lip} ,\infty ,\text {c} \}$, the $C^k$-topology on $C^k([a,b],F)$ is the Hausdorff locally convex topology that is generated by the seminorms $\textsf {q} _\infty ^\text {s}$ for $\textsf {q} \in \textsf {Sem}(F)$ and $0\le \text {s} \preceq k$.

By an element in $\text {CP} ^0([a,b],F)$ (piecewise continuous curves), we understand a decomposition $a= t_0<{\dots }<t_n=b$ (for $n\ge 1$) together with a collection $\gamma \equiv \{\gamma [p]\}_{0\le p\le n-1}$ of curves $\gamma [p]\in C^0([t_p,t_{p+1}],F)$ for $0\le p\le n-1$. Linear combinations as well as restrictions to compact intervals of elements in $\text {CP} ^0([a,b],F)$ are defined in the obvious way. We equip $\text {CP} ^0([a,b],F)$ with the $C^0$-topology, i.e., the Hausdorff locally convex topology that is generated by the seminorms

$$\begin{aligned} \textsf {q} _\infty (\gamma )=\max \{\textsf {q} _\infty (\gamma [p])\,|\, 0\le p\le n-1 \} \end{aligned}$$

for $\textsf {q} \in \textsf {Sem}(F)$, and $\gamma \equiv \{\gamma [p]\}_{0\le p\le n-1}$ as above.

2.2.2 Mackey convergence

Let $F\in \textsf {hlcVect}$. A subsystem $\mathfrak {H}\subseteq \textsf {Sem}(F)$ is said to be a fundamental system if $\{\text {B} _{\textsf {h} ,\,\varepsilon }(0)\}_{\textsf {h} \in \mathfrak {H},\varepsilon >0}$ is a local base of zero in F. We recall the following standard result [17]:

Lemma 1

Let $\mathfrak {H}\subseteq \textsf {Sem}(F)$ be a fundamental system, and $\mathfrak {S}\subseteq \textsf {Sem}(F)$ a subsystem. Then, $\mathfrak {S}$ is a fundamental system if and only if to each $\textsf {h} \in \mathfrak {H}$ there exist $c>0$ and $\textsf {s} \in \mathfrak {S}$ with $\textsf {h} \le c\cdot \textsf {s}$.

Proof

Confer, e.g., Lemma 20 in [11]. $\square$

Let $\mathfrak {H}\subseteq \textsf {Sem}(F)$ be a fundamental system. We write for $\{X_n\}_{n\in \mathbb {N}}\subseteq F$, $X\in F$ if

$$\begin{aligned} \textsf {h} (X-X_n)\le \mathfrak {c}_\textsf {h} \cdot \lambda _{n} \qquad \quad \forall \, n\in \mathbb {N},\,\,\textsf {h} \in \mathfrak {H}\end{aligned}$$

(12)

holds for certain $\{\mathfrak {c}_\textsf {h} \}_{\textsf {h} \in \mathfrak {H}}\subseteq \mathbb {R}_{\ge 0}$, and $\mathbb {R}_{> 0}\supseteq \{\lambda _{n}\}_{n\in \mathbb {N}}\rightarrow 0$. In this case, $\{X_n\}_{n\in \mathbb {N}}$ is said to be Mackey convergent to X.

Remark 1

• It is immediate from Lemma 1 that the definition made in (12) does not depend on the explicit choice of the fundamental system $\mathfrak {H}\subseteq \textsf {Sem}(F)$.

• Since we will make use of the differentiability results obtained in [8], we explicitly mention that in [8] an equivalent definition of Mackey convergence (more suitable for the technical argumentation there) was used. Specifically, (12) is equivalent to require that

  $$\begin{aligned} \textsf {h} (X-X_n)\le \mathfrak {c}_\textsf {h} \cdot \lambda _{n} \qquad \quad \forall \, n\ge \mathfrak {l}_\textsf {h} ,\,\,\textsf {h} \in \mathfrak {H}\end{aligned}$$

  (13)

  holds for certain $\{\mathfrak {c}_\textsf {h} \}_{\textsf {h} \in \mathfrak {H}}\subseteq \mathbb {R}_{\ge 0}$, $\{\mathfrak {l}_\textsf {h} \}_{\textsf {h} \in \mathfrak {H}}\subseteq \mathbb {N}$, and $\mathbb {R}_{\ge 0}\supseteq \{\lambda _{n}\}_{n\in \mathbb {N}}\rightarrow 0$.

  In other words, the indices $\{\mathfrak {l}_\textsf {h} \}_{\textsf {h} \in \mathfrak {H}}$ in (13) can be circumvented (can be set equal to zero) if $\{\lambda _n\}_{n\in \mathbb {N}}\subseteq \mathbb {R}_{>0}$ instead of $\{\lambda _n\}_{n\in \mathbb {N}}\subseteq \mathbb {R}_{\ge 0}$ is required. $\square$

Assume now $F= C^k([a,b],\mathfrak {g})$ for $k\in \mathbb {N}\cup \{\text {lip} ,\infty ,\text {c} \}$ and $a<b$. We write

(14)

if $\{\phi _n\}_{n\in \mathbb {N}}$ is Mackey convergent to $\phi$ w.r.t. to the $C^k$-topology.

2.2.3 The Riemann integral and completeness

Let $F\in \textsf {hlcVect}$. The Riemann integral of $\gamma \in C^0([a,b],F)$ for $a<b$ is denoted by $\int \gamma (s) \,\text {d} s\in \widehat{F}$.

Remark 2

The Riemann integral in complete Hausdorff locally convex vector spaces can be defined exactly as in the finite-dimensional case; namely, as the limit of Riemann sums (which form a Cauchy sequence in F), confer Sect. 2 in [16]. In particular, the following assertions hold:

• If F is sequentially complete, then $\int \gamma (s) \,\text {d} s\in F$ holds for $\gamma \in C^0([a,b],F)$.

• For $\gamma \in C^0([a,b],F)$ and $\gamma _-:[a,b]\ni t\mapsto \gamma (a+b-t)$, we have $\int \gamma _-(s)\,\text {d} s=\int \gamma (s)\, \text {d} s$. $\square$

Given $x<y$ and $\gamma \in C^0([x,y],F)$, then for $x\le a<b\le y$ and $x\le c\le y$, we set

$$\begin{aligned}&\textstyle \int _a^b \gamma (s)\,\text {d} s:= \int \gamma |_{[a,b]}(s) \,\text {d} s,\quad \,\,\int _b^a \gamma (s) \,\text {d} s:= - \int _a^b \gamma (s) \,\text {d} s,\quad \,\, \int _c^c \gamma (s)\, \text {d} s:=0. \end{aligned}$$

(15)

Clearly, the Riemann integral is linear with

$$\begin{aligned} \textstyle \int _a^c \gamma (s) \,\text {d} s&\textstyle =\int _a^b \gamma (s)\,\text {d} s+ \int _b^c \gamma (s)\,\text {d} s\qquad \quad \, \forall \, x\le a< b< c\le y. \end{aligned}$$

(16)

The Riemann integral admits the following additional properties:

• For $a<b$, we have

  $$\begin{aligned} \qquad \textstyle \gamma -\gamma (a)\textstyle =\int _a^\bullet {\dot{\gamma }}(s)\,\text {d} s&\qquad \,\,\quad \forall \, \gamma \in C^1([a,b],F), \end{aligned}$$

  (17)
  $$\begin{aligned} \textsf {q} (\gamma -\gamma (a))\textstyle \le \int _a^\bullet \textsf {q} ({\dot{\gamma }}(s))\, \text {d} s&\qquad \quad \forall \, \gamma \in C^1([a,b],F),\, \textsf {q} \in \textsf {Sem}(F), \end{aligned}$$

  (18)
  $$\begin{aligned} \textstyle \widehat{\textsf {q} }\big (\int _a^\bullet \gamma (s)\,\text {d} s\big )\textstyle \le \int _a^\bullet \textsf {q} (\gamma (s))\,\text {d} s&\qquad \quad \forall \, \gamma \in C^0([a,b],F),\, \textsf {q} \in \textsf {Sem}(F). \end{aligned}$$

  (19)

• For $a<b$ and $\gamma \in C^0([a,b],F)$, we have

  $$\begin{aligned} \textstyle C^1([a,b],\widehat{F})\ni \Gamma :[a,b]\ni t\mapsto \int _a^t \gamma (s)\,\text {d} s \in \widehat{F}\qquad \quad \text {with}\qquad \quad {\dot{\Gamma }}=\gamma . \end{aligned}$$

  (20)

• For $a<b$, $a'<b'$, $\gamma \in C^1([a,b],F)$, $\varrho :[a',b'] \rightarrow [a,b]$ of class $C^1$, and $t\in [a',b']$, we have the substitution formula

  $$\begin{aligned} \textstyle \int _a^{\varrho (t)} \gamma (s)\, \text {d} s=\int _{a'}^t {\dot{\varrho }}(s)\cdot (\gamma \circ \varrho )(s)\,\text {d} s + \int _a^{\varrho (a')} \gamma (s)\,\text {d} s. \end{aligned}$$

  (21)

• Let $\tilde{F}\in \textsf {hlcVect}$, $a<b$, and $\mathcal {L}:F\rightarrow \tilde{F}$ a continuous and linear map. Then, the following implication holds:

  $$\begin{aligned}&\textstyle \int _a^b \gamma (s)\,\text {d} s\in F \quad \text {for}\quad \gamma \in C^0([a,b],F)\nonumber \\&\qquad \Longrightarrow \qquad \textstyle \mathcal {L}\big (\int _a^b \gamma (s)\,\text {d} s\big )=\int _a^b \mathcal {L}(\gamma (s)) \, \text {d} s. \end{aligned}$$

  (22)

Next, let us recall the following definitions:

• We say that F is Mackey complete if $\int _0^1 \gamma (s)\,\text {d} s \in F$ exists for each $\gamma \in C^\infty ([0,1],F)$.

  This is equivalent to require that $\int _a^b \gamma (s)\,\text {d} s \in F$ exists for each $\gamma \in C^\text {lip} ([a,b],F)$ and $a<b$, confer Theorem 2.14 in [16] or Corollary 6 in [8].

• We say that F is integral complete if $\int _a^b \gamma (s)\,\text {d} s \in F$ exists for all $a<b$ and $\gamma \in C^0([a,b],F)$.

Evidently, F is Mackey complete if F is integral complete. Moreover, Remark 2 implies that F is integral complete (hence, Mackey complete) if F is sequentially complete.

Finally, for $\gamma \equiv \{\gamma [p]\}_{0\le p\le n-1}\in \text {CP} ^0([a,b],F)$ ($n\ge 1$), we define

$$\begin{aligned} \textstyle \int _a^{b}\!\gamma (s)\,\text {d} s:=\sum _{p=0}^{n-1}\int _{t_p}^{t_{p+1}} \gamma [p](s)\,\text {d} s. \end{aligned}$$

(23)

Clearly, the so-extended Riemann integral is linear, and $C^0$-continuous—specifically, (19) yields

$$\begin{aligned} \textstyle \widehat{\textsf {q} }\big (\int _a^t \gamma (s)\,\text {d} s\big )&\textstyle \le (t-a)\cdot \textsf {q} _\infty (\gamma )\qquad \quad \forall \, a<t\le b,\,\textsf {q} \in \textsf {Sem}(F). \end{aligned}$$

For $x<y$ and $\gamma \in \text {CP} ^0([x,y],F)$, we define $\int _a^b\gamma (s)\, \text {d} s$, $\int _b^a\gamma (s)\, \text {d} s$, $\int _c^c \gamma (s)\, \text {d} s$ for $x\le a<b\le y$ and $x\le c\le y$ as in (15). Clearly, then (16) also holds for the extended Riemann integral. We furthermore note that $[a,b]\ni t\mapsto \int _a^t \gamma (s)\,\text {d} s\in F$ is continuous for each $x\le a<b\le y$, as well as of class $C^1$ if $[a,b]\subseteq [t_p,t_{p+1}]$ holds for some $0\le p\le n-1$.

2.2.4 Extensions and completeness

Let $F\in \textsf {hlcVect}$ be given. In this subsection, we collect some statements that we shall need in Sect. 5. We recall the following result.⁷

Lemma 2

Let $D\subseteq \mathbb {R}$ be an interval, $k\in \mathbb {N}\cup \{\infty \}$, and $\gamma \in C^k(J,F)$ for $J:=\textsf {int} _{\mathbb {R}}(D)$. Let $\{\gamma _\ell \}_{0\le \ell \preceq k}\subseteq C^0(D,F)$ be given such that $\gamma ^{(\ell )}=\gamma _\ell |_{J}$ holds for all $0\le \ell \preceq k$. Then, there exists an open interval $I\subseteq \mathbb {R}$ with $D\subseteq I$ as well as $\tilde{\gamma }\in C^k(I,F)$ such that $\tilde{\gamma }|_D=\gamma$ holds.

Proof

This is immediate from Theorem 1 in [12]. $\square$

Lemma 3

If F is sequentially complete, then $C^k([a,b],F)$ is sequentially complete w.r.t. the $C^k$-topology for each $k\in \mathbb {N}\cup \{\infty \}$ and $a<b$.

Proof

Let $k\in \mathbb {N}\cup \{\infty \}$ be fixed; and let $\{\gamma _n\}_{n\in \mathbb {N}}\subseteq C^k([a,b],F)$ be a Cauchy sequence w.r.t. the $C^k$-topology. Then, $\{(\gamma _n)^{(\ell )}\}_{n\in \mathbb {N}}\subseteq C^0([a,b],F)$ is a Cauchy sequence w.r.t. the $C^0$-topology for $0\le \ell \preceq k$, and a standard $\frac{\varepsilon }{3}$-argument shows that it converges w.r.t. the $C^0$-topology to some $\gamma _\ell \in C^0([a,b],F)$. Let $\gamma :=\gamma _0|_{(a,b)}$. By Lemma 2, it suffices to show

$$\begin{aligned} \gamma \in C^k((a,b),F)\qquad \text {with}\qquad \gamma ^{(\ell )}=\gamma _\ell |_{(a,b)}\qquad \text {for}\qquad 0\le \ell \preceq k. \end{aligned}$$

(24)

Since the Riemann integral is $C^0$-continuous by (19), we have for $0\le \ell < k$ and $t\in [a,b]$

$$\begin{aligned} \textstyle \gamma _{\ell }(t) \textstyle = \lim _n(\gamma _n)^{(\ell )}(t) \textstyle {\mathop {=}\limits ^{(17)}} \lim _n\big ((\gamma _n)^{(\ell )}(a)+ \int _a^t (\gamma _n)^{(\ell +1)}(s) \,\text {d} s \big ) =\textstyle \gamma _\ell (a) + \int _{a}^t \gamma _{\ell +1}(s)\, \text {d} s. \end{aligned}$$

Then, (20) shows $\gamma _\ell |_{(a,b)}\in C^1((a,b),F)$ with $(\gamma _\ell |_{(a,b)})^{(1)}=\gamma _{\ell +1}|_{(a,b)}$ for $0\le \ell < k$.

Now, since $\gamma =\gamma _0|_{(a,b)}$ holds by definition, we can assume that there exists some $0\le q< k$ such that (24) holds for $k\equiv q$. We obtain

$$\begin{aligned} \gamma _{q+1}|_{(a,b)}=(\gamma _q|_{(a,b)})^{(1)}=(\gamma ^{(q)})^{(1)}=\gamma ^{(q+1)}, \end{aligned}$$

hence $\gamma \in C^{q+1}((a,b),F)$. $\square$

Corollary 1

Assume that F is sequentially complete, and let $\mathfrak {H}\subseteq \textsf {Sem}(F)$ be a fundamental system. Let $k\in \mathbb {N}\cup \{\infty \}$, $a<b$, and $\{\gamma _p\}_{p\in \mathbb {N}}\subseteq C^k([a,b],F)$ be given with

$$\begin{aligned} \textstyle \sum _{p=0}^\infty \textsf {q} ^\text {s} _\infty (\gamma _p)< \infty \qquad \quad \forall \, \textsf {q} \in \mathfrak {H},\, 0\le \text {s} \preceq k. \end{aligned}$$

Then, $\textstyle \gamma := \sum _{p=0}^\infty \gamma _p\in C^k([a,b],F)$ converges w.r.t. the $C^k$-topology, i.e., $\gamma ^{(\ell )}=\sum _{p=0}^\infty (\gamma _p)^{(\ell )}$ converges w.r.t. the $C^0$-topology for each $0\le \ell \preceq k$.

Proof

The assumptions imply that $\{\sum _{p=0}^n\gamma _p\}_{n\in \mathbb {N}}\subseteq C^k([a,b],F)$ is Cauchy w.r.t. the $C^k$-topology. The rest is clear from Lemma 3. $\square$

Corollary 2

Assume that F is sequentially complete, and let $\mathfrak {H}\subseteq \textsf {Sem}(F)$ be a fundamental system. Let $\{X_p\}_{p\in \mathbb {N}}\subseteq F$ and $\delta >0$ be given, such that

$$\begin{aligned} \textstyle \gamma [\textsf {q} ]:(-\delta ,\delta )\ni t\mapsto \sum _{p=0}^\infty t^p\cdot \textsf {q} (X_p)\in [0,\infty ) \end{aligned}$$

is defined for each $\textsf {q} \in \mathfrak {H}$. Define $\gamma _p:(-\delta ,\delta )\ni t\mapsto t^p\cdot X_p\in F$ for each $p\in \mathbb {N}$.

• (a) Let $-\delta<a<b<\delta$. Then, $\textstyle \gamma [a,b]:= \sum _{p=0}^\infty \gamma _p|_{[a,b]}\in C^\infty ([a,b],F)$ converges w.r.t. the $C^\infty$-topology, with

  $$\begin{aligned} \textstyle \gamma [a,b]^{(\ell )}=\sum _{p=0}^\infty (\gamma _p|_{[a,b]})^{(\ell )}:[a,b]\ni t\mapsto \sum _{p=\ell }^\infty \frac{p !}{(p-\ell )!}\cdot t^{p-\ell }\cdot X_p\in F\qquad \quad \forall \, \ell \in \mathbb {N}. \end{aligned}$$

• (b) We have $\gamma :=\sum _{p=0}^\infty \gamma _p\in C^\infty ((-\delta ,\delta ),F)$ with

  $$\begin{aligned} \textstyle \gamma ^{(\ell )}=\sum _{p=0}^\infty (\gamma _p)^{(\ell )}:(-\delta ,\delta )\ni t\mapsto \sum _{p=\ell }^\infty \frac{p !}{(p-\ell )!}\cdot t^{p-\ell }\cdot X_p\in F \qquad \quad \forall \, \ell \in \mathbb {N}. \end{aligned}$$

Proof

Point b) is clear from Point a). To prove Point a), we observe that $\gamma [\textsf {q} ]$ is smooth for each $\textsf {q} \in \mathfrak {H}$, with

$$\begin{aligned} \textstyle \gamma [\textsf {q} ]^{(\ell )}(t)= \sum _{p=\ell }^\infty \frac{p !}{(p-\ell )!}\cdot t^{p-\ell }\cdot \textsf {q} (X_p)\in [0,\infty ) \qquad \quad \forall \, t\in (-\delta ,\delta ),\, \ell \in \mathbb {N}. \end{aligned}$$

This implies $\sum _{p=0}^\infty \textsf {q} _\infty ^\text {s} (\gamma _p|_{[a,b]})<\infty$ for all $\textsf {q} \in \mathfrak {H}$, $\text {s} \in \mathbb {N}$, $a<b$, so that the claim is clear from Corollary 1. $\square$

2.2.5 Lie Algebras

Let $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ be a fixed Lie algebra (not necessarily the Lie algebra of a Lie group), i.e.,

• $\mathfrak {q}\in \textsf {hlcVect}$,

• $\varvec{[}\cdot , \cdot \varvec{]}:\mathfrak {q}\times \mathfrak {q}\rightarrow \mathfrak {q}$ is bilinear, antisymmetric, and continuous, with (Jacobi identity)

  $$\begin{aligned} \varvec{[}Z, \varvec{[}X, Y \varvec{]} \varvec{]}= \varvec{[}\varvec{[}Z, X \varvec{]}, Y \varvec{]} + \varvec{[}X, \varvec{[}Z, Y \varvec{]} \varvec{]}\qquad \quad \forall \, X,Y,Z\in \mathfrak {q}. \end{aligned}$$

  (25)

We shall need the following definitions:

• Given $X\in \mathfrak {q}$, we denote $\text {ad} _{X}^1\equiv \text {ad} _{X}:\mathfrak {q}\ni Y\mapsto \varvec{[}X, Y \varvec{]}\in \mathfrak {q}$,⁸ and define inductively

  $$\begin{aligned} \text {ad} _{X}^0:=\text {id} _\mathfrak {q}\qquad \quad \text {as well as}\qquad \quad \text {ad} _{X}^n:=\text {ad} _{X}\circ \text {ad} _{X}^{n-1}\qquad \forall \, n\ge 1. \end{aligned}$$

• Given a subset $S\subseteq \mathfrak {q}$, then $\varvec{\langle } S\varvec{\rangle } \subseteq \mathfrak {q}$ denotes the linear subspace that is generated by S. We define inductively

  $$\begin{aligned} \mathcal {V}_1(S):=\varvec{\langle } S\varvec{\rangle } \qquad \quad \text {as well as}\qquad \quad \mathcal {V}_{n+1}(S):=\varvec{\langle } \varvec{[}\mathcal {V}(S), \mathcal {V}_n(S) \varvec{]}\varvec{\rangle } \qquad \forall \, n\ge 1; \end{aligned}$$

  and set $\textstyle \mathcal {G}_n(S):= \varvec{\langle } \bigcup _{\ell = n}^{\infty } \mathcal {V}_\ell (S)\varvec{\rangle }$ for each $n\ge 1$.

A straightforward induction involving (25) shows (confer Appendix A.2 or Exercise 5.2.7 in [13])

$$\begin{aligned} \varvec{\langle } \varvec{[}\mathcal {V}_m(S), \mathcal {V}_n(S) \varvec{]}\varvec{\rangle } \subseteq \mathcal {V}_{m+n}(S)\qquad \quad \forall \,m,n\ge 1. \end{aligned}$$

(26)

For $n\ge 1$, we set $\overline{\mathcal {V}}_n:=\textsf {clos} _{\mathfrak {q}}(\mathcal {V}_n)$ as well as $\overline{\mathcal {G}}_n:=\textsf {clos} _{\mathfrak {q}}(\mathcal {G}_n)$. Since $\varvec{[}\cdot , \cdot \varvec{]}$ is continuous, (26) implies

$$\begin{aligned} \varvec{\langle } \varvec{[}\overline{\mathcal {V}}_m(S), \overline{\mathcal {V}}_n(S) \varvec{]}\varvec{\rangle } \subseteq \overline{\mathcal {V}}_{m+n}(S)\qquad \quad \forall \,m,n\ge 1. \end{aligned}$$

(27)

We observe that $\overline{\mathcal {G}}_{n+\ell }(S)\subseteq \overline{\mathcal {G}}_{n}(S)$ holds for all $n,\ell \in \mathbb {N}$. We shall need the following definitions: ⁹

• A subset $\text {M} \subseteq \mathfrak {q}$ is said to be an $\text {AE}$-set if to each $\textsf {v} \in \textsf {Sem}(\mathfrak {q})$, there exist $\textsf {v} \le \textsf {w} \in \textsf {Sem}(\textsf {q} )$ with

  $$\begin{aligned} \textsf {v} (\varvec{[}X_1,\varvec{[}X_2,\varvec{[}\dots ,\varvec{[}X_n,Y\varvec{]}{\dots }\varvec{]}\varvec{]}])\le \textsf {w} (X_1)\cdot {\dots }\cdot \textsf {w} (X_n)\cdot \textsf {w} (Y) \end{aligned}$$

  (28)

  for all $X_1,\dots ,X_n,Y\in \text {M}$, and $n\ge 1$. We say that $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ is asymptotic estimate if $\mathfrak {q}$ is an $\text {AE}$-set.

• A subset $\text {N} \subseteq \mathfrak {q}$ is said to be a $\text {Nil} _{q}$-set for $q\ge 2$ if

  $$\begin{aligned} \varvec{[}X_1,\varvec{[}X_2,\varvec{[}\dots ,\varvec{[}X_{q-1},X_q\varvec{]}{\dots }\varvec{]}\varvec{]}\varvec{]}=0 \qquad \quad \forall \, X_1,\dots ,X_q\in \text {N} . \end{aligned}$$

  (29)

  Evidently, each $\text {Nil} _{q}$-set (for $q\ge 2$) is an $\text {AE}$-set. We say that $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ is nilpotent if $\mathfrak {q}$ is a $\text {Nil} _{q}$-set for some $q\ge 2$.

In view of Sect. 4.2 (the proof of Lemma 16), we observe the following.

Remark 3

Let $\text {N} \subseteq \mathfrak {q}$ be a $\text {Nil} _{q}$-set for some $q\ge 2$. Then,

• 1) $\overline{\mathcal {G}}_{q+n}(\text {N} )=\{0\}$ holds for each $n\in \mathbb {N}$.

• 2) $\varvec{[}\overline{\mathcal {G}}_m(\text {N} ), \overline{\mathcal {G}}_n(\text {N} ) \varvec{]}\subseteq \overline{\mathcal {G}}_{m+n}(\text {N} )$ holds for $m,n\ge 1$, by (27).

• 3) $\overline{\mathcal {G}}_n(\text {N} )$ is a $\text {Nil} _{q}$-set for each $n\ge 1$, by the previous points 1) and 2).

• 4) Let $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ be given; and assume that $\mathfrak {q}$ is Mackey complete for $k\in \mathbb {N}_{\ge 1}\cup \{\text {lip} ,\infty \}$, as well as integral complete for $k\equiv 0$. Then, for $n\in \{1,\dots ,q\}$ and $\phi \in C^k([a,b],\mathfrak {q})$ with $\text {im} [\phi ]\subseteq \overline{\mathcal {G}}_n(\text {N} )$, we have

  $$\begin{aligned} \textstyle \int _s^t \phi (s)\,\text {d} s \in \overline{\mathcal {G}}_n(\text {N} )\qquad \quad \forall \, a\le s<t\le b. \qquad \qquad \quad \end{aligned}$$

  $\square$

2.2.6 Some properties of maps

Let $F_1,\dots ,F_n,F,E\in \textsf {hlcVect}$ be given.

Lemma 4

Let X be a topological space; and let $\Phi :X\times F_1\times {\dots }\times F_n\rightarrow F$ be continuous, such that $\Phi (x,\cdot )$ is n-multilinear for each $x\in X$. Then, to each compact $\text {K} \subseteq X$ and each $\textsf {q} \in \textsf {Sem}(F)$, there exist $\textsf {q} _1\in \textsf {Sem}(F_1),\dots ,\textsf {q} _n\in \textsf {Sem}(F_n)$ as well as $O\subseteq X$ open with $\text {K} \subseteq O$, such that

$$\begin{aligned} (\textsf {q} \circ \Phi )(y,X_1,\dots ,X_n) \le \textsf {q} _1(X_1)\cdot {\dots }\cdot \textsf {q} _n(X_n)\qquad \quad \forall \, y\in O,\, X_1\in F_1,\dots ,X_n\in F_n. \end{aligned}$$

Proof

Confer, e.g., Corollary 1 in [8]. $\square$

Lemma 5

Let $V\subseteq F$ be open with $0\in V$. Let furthermore $\Psi :V\times E\rightarrow E$ be smooth with $\Psi (0,\cdot )=\text {id} _{E}$, such that $\Psi (x,\cdot )$ is linear for each $x \in V$. Then, to each $\textsf {p} \in \textsf {Sem}(E)$, there exist $\textsf {q} \in \textsf {Sem}(F)$ and $\textsf {w} \in \textsf {Sem}(E)$ with

$$\begin{aligned} \textsf {p} (\Psi (x,Y)-Y)\le \textsf {q} (x)\cdot \textsf {w} (Y)\qquad \quad \forall \, x\in \text {B} _{\textsf {q} ,1}\subseteq V,\, Y\in E. \end{aligned}$$

Proof

Confer Appendix A.3. $\square$

Lemma 6

Let $F_1,F_2\in \textsf {hlcVect}$, and $f:F_1\supseteq U\rightarrow F_2$ of class $C^{2}$. Assume that $\gamma :D\rightarrow F_1$ is continuous at $t\in D$, such that $\lim _{h\rightarrow 0} 1/h \cdot (\gamma (t+h)-\gamma (t))=:X\in \widehat{F}_1$ exists. Then, we have

$$\begin{aligned} \textstyle \lim _{h\rightarrow 0} 1/h\cdot (f(\gamma (t+h))-f(\gamma (t)))=\widehat{\text {d} _{\gamma (t)}f}\,(X). \end{aligned}$$

Proof

Confer, e.g., Lemma 7 in [8]. $\square$

We close this subsection with the following convention concerning differentiable maps with values in Lie groups.

Convention 1

Let $F\in \textsf {hlcVect}$, $U\subseteq F$, G a Lie group modeled over $E\in \textsf {hlcVect}$. A map $f:U\rightarrow G$ is said to be

• differentiable at $x\in U$ if there exists a chart $(\Xi ',\mathcal {U}')$ of G with $f(x)\in \mathcal {U}'$, such that

  $$\begin{aligned} \textstyle (D^{\Xi '}_v f)(x):=\lim _{h\rightarrow 0} 1/h\cdot ((\Xi '\circ f)(x+h\cdot v)-(\Xi '\circ f)(x))\in E \end{aligned}$$

  (30)

  exists for all $v \in F$. Lemma 6 applied to coordinate changes shows that (30) holds for one chart around f(x) if and only if it holds for each chart around f(x), and that

  $$\begin{aligned} \text {d} _x f(v):= \big (\text {d} _{\Xi '(f(x))}\Xi '^{-1} \circ (D^{\Xi '}_v f)\big )(x)\in T_{f(x)}G \qquad \quad \forall \, v\in F \end{aligned}$$

  is independent of the explicit choice of $(\Xi ',U')$.

• differentiable if f is differentiable at each $x\in U$.       $\square$

2.3 Power series

In this subsection, we collect some statements concerning power series in (Banach) algebras that we shall need to work informally in Sect. 4.1. We set

$$\begin{aligned} U_{\varepsilon }(z):=\{w\in \mathbb {C}\, |\, |w-z|<\varepsilon \}\qquad \quad \forall \, \varepsilon >0,\, z\in \mathbb {C}, \end{aligned}$$

and let $\mathbb {K}\in \{\mathbb {R},\mathbb {C}\}$ be fixed.

• Let $(\mathcal {N},*_\mathcal {N},1_\mathcal {N})$ be a unital $\mathbb {K}$-algebra, and let $\mathcal {N}_q:=\{\mathfrak {n}\in \mathcal {N}\,|\, \mathfrak {n}^{q+1}=0\}$ for $q\ge 1$. Set furthermore $\mathfrak {n}^0:=1_\mathcal {N}$ and $\mathfrak {n}^1:=\mathfrak {n}$ for each $\mathfrak {n}\in \mathcal {N}$.

• Let $(\mathcal {A},*_\mathcal {A},1_\mathcal {A},\Vert \cdot \Vert _\mathcal {A})$ be a unital submultiplicative Banach algebra over $\mathbb {K}$, and set $\mathfrak {a}^0:=1_\mathcal {A}$ as well as $\mathfrak {a}^1:=\mathfrak {a}$ for each $\mathfrak {a}\in \mathcal {A}$.

Let $f:U_{R}(0)\ni z\mapsto \sum _{n=0}^\infty a_n\cdot z^n\in \mathbb {C}$ for $\{a_n\}_{n\in \mathbb {N}}\subseteq \mathbb {K}$ be a power series with radius of convergence $R >0$. We define

$$\begin{aligned} \textstyle |f|_r:=\sum _{n=0}^\infty |a_n|\cdot r^n\in [0,\infty )\qquad \quad \forall \, 0\le r<R, \end{aligned}$$

and set $\textstyle f_p:U_{R}(0)\ni z\mapsto \sum _{n=0}^p a_n\cdot z^n\in \mathbb {C}$ for each $p\in \mathbb {N}$. We furthermore let

$$\begin{aligned} f(\mathfrak {n})&\textstyle := \sum _{n=0}^q a_n\cdot \mathfrak {n}^n\in \mathcal {N}\qquad \,\,\,~\forall \, \mathfrak {n}\in \mathcal {N}_q\,\,~~\text {with}\,\, ~~q\ge 1\\ f(\mathfrak {a})&\textstyle :=\sum _{n=0}^\infty a_n\cdot \mathfrak {a}^n \in \mathcal {A}\qquad \quad \forall \, \mathfrak {a}\in \mathcal {A}~~~~\,\,\text {with}\,\,~~ \Vert \mathfrak {a}\Vert _\mathcal {A}<R. \end{aligned}$$

Let $g:U_{S}(0)\ni z\mapsto \sum _{n=0}^\infty b_n\cdot z^n\in \mathbb {C}$ be a power series with radius of convergence $S>0$.

• Assume $S=R$. The Cauchy product formula yields

  $$\begin{aligned} \textstyle f *g:U_R\ni z\mapsto \sum _{n=0}^\infty \big (\sum _{\ell =0}^n a_\ell \cdot b_{n-\ell }\big ) \cdot z^n=f(z)\cdot g(z)\in \mathbb {C}. \end{aligned}$$

$\circ$:

For $\mathfrak {n}\in \mathcal {N}_q$ with $q\ge 1$, we evidently have

$$\begin{aligned} \textstyle f(\mathfrak {n})*_\mathcal {N}g(\mathfrak {n}) = \sum _{n=0}^q \big (\sum _{\ell =0}^n a_\ell \cdot b_{n-\ell }\big ) \cdot \mathfrak {n}^n = (f*g)(\mathfrak {n})\in \mathcal {N}. \end{aligned}$$

$\circ$:

For $\mathfrak {a}\in \mathcal {A}$ with $\Vert \mathfrak {a}\Vert _\mathcal {A}< R$, we obtain

$$\begin{aligned} \textstyle f(\mathfrak {a})*_\mathcal {A}g(\mathfrak {a}) = \sum _{n=0}^\infty \big (\sum _{\ell =0}^n a_\ell \cdot b_{n-\ell }\big ) \cdot \mathfrak {a}^n = (f*g)(\mathfrak {a})\in \mathcal {A}. \end{aligned}$$

(Apply, e.g., Exercise 3.1.3 in [13], with $X,Y,Z\equiv \mathcal {A}$ and $\beta \equiv *_\mathcal {A}$ there.)

• Assume $|g|_s=\sum _{n=0}^\infty |b_n|\cdot s^n < R$ for all $0\le s< S$, thus $g(U_{S})\subseteq U_{R}(0)$. By analyticity, we have

  

  The following statements are verified in Appendix A.4:

$\circ$:

Assume $b_0=0$, and let $\mathfrak {n}\in \mathcal {N}_q$ with $q\ge 1$. We have $g(\mathfrak {n})\in \mathcal {N}_q$ with

$$\begin{aligned} \textstyle f(g(\mathfrak {n}))= \sum _{n=0}^q c_n \cdot \mathfrak {n}^n = (f\diamond g)(\mathfrak {n})\in \mathcal {N}. \end{aligned}$$

(31)

$\circ$:

For $\mathfrak {a}\in \mathcal {A}$ with $\Vert \mathfrak {a}\Vert _\mathcal {A}< S$, we have

$$\begin{aligned} \textstyle f(g(\mathfrak {a}))= \sum _{n=0}^\infty c_n \cdot \mathfrak {a}^n = (f\diamond g)(\mathfrak {a})\in \mathcal {A}. \end{aligned}$$

(32)

Example 1

Let $\mathbb {K}=\mathbb {R}$, and define the power series

In the context of the above notations, we have $b_0=0$, $R=1$, $S=\ln (2)$, with $|g|_s<\text {e} ^{S}-1=R$ for $0\le s<S$ as well as

$$\begin{aligned} \textstyle ((f\diamond g) *h)(t) =f(g(t))\cdot h(t)=1\qquad \quad \forall \, t\in (-\ln (2),\ln (2)). \end{aligned}$$

(33)

We consider the power series¹⁰

and obtain for $t\in (-\ln (2),\ln (2))$ that

$$\begin{aligned} \textstyle (\tilde{f}\diamond \tilde{g})(t)=\tilde{f}(\tilde{g}(t))=\tilde{f}(e^{-t}-1)= e^{-t}\cdot \frac{-t}{e^{-t}-1}=\frac{t}{e^t-1}=f(g(t))=(f\diamond g)(t). \end{aligned}$$

(34)

Let now $Z\in \mathfrak {g}$ be fixed, and assume that one of the following situations hold:

• A) Let $\text {N} \subseteq \mathfrak {g}$ be a $\text {Nil} _{q}$-set for $q\ge 1$. We define $V:=\overline{\mathcal {G}}_1(\text {N} )$, and let $\text {End} (V)$ denote the set of all linear maps $V\rightarrow V$. Then, $(\mathcal {N},*_\mathcal {N},1_\mathcal {N})\equiv (\text {End} (V),\circ ,\text {id} _{V})$ is a unital $\mathbb {R}$-algebra, with $\text {ad} _{Z}\in \mathcal {N}_{q-1}$ for each $Z\in V$ by Remark 3.3).

• B) G is a Banach Lie group with

  $$\begin{aligned} \Vert \varvec{[}X, Y \varvec{]}\Vert \le \Vert X\Vert \cdot \Vert Y\Vert \qquad \quad \forall \, X,Y\in \mathfrak {g}, \end{aligned}$$

  and we have $\Vert Z\Vert <\ln (2)$. We define $V:=\mathfrak {g}$, and let $\text {End} ^{\textrm{c}}(V)$ denote the set of all continuous linear maps $V\rightarrow V$. Then, $(\mathcal {A},*_\mathcal {A},1_\mathcal {A},\Vert \cdot \Vert _\mathcal {A})\equiv (\text {End} ^{\textrm{c}}(V),\circ ,\text {id} _{V},\Vert \cdot \Vert _\text {op} )$ is a unital submultiplicative Banach algebra over $\mathbb {R}$, and we have $\text {ad} _{Z}\in \text {End} ^{\textrm{c}}(V)$ with $\Vert \text {ad} _{Z}\Vert _\text {op} < \ln (2)$.

In both situations A) and B), the above discussions together with (33) and (34) show

$$\begin{aligned} f(g(\text {ad} _{Z}))\circ h(\text {ad} _{Z})&=((f\diamond g) *h)(\text {ad} _{Z})=\text {id} _V\\ \tilde{f}(\tilde{g}(\text {ad} _{Z}))&=f(g(\text {ad} _{Z})) \end{aligned}$$

for $Z\in V$. This can be rewritten as

$$\begin{aligned} \begin{aligned} \textstyle \Psi \big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{Z}^n\big )(\Phi (\text {ad} _{Z})(Y))&\textstyle =Y\\ \textstyle \widetilde{\Psi }\big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{-Z}^n\big )(Y)&\textstyle =\Psi \big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{Z}^n\big )(Y) \end{aligned} \end{aligned}$$

(35)

for each $Y\in V$, whereby for maps $\xi ,\zeta :\mathfrak {g}\rightarrow \mathfrak {g}$ and $X\in \mathfrak {g}$ we set (convergence presumed)

$$\begin{aligned} \textstyle \Psi (\xi )(X)&\textstyle := \sum _{n=1}^\infty \frac{(-1)^{n-1}}{n}\cdot (\xi -\text {id} _\mathfrak {g})^{n-1}(X)\\ \textstyle \widetilde{\Psi }(\xi )(X)&\textstyle := \sum _{n=1}^\infty \frac{(-1)^{n-1}}{n}\cdot \big (\xi \circ (\xi -\text {id} _\mathfrak {g})^{n-1}\big )(X)\\ \textstyle \Phi (\zeta )(X)&\textstyle := \sum _{n=0}^\infty \frac{1}{(n+1)!}\cdot \zeta ^{n}(X). \end{aligned}$$

Equation (35) will be relevant for our discussions in Sect. 4.1. $\square$

2.4 The evolution map

The subject of this section is the evolution map. We recall its elementary properties (confer also [8]), as well as the differentiability results obtained in [11]. We furthermore introduce the notion of weak $C^k$-regularity (confer Definition 1).

2.4.1 Elementary definitions

The right logarithmic derivative is given by

for each interval $D\subseteq \mathbb {R}$. Notably, for $\mu \in C^1(D,G)$, $g\in G$, an interval $D'\subseteq D$, and $\varrho :D''\rightarrow D$ of class $C^1$ for $D''\subseteq \mathbb {R}$ an interval, we have

(36)

Moreover, for $\mu ,\nu \in C^1(D,G)$, it follows from the product rule (11) that

holds. For $a<b$ and $k\in \mathbb {N}\cup \{\text {lip} ,\infty ,\text {c} \}$, we define

Now, restricted to the set

$$\begin{aligned} C_*^1([a,b],G):=\{\mu \in C^1([a,b],G)\,|\, \mu (a)=e\} \end{aligned}$$

is injective for $a<b$ (confer, e.g., Lemma 9 in [8]). We thus obtain a map

$$\begin{aligned} \textstyle \text {Evol} :\mathfrak {D}:=\bigcup _{a<b}\mathfrak {D}_{[a,b]}\rightarrow \bigcup _{a<b}C_*^1([a,b],G), \end{aligned}$$

if for $a<b$ we define

Notably, for $a<b$ and $k\in \mathbb {N}\cup \{\text {lip} ,\infty ,\text {c} \}$, we have (confer, e.g., Lemma 10 in [8])

$$\begin{aligned} \text {Evol} |_{\mathfrak {D}_{[a,b]}^k}:\mathfrak {D}_{[a,b]}^k\rightarrow C^{k+1}([a,b],G). \end{aligned}$$

(37)

Moreover, for $a<b$ and $\phi \in \mathfrak {D}_{[a,b]}^k$, we have

$$\begin{aligned} \phi |_{[a',b']}\in \mathfrak {D}_{[a',b']}^k\qquad \quad \forall \, a\le a'<b'\le b \end{aligned}$$

by the second equality in (36) as well as Lemma 10 in [8].

Remark 4

Given $X\in \mathfrak {g}$, there exist $\varepsilon >0$ and $\phi _X\in \mathfrak {D}^\infty _{[0,\varepsilon ]}$ with $\phi _X(0)=X$. In fact, fix $\varepsilon >0$ with $(-2\varepsilon ,2\varepsilon )\cdot \text {d} _e\Xi (X)\subseteq \mathcal {V}$, and define

$$\begin{aligned} \mu _X:(-\varepsilon ,\varepsilon )\ni t\mapsto \Xi ^{-1}(t\cdot \text {d} _e\Xi (X))\in \mathcal {U}. \end{aligned}$$

Then, $\mu _X$ is of class $C^\infty$, and has the desired properties. $\square$

Example 2

(The Riemann Integral) Assume $(G,\cdot )\equiv (F,+)$ equals the additive group of some $F\in \textsf {hlcVect}$. We have for each $a<b$, hence

$$\begin{aligned} \textstyle \mathfrak {D}_{[a,b]}=\big \{\gamma \in C^0([a,b],\mathfrak {g})\,\big |\, \int _a^t \gamma (s)\,\text {d} s \in \mathfrak {g}\,\text { for each }\, t\in [a,b]\big \} \end{aligned}$$

as well as $\text {Evol} (\gamma ):[a,b]\ni t\mapsto \int _a^t \gamma (s)\,\text {d} s$ for each $\gamma \in \mathfrak {D}_{[a,b]}$. $\square$

2.4.2 The product integral

The product integral is defined by

We set as well as for $a<b$, $\phi \in \mathfrak {D}_{[a,b]}$, $c\in [a,b]$, and define

We let ${\text {evol} }_\text {k} \equiv {\text {evol} }_{[0,1]}^k$ as well as $\text {D} _\text {k} \equiv \mathfrak {D}^k_{[0,1]}$ for each $k\in \mathbb {N}\cup \{\text {lip} ,\infty ,\text {c} \}$. We furthermore let

$$\begin{aligned} {\text {evol} }\equiv {\text {evol} }_{\textrm{0}}:\text {D} \equiv \text {D} _0\rightarrow G. \end{aligned}$$

The following elementary identities hold for $a<b$, confer [4, 15] or Sect. 3.5.2 in [8]:

• (a) For each $\phi ,\psi \in \mathfrak {D}_{[a,b]}$, we have with

  

• (b) For each $\phi ,\psi \in \mathfrak {D}_{[a,b]}$, we have with

  

• (c) For each $\phi \in \mathfrak {D}_{[a,b]}$, we have with

  

• (d) For $a=t_0<{\dots }<t_n=b$ and $\phi \in \mathfrak {D}_{[a,b]}$, we have

  

• (e) For $\varrho :[a',b']\rightarrow [a,b]$ of class $C^1$ and $\phi \in \mathfrak {D}_{[a,b]}$, we have ${\dot{\varrho }}\cdot (\phi \circ \varrho )\in \mathfrak {D}_{[a',b']}$ with

  

• (f) For each homomorphism $\Psi :G\rightarrow H$ between Lie groups G and H that is of class $C^1$, we have

  

Remark 5

Let $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ and $a<b$ be given. We have by (a), (c), and Lemma 13 in [8] (confer also Lemma 8) that

holds for each $\phi ,\psi \in \mathfrak {D}^k_{[a,b]}$. It is then not hard to see that $(\mathfrak {D}^k_{[a,b]},\star ,\cdot ^{-1},0)$ is a group:

• We have $(\psi ^{-1})^{-1}=\psi$ for each $\psi \in \mathfrak {D}^k_{[a,b]}$. In fact, applying (c) twice, we obtain

  

  The claim is thus clear from injectivity of .

• We have $\psi \star \psi ^{-1} =0=\psi ^{-1}\star \psi$ for each $\psi \in \mathfrak {D}^k_{[a,b]}$. In fact, it is clear that

  $$\begin{aligned} \psi \star \psi ^{-1} =0\qquad \quad \forall \, \psi \in \mathfrak {D}^k_{[a,b]}. \end{aligned}$$

  Then, we obtain from the previous point that

  $$\begin{aligned} \psi ^{-1}\star \psi =\psi ^{-1}\star (\psi ^{-1})^{-1}=0\qquad \quad \forall \, \psi \in \mathfrak {D}^k_{[a,b]}. \end{aligned}$$

• We have $\phi \star (\psi \star \chi ) = (\phi \star \psi )\star \chi$ for all $\phi ,\psi ,\chi \in \mathfrak {D}^k_{[a,b]}$. In fact, we obtain from (a) that

  

  The claim is thus clear from injectivity of .

We will reconsider this group structure in Sects. 4.2.1 and 5. $\square$

Example 3

(The Riemann Integral) Assume we are in the situation of Example 2. Then,

holds; and, the identities (a), (d), (e), and (f) encode (in the given order) the additivity of the Riemann integral, (16), (21), and (22) (for if in (f), $(H,\cdot )\equiv (\tilde{F},+)$ is the additive group of some further $\tilde{F}\in \textsf {hlcVect}$), respectively. $\square$

Example 4

For $a<b$, we define

$$\begin{aligned} \begin{aligned} {\mathfrak {inv}}:C^0([a,b],\mathfrak {g})&\rightarrow C^0([a,b],\mathfrak {g})\\ \phi&\mapsto [t\mapsto -\phi (a+b-t)]. \end{aligned} \end{aligned}$$

(38)

Let $a<b$ be fixed. Then, ${\mathfrak {inv}}|_{C^0([a,b],\mathfrak {g})}$ is linear. Moreover, for $\phi \in C^0([a,b],\mathfrak {g})$ and $\varrho :[a,b]\ni t\mapsto a +b -t\in [a,b]$, we have ${\mathfrak {inv}(\phi )}={\dot{\varrho }}\cdot \phi \circ \varrho$. Then, (e) shows that ${\mathfrak {inv}(\phi )}\in \mathfrak {D}^k_{[a,b]}$ holds for each $\phi \in \mathfrak {D}^k_{[a,b]}$ and $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, with

(39)

For instance, in the situation of Example 3, the right side of (39) reads

$$\begin{aligned} \textstyle -\int _a^{b} \phi (s)\,\text {d} s = \int _a^{b} -\phi (a+b-s)\,\text {d} s, \end{aligned}$$

which is in line with the second point in Remark 2. The relation will be useful for our argumentation in Sect. 3.2. $\square$

Example 5

(The Lie bracket and Homomorphisms) Assume that we are in the situation of (f), and let $(\mathfrak {h},\varvec{[}\cdot , \cdot \varvec{]}_\mathfrak {h})$ denote the Lie algebra of H. Then, we have

$$\begin{aligned} \textstyle \text {d} _e\Psi (\varvec{[}X, Y \varvec{]})=\varvec{[}\text {d} _e\Psi (X), \text {d} _e\Psi (Y) \varvec{]}_\mathfrak {h}\qquad \quad \forall \, X,Y\in \mathfrak {g}. \end{aligned}$$

(40)

Proof of Equation (40)

Let $X,Y\in \mathfrak {g}$ be fixed, and choose $\phi _{X}:[0,\varepsilon _1]\rightarrow \mathfrak {g}$ as well as $\phi _{Y}:[0,\varepsilon _2]\rightarrow \mathfrak {g}$ as in Remark 4. We obtain

which shows (40). $\square$

For instance, let $\Psi \equiv \text {Conj} _g:G\rightarrow H= G$ with $g\in G$. We obtain

$$\begin{aligned} \text {Ad} _g(\varvec{[}X, Y \varvec{]})=\varvec{[}\text {Ad} _g(X), \text {Ad} _g(Y) \varvec{]}\qquad \quad \forall \, g\in G,\, X,Y\in \mathfrak {g}. \end{aligned}$$

(41)

Then, given $X,Y,Z\in \mathfrak {g}$, fix $\mu :(-\varepsilon ,\varepsilon )\rightarrow G$ ($\varepsilon >0$) of class $C^1$, with $\mu (0)=e$ and ${\dot{\mu }}(0)=Z$. We obtain from (41) and the parts b), d), e) of Proposition A.1 that

$$\begin{aligned} \textstyle \varvec{[}Z, \varvec{[}X, Y \varvec{]} \varvec{]}&\textstyle = \frac{\text {d} }{\text {d} h}\big |_{h=0} \text {Ad} _{\mu (h)}(\varvec{[}X, Y \varvec{]}) \textstyle = \frac{\text {d} }{\text {d} h}\big |_{h=0} [\text {Ad} _{\mu (h)}(X),\text {Ad} _{\mu (h)}(Y)] \\&\textstyle = \varvec{[}\varvec{[}Z, X \varvec{]}, Y \varvec{]}+ \varvec{[}X, \varvec{[}Z, Y \varvec{]} \varvec{]} \end{aligned}$$

holds, which is the Jacobi identity (10). $\square$

2.4.3 Weak regularity

In this section, we recall certain differentiation results from [11] and introduce the notion of weak $C^k$-regularity for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ (cf. Definition 1). We say that G is $C^k$-semiregular [4, 8, 11] for $k\in \mathbb {N}\cup \{\text {lip} ,\infty ,\text {c} \}$ if $\text {D} _\text {k} =C^k([0,1],\mathfrak {g})$ holds.

Remark 6

It follows from (e) when applied to affine transformations that G is $C^k$-semiregular if and only if $\mathfrak {D}^k_{[a,b]}=C^k([a,b],\mathfrak {g})$ holds for all $a<b$ (confer, e.g., Lemma 12 in [8]).$\square$

We write $\lim ^\infty _{n}\mu _n=\mu$ for $\{\mu _n\}_{n\in \mathbb {N}}\subseteq C^0([a,b],G)$ and $\mu \in C^0([a,b],G)$ if $\{\mu _n\}_{n\in \mathbb {N}}$ converges uniformly to $\mu$, i.e., if to each open $U\subseteq G$ with $e\in U$, there exists $N_U\in \mathbb {N}$ with

$$\begin{aligned} \mu _n(t)\in \mu (t)\cdot U\qquad \quad \forall \, t\in [a,b]. \end{aligned}$$

We recall (14), and say that G is Mackey k-continuous for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ if

Lemma 13 in [11] shows:

Lemma A

G is Mackey k-continuous for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ if and only if for each $a<b$ the following implication holds:

Theorem 1 in [11] states:

Theorem B

If G is $C^k$-semiregular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, then G is Mackey k-continuous.

Theorem 3 in [11] (in particular) states:

Theorem C

Assume that G is Mackey k-continuous for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$. Let $\Phi :I\times [a,b]\rightarrow \mathfrak {g}$ ($I\subseteq \mathbb {R}$ open) be given with $\Phi (z,\cdot )\in \mathfrak {D}^k_{[a,b]}$ for each $z\in I$. Then, for $x\in I$ we have

provided that the following conditions hold:

• (i) We have $(\partial _1 \Phi )(x,\cdot )\in C^k([a,b],\mathfrak {g})$.

• (ii) To $\textsf {p} \in \textsf {Sem}(E)$ and $\text {s} \preceq k$, there exists $L_{\textsf {p} ,\text {s} }\ge 0$ as well as $I_{\textsf {p} ,\text {s} }\subseteq I$ open with $x\in I_{\textsf {p} ,\text {s} }$, such that

  $$\begin{aligned} \textstyle \textsf {p} ^\text {s} _\infty (\Phi (x+h,\cdot )-\Phi (x,\cdot ))\le |h|\cdot L_{\textsf {p} ,\text {s} }\qquad \quad \forall \, h\in \mathbb {R}_{\ne 0}\,\text { with }\, x+h\in I_{\textsf {p} ,\text {s} }. \end{aligned}$$

In particular, we have (recall Convention 1)

if and only if the Riemann integral on the right side exists in $\mathfrak {g}$.

Recall from the end of Sect. 2.2.3 that the last condition in Theorem C concerning the Riemann integral is always fulfilled

• for $k\in \mathbb {N}_{\ge 1}\cup \{\text {lip} ,\infty \}$ if $\mathfrak {g}$ is Mackey complete,

• for $k=0$ if $\mathfrak {g}$ integral complete for $k=0$.

In particular, this implies the following statement (cf. Theorem 2 in [11] and Corollary 3 in [11]):

Proposition D

Assume that G is $C^k$-semiregular for $k\in \mathbb {N}\cup \{\text {lip} , \infty \}$. Then, ${\text {evol} }_\text {k}$ is differentiable if and only if $\mathfrak {g}$ is Mackey complete for $k\in \mathbb {N}_{\ge 1}\cup \{\text {lip} ,\infty \}$ as well as integral complete for $k=0$. In this case, ${\text {evol} }^k_{[a,b]}$ is differentiable for each $a<b$, with

Moreover, for $a<b$ and $\phi ,\psi \in C^k([a,b],\mathfrak {g})$, we have

Proof

Clear from Theorem 2 in [11] and Corollary 3 in [11]. $\square$

Proposition D motivates the following definition¹¹:

Definition 1

G is said to be weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ if G is $C^k$-semiregular, and $\mathfrak {g}$ is Mackey complete for $k\in \mathbb {N}_{\ge 1}\cup \{\text {lip} ,\infty \}$ as well as integral complete for $k=0$. $\square$

Remark 7

Apart from $C^k$-semiregularity, the (standard) notion of $C^k$-regularity involves smoothness (and continuity) of the evolution map w.r.t. the $C^k$-topology. These additional assumptions, however, are unnecessarily strong for our purposes. This is because, due to the results stated above, the usual differentiability properties of the evolution map are already available in the weakly $C^k$-regular context. $\square$

Remark 8

Theorem C will in particular be applied to the situation in Example 2, i.e., where $(G,\cdot )\equiv (F,+)$ equals the additive group of some given $F\in \textsf {hlcVect}$. For this observe that is $C^0$-continuous, hence Mackey 0-continuous. Moreover, $G\equiv F$ is

• $C^0$-semiregular if F is integral complete,

• $C^\text {lip}$-semiregular if F is Mackey complete.$\square$

For instance, we obtain the following statements that we shall need in Sect. 3.1.

Corollary 3

Let $F\in \textsf {hlcVect}$, $k\in \mathbb {N}$, $a<b$, and $I\subseteq \mathbb {R}$ an open interval. Let furthermore $\Theta ^0:I\times [a,b] \rightarrow F$ be a map with $\alpha _x:=\Theta ^0(\cdot ,x)\in C^k(I,F)$ for each $x\in [a,b]$, and assume that for $0\le \ell \le k$ the map

$$\begin{aligned} \textstyle \Theta ^\ell :I\times [a,b]\ni (t,x)\mapsto \alpha _x^{(\ell )}(t)\in F \end{aligned}$$

is continuous such that

$$\begin{aligned} \textstyle \varvec{\gamma }^\ell :I\ni t \mapsto \int _a^b \Theta ^\ell (t,x) \,\text {d} x\in F \end{aligned}$$

is defined. Then, $\varvec{\gamma }:=\varvec{\gamma }^0\in C^k(I,F)$ holds with $\varvec{\gamma }^{(\ell )}=\varvec{\gamma }^\ell$ for $0\le \ell \le k$.

Proof

For each $0\le \ell \le k$, we have by (19), compactness of [a, b], and continuity of $\Theta ^\ell$ that $\varvec{\gamma }^\ell \in C^0(I,F)$ holds. Let now $0\le \ell < k$ be given, and set $\Phi \equiv \Theta ^\ell$. We observe the following:

• For $t\in I$, we have $\Phi (t,\cdot )=\Theta ^\ell (t,\cdot )\in C^0([a,b],F)$.

• For $t\in I$, we have $\partial _1\Phi (t,\cdot )\in C^0([a,b],F)$, as

  $$\begin{aligned} \partial _1\Phi (t,x)=\partial _1\Theta ^\ell (t,x)=\alpha _x^{(\ell +1)}(t)=\Theta ^{\ell +1}(t,x)\qquad \quad \forall \, t\in I,\, x\in [a,b]. \end{aligned}$$

• For $\textsf {q} \in \textsf {Sem}(F)$, $x\in [a,b]$, $t\in I$, and $\tau >0$ with $t+[-\tau ,\tau ]\subseteq I$, we have by (18)

  $$\begin{aligned} \textstyle 1/|h|\cdot \textsf {q} (\Phi (t+h,x)&-\Phi (t,x))\\&\textstyle = 1/|h|\cdot \textsf {q} \big (\alpha ^{(\ell )}_x(t+h)-\alpha _x^{(\ell )}(t)\big )\\&\le \sup \big \{-\tau \le s\le \tau \,|\, \textsf {q} \big (\alpha _x^{(\ell +1)}(t+s)\big )\big \}\\&\le \sup \big \{-\tau \le s\le \tau ,\,a\le y\le b\,|\, \textsf {q} \big (\Theta ^{\ell +1}(t+s,y)\big )\big \} \end{aligned}$$

  for each $-\tau \le h\le \tau$.

Theorem C (Remark 8) shows that for each $t\in I$, we have

$$\begin{aligned} \textstyle \frac{\text {d} }{\text {d} h}\big |_{h=0} \,\varvec{\gamma }^\ell (t+h)&\textstyle =\frac{\text {d} }{\text {d} h}\big |_{h=0} \, \int _a^b \Theta ^\ell (t+h,x)\,\text {d} x\\&\textstyle =\frac{\text {d} }{\text {d} h}\big |_{h=0} \, \int _a^b \Phi (t+h,x)\,\text {d} x\\&\textstyle =\int _a^b\partial _1 \Phi (t,x)\,\text {d} x\\&\textstyle =\int _a^b\partial _1 \Theta ^\ell (t,x)\,\text {d} x\\&\textstyle = \int _a^b \alpha _x^{(\ell +1)}(t)\,\text {d} x\\&\textstyle = \int _a^b \Theta ^{\ell +1}(t,x)\,\text {d} x\\&\textstyle = \varvec{\gamma }^{\ell +1}(t). \end{aligned}$$

Since this holds for each $0\le \ell <k$, the claim follows from $\varvec{\gamma }^{(0)}=\varvec{\gamma }=\varvec{\gamma }^{0}$ by induction. $\square$

Lemma 7

Let $F\in \textsf {hlcVect}$ be Mackey complete, $a<b$, $I\subseteq \mathbb {R}$ an open interval, $\Theta ^0:I\times [a,b] \rightarrow F$ continuous, and $\Omega :F\times F\rightarrow F$ smooth. Assume that the following two conditions are fulfilled:

• (a) $\Theta ^0(t,\cdot )~~\in C^1([a,b],F)$ holds for each $t\in I$.

• (b) $\alpha _x:=\Theta ^0(\cdot ,x)\in C^1(I,F)$ holds for each $x\in [a,b]$, with

  $$\begin{aligned} \textstyle {\dot{\alpha }}_x(t)=\Omega \big (\int _x^b\alpha _y(t)\,\text {d} y,\alpha _x(t)\big )\qquad \quad \forall \, t\in I. \end{aligned}$$

Then, $\varvec{\gamma }[z]:I\ni t\mapsto \int _z^b \Theta ^0(t,x)\,\text {d} x\in F$ is smooth for each $z\in [a,b]$.

Proof

It suffices to prove the following statement. $\square$

Statement 1

Let $k\in \mathbb {N}$ be given. Then, $\alpha _x\in C^k(I,F)$ holds for each $x\in [a,b]$. Moreover,

$$\begin{aligned} \textstyle \Theta ^\ell :I\times [a,b]\ni (t,x)\mapsto \alpha _x^{(\ell )}(t)\in F \end{aligned}$$

is continuous for $0\le \ell \le k$, with $\Theta ^\ell (t,\cdot )\in C^1([a,b],F)$ for each $t\in I$.

In fact, let $k\in \mathbb {N}$ be given. Since F is Mackey complete, it follows from Statement 1 that

$$\begin{aligned} \textstyle \varvec{\gamma }[z]^\ell :I\ni t \mapsto \int _z^b \Theta ^\ell (t,x)\,\text {d} x\in F \end{aligned}$$

exists for each $z\in [a,b]$ and $0\le \ell \le k$. Moreover, Corollary 3 shows that $\varvec{\gamma }[z]\in C^k(I,F)$ holds for each $z\in [a,b]$, with

$$\begin{aligned} \textstyle \varvec{\gamma }[z]^{(\ell )}= \varvec{\gamma }[z]^\ell =\int _z^b \Theta ^\ell (\cdot ,x) \,\text {d} x\qquad \quad \forall \, 0\le \ell \le k. \end{aligned}$$

(42)

Since $k\in \mathbb {N}$ was arbitrary, Lemma 7 follows.

It thus remains to prove Statement 1:

Proof of Statement 1 We first discuss the cases $k\in \{0,1\}$ and then argue by induction:

• $k=0$: Clear from the assumptions.

• $k=1$: By (b), we have $\alpha _x\in C^1(I,F)$ for each $x\in [a,b]$. Since $\Theta ^0$ is continuous,

  $$\begin{aligned} \textstyle \Gamma ^0:I\times [a,b]\ni (t,z) \mapsto \int _z^b \Theta ^0(t,x)\,\text {d} x\in F \end{aligned}$$

  is continuous. It is clear that $\Gamma ^0(t,\cdot )\in C^1([a,b],F)$ holds for each $t\in I$; and we have $\Theta ^0(t,\cdot )\in C^1([a,b],F)$ for each $t\in I$ by (a). Now, (b) yields

  $$\begin{aligned} \textstyle \Theta ^1(t,x)={\dot{\alpha }}_x(t)=\Omega (\Gamma ^0(t,x),\Theta ^0(t,x))\qquad \quad \forall \, t\in I,\, x\in [a,b]. \end{aligned}$$

  Since $\Omega$ is smooth, $\Theta ^1$ is continuous with $\Theta ^1(t,\cdot )\in C^1([a,b],F)$ for each $t\in I$.

Assume now that Statement 1 holds for some $k\ge 1$. We observe the following:

• (i) $\Theta ^\ell$ is continuous for $0\le \ell \le k$, hence

  $$\begin{aligned} \textstyle \Gamma ^\ell :I\times [a,b]\ni (t,z) \mapsto \int _z^b \Theta ^\ell (t,x)\,\text {d} x\in F \end{aligned}$$

  is continuous for $0\le \ell \le k$.

• (ii) For $z\in [a,b]$, we have $\alpha _z,\varvec{\gamma }[z]\in C^k(I,F)$, with

  $$\begin{aligned} \textstyle \alpha _z^{(\ell )}=\Theta ^\ell (\cdot ,z)\qquad \text {and}\qquad \varvec{\gamma }[z]^{(\ell )}\textstyle {\mathop {=}\limits ^{(42)}} \Gamma ^\ell (\cdot ,z)\qquad \text {for}\qquad 0\le \ell \le k. \end{aligned}$$

  (43)

  Then, i) implies that the following maps are continuous:

  $$\begin{aligned} \begin{aligned} I\times [a,b]\ni (t,z)&\mapsto ~~\alpha ^{(\ell )}_z(t)\in F\\ I\times [a,b]\ni (t,z)&\mapsto \varvec{\gamma }[z]^\ell (t)\in F \end{aligned} \end{aligned}$$

  (44)

  for $0\le \ell \le k$

• (iii) For $t\in I$ and $0\le \ell \le k$, the maps

  $$\begin{aligned} \begin{aligned} {[}a,b]\ni z&\mapsto ~~\alpha _z^{(\ell )}(t)\in F\\ {[}a,b]\ni z&\mapsto \varvec{\gamma }[z]^{(\ell )}(t)\in F \end{aligned} \end{aligned}$$

  (45)

  are of class $C^1$ (use (43)).

Then, by (b) we have

$$\begin{aligned} {\dot{\alpha }}_x(t)=\Omega (\varvec{\gamma }[x](t),\alpha _x(t))\qquad \quad \forall \, t\in I,\, x\in [a,b]. \end{aligned}$$

(46)

Together with ii), this shows $\alpha _x\in C^{k+1}(I,F)$ for all $x\in [a,b]$. Moreover, set $\gamma [x]_1:=\alpha _x$ and $\gamma [x]_2:=\varvec{\gamma }[x]$ for each $x\in [a,b]$. It follows from (46) that¹²

$$\begin{aligned} \Theta ^{k+1}:I\times [a,b]\ni (t,x)=\alpha _x^{(k+1)}(t)\in F \end{aligned}$$

is a sum of maps of the form

$$\begin{aligned} I\times [a,b]\ni (t,x)\mapsto \Psi \big (\gamma [x]^{(\ell _1)}_{i_1}(t),\dots ,\gamma [x]_{i_m}^{(\ell _m)}(t)\big )\in F \end{aligned}$$

for certain $0\le \ell _1,\dots ,\ell _m\le k$, $1\le i_1,\dots ,i_m \le 2$, $m\ge 2$, where $\Psi :F^{m}\rightarrow F$ is smooth. Then, (44) implies that $\Theta ^{k+1}$ is continuous, and (45) implies that $\Theta ^{k+1}(t,\cdot )\in C^1([a,b],F)$ holds for each $t\in I$. This establishes the claim for $k+1$, so that Statement 1 follows by induction. $\square$

This proves the Lemma 7. $\square$

2.4.4 The exponential map

We let $\mathfrak {i}:\mathfrak {g}\ni X\rightarrow \mathcal {C}_X|_{[0,1]}\in C^\text {c} ([0,1],\mathfrak {g})$, hence

$$\begin{aligned} \mathfrak {i}^{-1}:C^\text {c} ([0,1],\mathfrak {g}) \rightarrow \mathfrak {g},\qquad \chi \mapsto \chi (0). \end{aligned}$$

(47)

The exponential map is given by

• Instead of saying that G is $C^\text {c}$-semiregular, in the following we rather say that G admits an exponential map.

• The relation (e) implies $\mathbb {R}\cdot \text {dom} [\exp ]\subseteq \text {dom} [\exp ]$, as well as that $t\mapsto \exp (t\cdot X)$ is a 1-parameter group for each $X\in \text {dom} [\exp ]$ with

  

  (48)

  confer, e.g., Remark 2.1) in [8].

Finally, Theorem 1 in [11] (cf. Theorem B), Corollary 6 in [11], and Remark 9.1 in [11] provide the following statement.

Corollary 4

Let $\mathfrak {X}:I\rightarrow \text {dom} [\exp ]\subseteq \mathfrak {g}$ ($I\subseteq \mathbb {R}$ an open interval) be of class $C^1$, and set $\alpha :=\exp \circ \,\mathfrak {X}$. Assume that G is weakly $C^\infty$-regular, or that $\exp :\mathfrak {g}\rightarrow G$ is defined and of class $C^1$. Then, $\alpha$ is of class $C^1$, with

$$\begin{aligned} \textstyle {\dot{\alpha }}(t)=\text {d} _e\text {L} _{\exp (\mathfrak {X}(t))}\big (\int _0^1 \text {Ad} _{\exp (-s\cdot \mathfrak {X}(t))}({\dot{\mathfrak {X}}}(t)) \,\text {d} s \, \big )\qquad \quad \forall \, t\in I. \end{aligned}$$

Proof

If G is weakly $C^\infty$-regular, then the claim is clear from Theorem 1 in [11] and Corollary 6 in [11]. Assume now that $\exp$ is of class $C^1$. Then, Remark 9.1 in [11] shows that

$$\begin{aligned} \textstyle \text {d} _X\exp (Y)= \text {d} _e\text {L} _{\exp (X)}\big (\int _0^1 \text {Ad} _{\exp (-s\cdot X)}(Y) \,\text {d} s\big ) \end{aligned}$$

holds, so that the claim is clear from Part d of Proposition A.1. $\square$

3 Preliminary results

In this section, we derive some elementary results from Theorem C (Proposition 1 in Sect. 3.1 and Proposition 2 in Sect. 3.2) and provide an integral expansion for the adjoint action (Lemma 13 in Sect. 3.3). Proposition 1 and Lemma 13 will be used in Sect. 4.2 to investigate the product integral of nilpotent curves. Also, Proposition 2 was supposed to be applied in this paper, but eventually turned out not to be necessary for our argumentation. We kept this result for academic reasons, and because it certainly will play a role in future applications. Our considerations in Sects. 3.1 and 3.2 furthermore serve as motivations for the constructions made in Sect. 5.

3.1 An integral transformation

Let $\mathfrak {R}:=\{\phi \in \mathfrak {D}\,|\, [0,1]\cdot \phi \subseteq \mathfrak {D}\}$. We define $\mathfrak {T}:\mathfrak {R}\rightarrow \textsf {Maps} ([0,1],\widehat{\mathfrak {g}})$ by

(49)

for each $a<b$. In view of Sect. 4.2 (the proof of Lemma 16), for $a<b$ and $\phi \in \mathfrak {R}\cap \mathfrak {D}_{[a,b]}$, we define

$$\begin{aligned} \mathfrak {T}(\phi |_{[a,a]}):[0,1]\ni t \mapsto 0\in \mathfrak {g}. \end{aligned}$$

(50)

In this section, we proof the following proposition.

Proposition 1

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$. Then, for $a<b$ and $\phi \in C^k([a,b],\mathfrak {g})$, we have $\mathfrak {T}(\phi )\in C^\infty ([0,1],\mathfrak {g})$ with

In particular, holds, and is smooth by (37).

Example 6

Assume that G is abelian, as well as weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$. Then, Proposition 1 recovers the well-known formula [18]

(51)

for each $a<b$ and $\phi \in C^k([a,b],\mathfrak {g})$. $\square$

Remark 9

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$; and recall the group structure discussed in Remark 5, as well as the map (38) introduced in Example 4. It is straightforward from Proposition 1 and the properties of the product integral that

$$\begin{aligned} \begin{aligned} \mathfrak {T}(\psi )^{-1}&=\mathfrak {T}({\mathfrak {inv}(\psi )})\\ \mathfrak {T}(\psi )&=\mathfrak {T}(\psi |_{[c,b]})\star \mathfrak {T}(\psi |_{[a,c]})\\ \mathfrak {T}(\psi )&=\mathfrak {T}({\dot{\varrho }}\cdot (\psi \circ \varrho )) \end{aligned} \end{aligned}$$

(52)

holds for $a<c<b$, $\psi \in C^k([a,b],\mathfrak {g})$, as well as $\varrho :[a',b']\rightarrow [a,b]$ ($a'<b'$) of class $C^1$ with ${\dot{\varrho }}|_{(a',b')}>0$, $\varrho (a')=a$, $\varrho (b')=b$:

• By (a) (first step), Proposition 1 (second step), and (39) (fourth step), we have

  

  for each $t\in [0,1]$. Then, injectivity of implies $\mathfrak {T}(\psi )\star \mathfrak {T}({\mathfrak {inv}(\psi )})=0$, which proves the first line in (52).

• By Proposition 1 (first and third step), (d) (second step), and (a) (fourth step), we have

  

  for each $t\in [0,1]$. The second line in (52) now follows from injectivity of .

• Since ${\dot{\varrho }}|_{(a',b')}>0$ holds, the substitution formula (e) applies to $\varrho |_{[x',y']}$ and $t\cdot \psi |_{[\varrho (x'),\varrho (y')]}$ for all $a'\le x'< y'\le b'$ and $t\in [0,1]$. We obtain

  

  where we have applied (21) in the third step. This proves the third line in (52). $\square$

Remark 10

The smoothness statement in Proposition 1, in particular, ensures that $\mathfrak {T}$ can be applied iteratively. We will use this fact in Sect. 4.2 to prove an identity for the product integral in the nilpotent context.$\square$

For the proof of Proposition 1, we shall need the following statements.

Lemma 8

Let $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, $a<b$, $\mu \in C^{k+1}([a,b],G)$, and $\phi \in C^k([a,b],\mathfrak {g})$. Then, we have $\text {Ad} _\mu (\phi )\in C^k([a,b],\mathfrak {g})$.

Proof

Confer, e.g., Lemma 13 in [8]. $\square$

Lemma 9

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $\phi \in C^k([a,b])$ ($a<b$) be given. Then, is continuous.

Proof

Confer Appendix A.5. $\square$

Lemma 10

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $a<b$ as well as $\phi \in C^k([a,b])$ be given. Then, is smooth.

Proof

We consider the map

and observe the following:

• Lemma 9 shows that $\Theta ^0$ is continuous, as we have (with $\kappa$ as in Lemma 9)

  $$\begin{aligned} \Theta ^0(t,x)= \text {Ad} _{\kappa (t,x)}(\phi (x))\qquad \quad \forall \, t\in \mathbb {R}, \, x\in [a,b]. \end{aligned}$$

• Given $t\in \mathbb {R}$, we have $\Theta ^0(t,\cdot )\in C^k([a,b],\mathfrak {g})\subseteq C^1([a,b],\mathfrak {g})$ by Lemma 8, (37), and

  

• Given $x\in [a,b]$, Proposition D yields

  

  Then, Lemma 8 shows

  $$\begin{aligned} \textstyle C^1(\mathbb {R},\mathfrak {g})\ni \alpha _x:= \Theta ^0(\cdot ,x) :\mathbb {R}\ni t \mapsto \text {Ad} _{\mu _x(t)}(\phi (x))\in \mathfrak {g}. \end{aligned}$$

  (53)

  Moreover, for $t,h\in \mathbb {R}$ we have

  

  (54)

  By Proposition D, the map is of class $C^1$ for $t\in \mathbb {R}$ (recall that $\Theta ^0(t,\cdot )\in C^k([a,b],\mathfrak {g})$ holds by the second point above), with

  

  (55)

  We obtain for $t\in \mathbb {R}$ that

  

  (56)

The claim is now clear from Lemma 7 (with $I=\mathbb {R}$, $\Omega =\varvec{[}\cdot , \cdot \varvec{]}$, and $z=b$ there). $\square$

We are ready for the proof of Proposition 1.

Proof of Proposition 1

By Lemma 10, we have $\mathfrak {T}(\phi )\in C^\infty ([0,1],\mathfrak {g})$. By Proposition D, we have with

In the third step, we have applied (22) with $\mathcal {L}\equiv \text {Ad} _{\mu (t)}$. This shows that holds, with $\mu (0)=e$. We obtain

which proves the claim. $\square$

The following corollary verifies the guess made in Remark 8.2) in [11].

Corollary 5

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $\phi ,\psi \in C^k([a,b],\mathfrak {g})$ be given. Then, is smooth.

Proof

We have

By Lemma 8, it thus suffices to prove the claim for $\phi =0$ and arbitrary $\psi \in C^k([a,b],\mathfrak {g})$. Let thus $\psi \in C^k([a,b],\mathfrak {g})$ be given. Then, we have

By Lemma 8, it thus suffices to prove smoothness of the map for each $\chi \in C^k([a,b],\mathfrak {g})$, so that the claim is clear from (the last statement in) Proposition 1. $\square$

3.2 Differentiation of parameter-dependent integrals

In this section, we prove the following differentiation result.

Proposition 2

Assume that G is Mackey k-continuous for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$. Let $a<b$, $\sigma >0$, $I\subseteq \mathbb {R}$ an open interval with $[a-\sigma ,b+\sigma ]\subseteq I$, as well as $\Phi :I\times [a-\sigma ,b+\sigma ]\rightarrow \mathfrak {g}$ a map with $\Phi (z,\cdot )\in \mathfrak {D}^k_{[a-\sigma ,b+\sigma ]}$ for each $z\in I$. Let $x\in [a,b]$ be given, such that the following conditions hold:

• (i) We have $(\partial _1 \Phi )(x,\cdot )\in C^k([a,b],\mathfrak {g})$ with .

• (ii) To $\textsf {p} \in \textsf {Sem}(E)$ and $\text {s} \preceq k$, there exists $L_{\textsf {p} ,\text {s} }\ge 0$ as well as $I_{\textsf {p} ,\text {s} }\subseteq I$ open with $x\in I_{\textsf {p} ,\text {s} }$, such that

  $$\begin{aligned} \textsf {p} ^\text {s} _\infty (\Phi (x+h,\cdot )-\Phi (x,\cdot ))\le |h|\cdot L_{\textsf {p} ,\text {s} }\qquad \quad \forall \, h\in \mathbb {R}_{\ne 0}\,\text { with }\, x+h\in I_{\textsf {p} ,\text {s} }. \end{aligned}$$

Then, we have

For the proof of Proposition 2, we shall need the following statements.

Lemma 11

To $\textsf {p} \in \textsf {Sem}(E)$, there exist $\textsf {p} \le \textsf {q} \in \textsf {Sem}(E)$ and $U\subseteq G$ open with $e\in U$, such that for each $a<b$ and $\chi \in \mathfrak {D}_{[a,b]}$ with we have

Proof

Confer Lemma 4 in [11]. $\square$

Corollary 6

For each compact $\text {C} \subseteq G$ and each $\textsf {q} \in \textsf {Sem}(E)$, there exists some $\textsf {q} \le \textsf {m} \in \textsf {Sem}(E)$, such that $\textsf {q} \circ \text {Ad} _g\le \textsf {m}$ holds for each $g\in \text {C}$.

Proof

This is clear from Lemma 4, because $\text {Ad} :G\times \mathfrak {g}\ni (g,X)\mapsto \text {Ad} _g(X)\in \mathfrak {g}$ is smooth as well as linear in the second argument. $\square$

Proof of Proposition 2

We consider the maps

and obtain

(57)

We furthermore observe the following:

(1):

The maps $\nu _+,\eta _-$ are continuous at $h=0$, with $\nu _+(0)=e=\eta _-(0)$. To show this, it suffices to prove the same property for the maps

First, it is clear that $\alpha _\pm$ is continuous (in particular at $h=0$) with $\alpha _\pm (0)=e$, as we have

Second, Condition ii) implies that for each sequence $[-\sigma ,0)\cup (0,\sigma ]\supseteq \{h_n\}_{n\in \mathbb {N}}\rightarrow 0$, we have . Since G is Mackey k-continuous, Lemma A yields the following:

• For each sequence $(0,\sigma ]\supseteq \{h_n\}_{n\in \mathbb {N}}\rightarrow 0$, we have

  

  This implies $\lim _{h\rightarrow 0}\beta _+(h) = e$.

• For each sequence $[-\sigma ,0)\supseteq \{h_n\}_{n\in \mathbb {N}}\rightarrow 0$, we have

  

  This implies $\lim _{h\rightarrow 0}\beta _-(h) = e$, as we have

  

(2):

We have

(58)

$$\begin{aligned} {\dot{\nu }}_-(0)&=\Phi (x,x) \end{aligned}$$

(59)

$$\begin{aligned} {\dot{\eta }}_+(0)&=\Phi (x,x). \end{aligned}$$

(60)

In fact, (60) is evident, (58) is clear from Theorem C, and (59) is obtained as follows:

$\square$

Proof of Equation (59). Given $-\sigma \le h<0$, we define $\varrho _h:[x,x-h]\ni t\mapsto t+h \in [x + h,x]$, and obtain

(61)

According to Example 4 (with $\varrho :[x+h,x]\ni t\mapsto 2x+h -t\in [x+h,x]$ there), we have

We obtain

which proves (59). $\square$

(3):

By 1), there exists $0<\delta \le \sigma$, such that the following maps are defined:

$$\begin{aligned} \Delta _+:(0,\delta ]\ni h\mapsto \textstyle \frac{1}{h}\cdot \Xi (\nu _+(h))\qquad \quad \text {and}\qquad \quad \Delta _-:[-\delta ,0)\ni h\mapsto \textstyle \frac{1}{h}\cdot \Xi (\eta _-(h)). \end{aligned}$$

Then, to establish the proof, it suffices to show

$$\begin{aligned} \textstyle \lim _{0<h\rightarrow 0}\Delta _+(h)=0\qquad \quad \text {as well as}\qquad \quad \lim _{0>h\rightarrow 0}\Delta _-(h)=0. \end{aligned}$$

(62)

In fact, let $(\Xi ',\mathcal {U}')$ be a chart around $\mu (0)$. Let furthermore $O\subseteq \mathcal {U}$ and $O'\subseteq \mathcal {U}'$ be open with $e\in O$ and $\mu (0)\in O'$, such that $O\cdot O\cdot O'\subseteq \mathcal {U}'$ holds. We set $W:=\Xi (O)$, $W':=\Xi '(O')$, $\mathcal {V}':=\Xi '(\mathcal {U}')$. We furthermore define

$$\begin{aligned} f:W\times W\times W'&\rightarrow \mathcal {V}'\\ (X,Y,Z)&\mapsto \Xi '(\Xi ^{-1}(X)\cdot \Xi ^{-1}(Y)\cdot \Xi '^{-1}(Z)), \end{aligned}$$

as well as (shrink $\delta >0$ if necessary, to ensure $\text {im} [\eta _\pm ]\subseteq O$, $\text {im} [\nu _\pm ]\subseteq O$, $\text {im} [\mu ]\subseteq O'$)

Then, Lemma 6 (with $F_1\equiv E\times E\times E$, $F_2\equiv E$, and $\gamma \equiv \gamma _\pm$ there) together with Part e) of Proposition A.1 (cf. also (11)), (57), (58), (59), (60), (62) implies the claim.

Finally, to prove (62), let $\textsf {p} \in \textsf {Sem}(E)$ be fixed, and choose $\textsf {p} \le \textsf {q} \in \textsf {Sem}(E)$ as well as $U\subseteq G$ as in Lemma 11. Since both $\nu _+,\eta _-$ are continuous by 1), there exists $0<\delta _U\le \delta$ ($\delta$ as in 3)) with $\nu _+([0,\delta _U])\subseteq U$ and $\eta _-([-\delta _U,0]))\subseteq U$. We observe that

holds by (b), and obtain from Lemma 11 that

$$\begin{aligned} \textsf {p} (\Delta _+(h))&\textstyle \le \frac{1}{|h|}\cdot \int _x^{x+h} \textsf {q} \big (\text {Ad} _{\alpha _+(\cdot )^{-1}}(\Phi (x+h,\cdot )-\Phi (x,\cdot )\big )\,\text {d} s\qquad \quad \forall \, h\in (0,\delta _U]\\ \textsf {p} (\Delta _-(h))&\textstyle \le \frac{1}{|h|}\cdot \int _x^{x+h} \textsf {q} \big (\text {Ad} _{\beta _-(\cdot )^{-1}}(\Phi (x,\cdot )-\Phi (x+h,\cdot )\big )\,\text {d} s\qquad \quad \forall \, h\in [-\delta _U,0). \end{aligned}$$

Let $\textsf {q} \le \textsf {m} \in \textsf {Sem}(E)$ be as in Corollary 6, for $\text {C} \equiv \textsf {inv} (\text {im} [\alpha _+])\cup \textsf {inv} (\text {im} [\beta _-])$ there. We shrink $0<\delta _U\le \delta$ such that $x+ (-\delta _U,\delta _U)\subseteq I_{\textsf {m} ,0}$ holds, for $I_{\textsf {m} ,0}$ as in Condition ii) (with $\text {s} =0$ there). Then, Condition ii) yields the following:

• For $0< h<\delta _U$, we have

  $$\begin{aligned} \textstyle \textsf {p} (\Delta _+(h))&\textstyle \le \frac{1}{|h|}\cdot \int _x^{x+h}\textsf {q} \big (\text {Ad} _{\alpha _+(\cdot )^{-1}}(\Phi (x+h,s)-\Phi (x,s)\big )\,\text {d} s \\&\textstyle \le \sup \{x\le s\le x+h\,|\,\textsf {m} (\Phi (x+h,s)-\Phi (x,s))\} \\&\textstyle \le |h|\cdot L_{\textsf {m} ,0}. \end{aligned}$$

• For $-\delta _U< h<0$, we have

  $$\begin{aligned} \textstyle \textsf {p} (\Delta _-(h))&\textstyle \le \frac{1}{|h|}\cdot \int _{x+h}^{x}\textsf {q} \big (\text {Ad} _{\beta _-(\cdot )^{-1}}(\Phi (x,s)-\Phi (x+h,s)\big )\,\text {d} s \\&\textstyle \le \sup \{x+h\le s\le x\,|\, \textsf {m} (\Phi (x,s)-\Phi (x+h,s))\} \\&\textstyle \le |h|\cdot L_{\textsf {m} ,0}. \end{aligned}$$

This proves (62), hence the claim. $\square$

3.3 An integral expansion for the adjoint action

Let $a<b$ and $\psi \in \mathfrak {D}_{[a,b]}$ be given. For $X\in \mathfrak {g}$, we define

(63)

We furthermore define

$$\begin{aligned} {\textbf {Ad}} _{\psi }^\pm [t]:\mathfrak {g}\ni X&\mapsto {\textbf {Ad}} _\psi ^\pm [X](t)\in \mathfrak {g}\qquad \quad \forall \, t\in [a,b]\\ {\textbf {Ad}} _\psi ^\pm&:= {\textbf {Ad}} _{\psi }^\pm [b]. \end{aligned}$$

For $\chi \in C^0([a,b],\mathfrak {g})$, we set

$$\begin{aligned} \varvec{\eta }^\pm (\psi ,\chi ):[a,b]\ni t \mapsto {\textbf {Ad}} _\psi ^\pm [\chi (t)](t)\in \mathfrak {g}. \end{aligned}$$

The following assertions are immediate from the definitions:

• We have $\varvec{\eta }^\pm (\psi ,\chi )\in C^{k+1}([a,b],\mathfrak {g})$ for $k\in \mathbb {N}\cup \{\infty \}$, for each $\psi \in \mathfrak {D}^k_{[a,b]}$ and $\chi \in C^{k+1}([a,b],\mathfrak {g})$ (by (37) and smoothness of the group operations).

• We have $\varvec{\eta }^\pm (\psi ,\varvec{\eta }^\mp (\psi ,\chi ))=\chi$ for each $\psi \in \mathfrak {D}_{[a,b]}$ and $\chi \in C^0([a,b],\mathfrak {g})$.

We furthermore observe the following:

Lemma 12

Let $a<b$, $\psi \in \mathfrak {D}_{[a,b]}$, and $X\in \mathfrak {g}$ be given. Then, ${\textbf {Ad}} ^\pm _\psi [X]\in C^1([a,b],\mathfrak {g})$ holds, with ${\textbf {Ad}} ^\pm _\psi [X](a)=X$ as well as

$$\begin{aligned} \partial _t {\textbf {Ad}} ^+_\psi [X](t) \textstyle&=\varvec{[}\psi (t), {\textbf {Ad}} ^+_\psi [X](t) \varvec{]}\nonumber \\ \qquad \,\,\text {and}\qquad \,\, \partial _t {\textbf {Ad}} ^-_\psi [X](t) \textstyle&=-{\textbf {Ad}} ^-_\psi [\varvec{[}\psi (t), X \varvec{]}](t) \end{aligned}$$

(64)

for each $t\in [a,b]$. In particular, for $\chi \in C^1([a,b],\mathfrak {g})$, we have

$$\begin{aligned} \begin{aligned} {\dot{\varvec{\eta }}}^+(\psi ,\chi )&= \varvec{\eta }^+(\psi ,{\dot{\chi }})+\varvec{[}\psi , \varvec{\eta }^+(\psi ,\chi ) \varvec{]}\\ {\dot{\varvec{\eta }}}^-(\psi ,\chi )&= \varvec{\eta }^-(\psi ,{\dot{\chi }})-\varvec{\eta }^-(\psi ,\varvec{[}\psi , \chi \varvec{]}). \end{aligned} \end{aligned}$$

(65)

Proof

Equation (64) is verified in Appendix A.6; and (65) is clear from (64) as well as the parts b), d), e) of Proposition A.1. $\square$

For the sake of completeness, we want to mention the following well-known result:

Corollary 7

Let $a<b$, $\psi \in \mathfrak {D}_{[a,b]}$, $X\in \mathfrak {g}$. Then, ${\textbf {Ad}} ^+_\psi [X]$ is the unique solution $\alpha \in C^1([a,b],\mathfrak {g})$ to the differential equation (Lax equation) ${\dot{\alpha }}=\varvec{[}\psi , \alpha \varvec{]}$, with the initial condition $\alpha (a)=X$.

Proof

By Lemma 12, it remains to show uniqueness. Let thus $\alpha \in C^1([a,b],\mathfrak {g})$ be given, with ${\dot{\alpha }}=\varvec{[}\psi , \alpha \varvec{]}$ and $\alpha (a)=X$. Then, ${\dot{\varvec{\eta }}}^-(\psi ,\alpha )=0$ holds by (65), with $\varvec{\eta }^-(\psi ,\alpha )(a)=X$. Then, (17) yields $\varvec{\eta }^-(\psi ,\alpha )=X$, hence

$$\begin{aligned} {\textbf {Ad}} ^+_\psi [X]= \varvec{\eta }^+(\psi ,\mathcal {C}_X|_{[a,b]})=\varvec{\eta }^+(\psi ,\varvec{\eta }^-(\psi ,\alpha ))=\alpha . \end{aligned}$$

This proves the claim. $\square$

Let $\widehat{\mathfrak {g}}$ denote the completion of $\mathfrak {g}$. Let $\psi \in C^0([a,b],\mathfrak {g})$, $X\in \mathfrak {g}$, $n\in \mathbb {N}$ be given. We set

$$\begin{aligned} \varvec{\lambda }^\pm _{0,\psi }[X]:[a,b]\rightarrow \widehat{\mathfrak {g}},\qquad t\mapsto X, \end{aligned}$$

and define for $\ell \ge 1$

For $\ell \ge 1$, we set

To simplify the notations, we define

We obtain the following statement.

Lemma 13

Let $\psi \in \mathfrak {D}_{[a,b]}$, $X\in \mathfrak {g}$, and $n\in \mathbb {N}$ be given. Then, for $t\in [a,b]$ we have

$$\begin{aligned} \textstyle {\textbf {Ad}} ^+_\psi [X](t)&\textstyle = \sum _{\ell =0}^n \varvec{\lambda }^+_{\ell ,\psi }[X](t) + \varvec{\mathfrak {R}}^{+}_{n+1,\psi }[X](t)\\ \textstyle {\textbf {Ad}} ^-_\psi [X](t)&\textstyle = \sum _{\ell =0}^n \varvec{\lambda }^-_{\ell ,\psi }[X](t) + \varvec{\mathfrak {R}}^{-}_{n+1,\psi }[X](t). \end{aligned}$$

Proof

By Lemma 12 and (17), we have

$$\begin{aligned} {\textbf {Ad}} ^+_\psi [X](t)\textstyle&\textstyle =X+\int _a^t\, \varvec{[}\psi (s), {\textbf {Ad}} ^+_\psi [X](s) \varvec{]} \,\text {d} s\textstyle = \varvec{\lambda }^+_{0,\psi }[X](t) + \varvec{\mathfrak {R}}^+_{1,\psi }[X](t) \end{aligned}$$

(66)

$$\begin{aligned} {\textbf {Ad}} ^-_\psi [X](t)\textstyle&\textstyle =X-\int _a^t\, {\textbf {Ad}} ^-_\psi [\varvec{[}\psi (s), X \varvec{]}](s) \,\text {d} s\textstyle = \varvec{\lambda }^-_{0,\psi }[X](t) + \varvec{\mathfrak {R}}^-_{1,\psi }[X](t). \end{aligned}$$

(67)

We thus can assume that the claim holds for some $n\in \mathbb {N}$. We obtain from (22) (third steps) that

$$\begin{aligned} {\textbf {Ad}} ^+_\psi [X](t)&\textstyle = \sum _{\ell =0}^n \varvec{\lambda }^+_{\ell ,\psi }[X](t) + \varvec{\mathfrak {R}}^+_{n+1,\psi }[X](t)\\&\textstyle = \sum _{\ell =0}^n \varvec{\lambda }^+_{\ell ,\psi }[X](t)\\&\quad \,\,\,\,\textstyle + \int _a^{t}\text {d} s_1\int _a^{s_1}\text {d} s_2 \, {\dots } \int _a^{s_{n}}\text {d} s_{n+1} \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{n+1})})({\textbf {Ad}} ^+_\psi [X](s_{n+1}))\\&\textstyle {\mathop {=}\limits ^{(66)}} \sum _{\ell =0}^n \varvec{\lambda }^+_{\ell ,\psi }[X](t) + \big (\varvec{\lambda }^{+}_{n+1,\psi }[X](t) + \varvec{\mathfrak {R}}^+_{n+2,\psi }[X](t)\big ) \\&\textstyle = \sum _{\ell =0}^{n+1} \varvec{\lambda }^+_{\ell ,\psi }[X](t) + \varvec{\mathfrak {R}}^+_{n+2,\psi }[X](t) \end{aligned}$$

as well as

$$\begin{aligned} {\textbf {Ad}} ^-_\psi [X](t)&\textstyle = \sum _{\ell =0}^n \varvec{\lambda }^-_{\ell ,\psi }[X](t) + \varvec{\mathfrak {R}}^-_{n+1,\psi }[X](t)\\&\textstyle = \sum _{\ell =0}^n \varvec{\lambda }^-_{\ell ,\psi }[X](t)\\&\quad \,\,\,\,\textstyle + (-1)^{n+1}\cdot \int _a^{t}\text {d} s_1\int _a^{s_1}\text {d} s_2 \, {\dots } \\&\quad \textstyle \int _a^{s_{n}}\text {d} s_{n+1} \, {\textbf {Ad}} ^-_\psi [(\text {ad} _{\psi (s_{n+1})}\circ \dots \circ \text {ad} _{\psi (s_{1})})(X)](s_{n+1})\\&\textstyle {\mathop {=}\limits ^{(67)}} \sum _{\ell =0}^n \varvec{\lambda }^-_{\ell ,\psi }[X](t) + \big ( \varvec{\lambda }^{-}_{n+1,\psi }[X](t) + \varvec{\mathfrak {R}}^-_{n+2,\psi }[X](t)\big ) \\&\textstyle = \sum _{\ell =0}^{n+1} \varvec{\lambda }^-_{\ell ,\psi }[X](t) + \varvec{\mathfrak {R}}^-_{n+2,\psi }[X](t). \end{aligned}$$

The claim now follows by induction. $\square$

We recall the definition of an $\text {AE}$-set in (28), and obtain the following corollary.

Corollary 8

Let $\text {M} \subseteq \mathfrak {g}$ be an $\text {AE}$-set, $a<b$, and $\psi \in \mathfrak {D}_{[a,b]}$ with $\text {im} [\psi ]\subseteq \text {M}$ be given. Then,

$$\begin{aligned} \textstyle {\textbf {Ad}} ^\pm _\psi [X]= \lim _n \sum _{\ell =0}^n \varvec{\lambda }^\pm _{\ell ,\psi }[X]= \sum _{\ell =0}^\infty \varvec{\lambda }^\pm _{\ell ,\psi }[X] \end{aligned}$$

(68)

converges uniformly for each fixed $X\in \text {M}$. In particular, the following assertions hold:

• (1) For $\textsf {v} \le \textsf {w}$ as in (28), we have

  $$\begin{aligned} \textsf {v} \circ {\textbf {Ad}} ^\pm _\psi [X]\le \textsf {w} (X)\cdot \text {e} ^{\int _a^\bullet \textsf {w} (\psi (s))\,\text {d} s}\qquad \quad \forall \, X\in \text {M} . \end{aligned}$$

• (2) By (48), for $Z\in \text {dom} [\exp ]\cap \text {M}$ we have

  $$\begin{aligned} \textstyle \text {Ad} _{\exp (t\cdot Z)^{\pm 1}}(X)={\textbf {Ad}} ^\pm _{\mathcal {C}_Z|_{[0,1]}}[X](t)= \sum _{\ell =0}^\infty \frac{t^\ell }{\ell !}\cdot \text {ad} _{\pm Z}^\ell (X)\qquad \quad \forall \, X\in \text {M} ,\, t\in [0,1]. \end{aligned}$$

Proof

Equation (68) is clear from the following two estimates:

• Let $\textsf {v} \le \textsf {w}$ be as in (28), as well as $\textsf {w} \le \textsf {m}$ be as in Corollary 6 for $\textsf {q} \equiv \textsf {w}$ and there. We obtain from Lemma 13 that

  $$\begin{aligned} \textstyle \textsf {v} \big ({\textbf {Ad}} ^+_\psi [X](t)\textstyle -\sum _{\ell =0}^n \varvec{\lambda }^+_{\ell ,\psi }[X](t)\big )&\textstyle =\textsf {v} \big (\varvec{\mathfrak {R}}^+_{n+1,\psi }[X](t)\big )\\&\textstyle \le \frac{(b-a)^{n+1}}{(n+1)!}\cdot \textsf {w} _\infty (\psi )^{n+1}\cdot \textsf {w} _\infty ({\textbf {Ad}} ^+_\psi [X])\\&\textstyle \le \frac{(b-a)^{n+1}}{(n+1)!}\cdot \textsf {w} _\infty (\psi )^{n+1}\cdot \textsf {m} (X) \end{aligned}$$

  holds for each $t\in [a,b]$, $X\in \mathfrak {g}$, and $n\in \mathbb {N}$.

• Let $\textsf {q} \le \textsf {m}$ be as in Corollary 6 for there. Let furthermore $\textsf {m} \le \textsf {w}$ be as in (28) for $\textsf {v} \equiv \textsf {m}$ there. We obtain from Lemma 13 that

  $$\begin{aligned} \textstyle \textsf {q} \big ({\textbf {Ad}} ^-_\psi [X](t)\textstyle -\sum _{\ell =0}^n \varvec{\lambda }^-_{\ell ,\psi }[X](t)\big ) \textstyle =\textsf {q} \big ( \varvec{\mathfrak {R}}^-_{n+1,\psi }[X](t) \big )\le \frac{(b-a)^{n+1}}{(n+1)!}\cdot \textsf {w} _\infty (\psi )^{n+1}\cdot \textsf {w} (X) \end{aligned}$$

  holds for each $t\in [a,b]$, $X\in \mathfrak {g}$, and $n\in \mathbb {N}$.

Now, Part 2) is just clear from (68), and Part 1) is clear from (68) as well as

$$\begin{aligned} \textstyle \textsf {v} \big (\varvec{\lambda }^\pm _{\ell ,\psi }[X](t)\big )&\textstyle \le \int _a^{t}\text {d} s_1\int _a^{s_1}\text {d} s_2 \, {\dots } \int _a^{s_{\ell -1}}\text {d} s_\ell \, \textsf {w} (\psi (s_1))\cdot {\dots } \cdot \textsf {w} (\psi (s_{\ell }))\cdot \textsf {w} (X) \\&\textstyle = \frac{1}{\ell !}\cdot (\int _a^t \textsf {w} (\psi (s))\,\text {d} s)^\ell \cdot \textsf {w} (X) \end{aligned}$$

for each $\ell \ge 1$. (The equality in the second step just follows by induction over $\ell \ge 1$, via taking the derivative of both expressions.) $\square$

Example 7

Let $\text {N} \subseteq \mathfrak {g}$ be a $\text {Nil} _{q}$-set for some $q\ge 2$. Then, $\text {N}$ is an $\text {AE}$-set, and Corollary 8 shows

$$\begin{aligned} \textstyle {\textbf {Ad}} ^\pm _\psi [X]=\sum _{\ell =0}^{q-2} \varvec{\lambda }^\pm _{\ell ,\psi }[X] \end{aligned}$$

for each $X\in \text {N}$ and $\psi \in \mathfrak {D}$ with $\text {im} [\psi ]\subseteq \text {N}$. In particular, we have

$$\begin{aligned} \textstyle \text {Ad} _{\exp (t\cdot Z)^{\pm }}(X)= \sum _{\ell =0}^{q-2} \frac{t^\ell }{\ell !}\cdot \text {ad} _{\pm Z}^\ell (X)\qquad \quad \forall \, t\in [0,1] \end{aligned}$$

for each $X\in \text {N}$ and $Z\in \text {dom} [\exp ]\cap \text {N}$. $\square$

4 A generalized BCDH formula

In the first part of this section, we generalize the Baker–Campbell–Dynkin–Hausdorff formula (for the exponential map) to the product integral (cf. Proposition 3). In the second part, we apply this formula (together with the integral transformation introduced in Sect. 3.1) to express the product integral of nilpotent curves¹³ (cf. Theorem 1) via the exponential map. Various applications of the derived formula are discussed in Sect. 4.2.1.

4.1 A BCDH formula for the product integral

Let $\mathfrak {X}\in C^1([a,b],\mathfrak {g})$ for $a<b$ be given, with $\text {im} [\mathfrak {X}]\subseteq \text {dom} [\exp ]$. In this section, we consider the following two situations:

• A) G is a Banach Lie group with $\Vert [X,Y]\Vert \le \Vert X\Vert \cdot \Vert Y\Vert$ for all $X,Y\in \mathfrak {g}$, and we have $\Vert \mathfrak {X}\Vert _\infty <\ln (2)$. We set $V:=\mathfrak {g}$ as well as ${\textbf {I}} ,{\textbf {J}} :=\infty$, and observe the following:

  • − $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is asymptotic estimate.

  • − $\exp :\mathfrak {g}\rightarrow \mathfrak {g}$ is defined and smooth, and a local diffeomorphism. In particular, there exists an identity neighborhood $\mathcal {O}\subseteq G$ on which $\exp$ is a diffeomorphism, such that $\Vert \exp ^{-1}(g)\Vert <\ln (2)$ holds for each $g\in \mathcal {O}$.

  • − G is $C^0$-regular (even $L^1$-regular by Theorem C in [5]). In particular, there exists some $\mathfrak {r}>0$, such that for all $\phi \in C^0([a,b],\mathfrak {g})$ ($a<b$) and $\psi \in C^0([a',b'],\mathfrak {g})$ ($a'<b'$) with $\int _a^b \Vert \phi (s)\Vert \, \text {d} s + \int _{a'}^{b'} \Vert \psi (s)\Vert \, \text {d} s< \mathfrak {r}$, we have¹⁴

    

    (69)

• B) G is weakly $C^\infty$-regular, or $\exp :\mathfrak {g}\rightarrow G$ is defined ($\text {dom} [\exp ]=\mathfrak {g}$) and of class $C^1$. Moreover, $\text {N} :=\text {im} [\mathfrak {X}]\cup \text {im} [{\dot{\mathfrak {X}}}]$ is a $\text {Nil} _{q}$-set for some $q\ge 2$. We set $V:=\overline{\mathcal {G}}_1(\text {N} )$ as well as ${\textbf {I}} :=q-1$ and ${\textbf {J}} :=q-2$, and recall that V is a $\text {Nil} _{q}$-set by Remark 3.3).

In both situations, the presumptions in Corollary 4 are fulfilled. Moreover, by Example 1, for $t\in [a,b]$ and $Y\in V$, we have

$$\begin{aligned} \textstyle \Psi \big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{\mathfrak {X}(t)}^n\big )(\Phi (\text {ad} _{\mathfrak {X}(t)})(Y))&\textstyle =Y \end{aligned}$$

(70)

$$\begin{aligned} \textstyle \widetilde{\Psi }\big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{-\mathfrak {X}(t)}^n\big )(Y)&\textstyle =\Psi \big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{\mathfrak {X}(t)}^n\big )(Y) \end{aligned}$$

(71)

with $\Phi (\text {ad} _{\mathfrak {X}(t)})(Y)\in V$ (by Remark 3 in the situation of B)). Here, for maps $\xi ,\zeta :\mathfrak {g}\rightarrow \mathfrak {g}$ as well as $X\in \mathfrak {g}$, we set (convergence presumed)

$$\begin{aligned} \textstyle \Psi (\xi )(X)&\textstyle := \sum _{n=1}^\infty \frac{(-1)^{n-1}}{n}\cdot (\xi -\text {id} _\mathfrak {g})^{n-1}(X)\\ \textstyle \widetilde{\Psi }(\xi )(X)&\textstyle := \sum _{n=1}^\infty \frac{(-1)^{n-1}}{n}\cdot \big (\xi \circ (\xi -\text {id} _\mathfrak {g})^{n-1}\big )(X)\\ \textstyle \Phi (\zeta )(X)&\textstyle := \sum _{n=0}^\infty \frac{1}{(n+1)!}\cdot \zeta ^{n}(X). \end{aligned}$$

We obtain the following proposition:

Proposition 3

Let $\phi \in \mathfrak {D}_{[a,b]}$, $g\in G$, $\mathfrak {X}\in C^1([a,b],\mathfrak {g})$ be given, with $\text {im} [\mathfrak {X}]\subseteq \text {dom} [\exp ]$ as well as

(72)

Assume furthermore that we are in the situation of A) or B). Then, we have $\text {im} [\phi ]\subseteq V$, and the following identities hold for each $t\in [a,b]$:

(73)

(74)

(75)

Proof

We observe the following:

• For $t\in [a,b]$, we have by Corollary 8.2) (and Example 7)

  $$\begin{aligned} \textstyle \text {Ad} _{\exp (\lambda \cdot \mathfrak {X}(t))^{\pm 1}}|_V \textstyle = \sum _{n=0}^{\textbf {J}} \frac{\lambda ^n}{n!}\cdot \text {ad} _{\pm \mathfrak {X}(t)}^n|_V. \end{aligned}$$

  (76)

• For $t\in [a,b]$, we have

  $$\begin{aligned} \Phi (\text {ad} _{\mathfrak {X}(t)})(\dot{\mathfrak {X}}(t))&\textstyle = \sum _{n=0}^{{\textbf {J}} } \frac{1}{(n+1)!} \cdot \text {ad} _{\mathfrak {X}(t)}^n(\dot{\mathfrak {X}}(t))\in V. \end{aligned}$$

  (77)

• For $t\in [a,b]$, we have by (70), (71), and (77) that

  $$\begin{aligned} \begin{aligned} \dot{\mathfrak {X}}(t)&= \textstyle \Psi \big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{\mathfrak {X}(t)}^n\big )(\Phi (\text {ad} _{\mathfrak {X}(t)})(\dot{\mathfrak {X}}(t)))\\&= \textstyle \widetilde{\Psi }\big (\sum _{n=0}^\infty \frac{1}{n!}\cdot \text {ad} _{-\mathfrak {X}(t)}^n\big )(\Phi (\text {ad} _{\mathfrak {X}(t)})(\dot{\mathfrak {X}}(t))). \end{aligned} \end{aligned}$$

  (78)

Set $\alpha :=\exp \circ \mathfrak {X}\in C^1([a,b],\mathfrak {g})$. Then, given $t\in [a,b]$, we obtain from (72) (first step), Corollary 4 (second step), (22) (third step), (76) (fourth step), $C^0$-continuity of the Riemann integral in the situation of A) (fifth step), as well as (77) (fifth step) that

$$\begin{aligned} \textstyle \text {d} _e\text {R} _{\alpha (t)}(\phi (t))&={\dot{\alpha }}(t)\\&\textstyle = \text {d} _e\text {L} _{\alpha (t)}\big (\int _0^1 \text {Ad} _{\exp (-s\cdot \mathfrak {X}(t))}(\dot{\mathfrak {X}}(t)) \,\text {d} s\big )\\&\textstyle = \text {d} _e\text {R} _{\alpha (t)}\big (\int _0^1 \text {Ad} _{\exp ((1-s)\cdot \mathfrak {X}(t))}(\dot{\mathfrak {X}}(t)) \,\text {d} s\big )\\&\textstyle = \text {d} _e\text {R} _{\alpha (t)}\big (\int _0^1 \sum _{n=0}^{\textbf {J}} \frac{(1-s)^n}{n!} \cdot \text {ad} _{\mathfrak {X}(t)}^n(\dot{\mathfrak {X}}(t))\,\text {d} s\big )\\&\textstyle = \text {d} _e\text {R} _{\alpha (t)}\big (\sum _{n=0}^{\textbf {J}} \frac{1}{(n+1)!} \cdot \text {ad} _{\mathfrak {X}(t)}^n(\dot{\mathfrak {X}}(t))\big ). \end{aligned}$$

Together with (77), this shows

$$\begin{aligned} \phi (t)=\Phi (\text {ad} _{\mathfrak {X}(t)})(\dot{\mathfrak {X}}(t))\in V\qquad \quad \forall \, t\in [a,b]. \end{aligned}$$

(79)

We obtain from (17), (78), (79), and (76) the following identities for each $t\in [a,b]$:

$$\begin{aligned} \begin{aligned} \mathfrak {X}(t)-\mathfrak {X}(a)&\textstyle =\int _a^t \Psi (\text {Ad} _{\exp (\mathfrak {X}(s))})(\phi (s))\,\text {d} s\\ \mathfrak {X}(t)-\mathfrak {X}(a)&\textstyle = \int _a^t \widetilde{\Psi }(\text {Ad} _{\exp (-\mathfrak {X}(s))})(\phi (s))\,\text {d} s\\&\textstyle = \int _a^t \widetilde{\Psi }(\text {Ad} _{\exp (\mathfrak {X}(s))^{-1}})(\phi (s))\,\text {d} s. \end{aligned} \end{aligned}$$

(80)

Together with (72), this implies (73) and (74). We now finally have to prove (75):

• Assume that we are in the situation of B). Since $\text {im} [\phi ]\subseteq V$ holds by (79), we obtain from (17), (78), (79) (first step), $\text {im} [\phi ]\subseteq V$ (second step), linearity of the Riemann integral (third step), (76) (fourth step), and (72) (fifth step) that

  

• Assume that we are in the situation of A). Then, (76) (first step) together with $\Vert \mathfrak {X}\Vert _\infty <\ln (2)$ (last step) implies

  $$\begin{aligned} \begin{aligned} \textstyle \Vert (\text {Ad} _{\exp (\mathfrak {X}(t))}-\text {id} _\mathfrak {g})^k(\phi (t))\Vert&\textstyle \le (\Vert \sum _{\ell =1}^\infty \frac{1}{\ell !}\cdot \text {ad} _{\mathfrak {X}(t)}^\ell \Vert _\text {op} )^k\cdot \Vert \phi \Vert _\infty \\&\textstyle \le (\sum _{\ell =1}^\infty \frac{1}{\ell !}\cdot \Vert \text {ad} _{\mathfrak {X}(t)}^\ell \Vert _\text {op} )^k \cdot \Vert \phi \Vert _\infty \\&\textstyle \le \big (\text {e} ^{\Vert \mathfrak {X}\Vert _\infty }-1\big )^k \cdot \Vert \phi \Vert _\infty \\&\textstyle <\Vert \phi \Vert _\infty \end{aligned} \end{aligned}$$

  (81)

  for each $t\in [a,b]$ and $k \ge 1$. We obtain from (80) (first step), $C^0$-continuity of the Riemann integral and (81) (third step), as well as (72) (fourth step) that

  

  holds for each $t\in [a,b]$.$\square$

Proposition 4

Assume that G is a Banach Lie group, and let $\mathfrak {r}>0$ be as in A). Let $\phi \in C^0([a,b],\mathfrak {g})$ ($a<b$) and $\psi \in C^0([a',b'])$ ($a'<b'$) be given, with $\int _a^b \Vert \phi (s)\Vert \, \text {d} s+ \int _{a'}^{b'} \Vert \psi (s)\Vert \, \text {d} s< \mathfrak {r}$. Then, we have

provided that for $t'\in [a',b']$ and $t\in [a,b]$ we set

$$\begin{aligned} \mathfrak {X}_\psi (t')&:= \textstyle \sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{n}\cdot \int _{a'}^{t'} \big (\sum _{\ell =1}^{\infty } \varvec{\lambda }^+_{\ell ,\psi }[s]\big )^{n-1} (\psi (s))\, \text {d} s\\ \mathfrak {X}_{\phi ,\psi }(t)&:=\textstyle \sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{n}\cdot \int _a^t \big (\big (\sum _{\ell =0}^{\infty } \varvec{\lambda }^+_{\ell ,\phi }[s]\big )\circ \big (\sum _{\ell =0}^{\infty } \varvec{\lambda }^+_{\ell ,\psi }[b']\big )-\text {id} _\mathfrak {g}\big )^{n-1}(\phi (s))\, \text {d} s. \end{aligned}$$

Proof

By definition, there exist open intervals $I,I'\subseteq \mathbb {R}$ with $[a,b]\subseteq I$ and $[a',b']\subseteq I'$, as well as $\mu \in C^1(I,G)$ and $\nu \in C^1(I',G)$, with and . We make the following modifications to these curves:

• We replace $\mu$ by $\mu \cdot \mu (a)^{-1}$ to ensure .

• We replace $\nu$ by $\nu \cdot \nu (a')^{-1}$ to ensure .

• We define $\alpha :=\mu \cdot \nu (b')$, and observe .

• We shrink I around [a, b] and $I'$ around $[a',b']$, to ensure $\mu (I)\cdot \nu (I')\subseteq \mathcal {O}$ (recall (69)).

Then, the maps

$$\begin{aligned} \mathfrak {X}:[a,b]\ni t\mapsto \exp ^{-1}(\alpha (t))\in \mathfrak {g}\qquad \quad \text {and}\qquad \quad \mathfrak {X}':[a',b']\ni t'\mapsto \exp ^{-1}(\nu (t'))\in \mathfrak {g}\end{aligned}$$

are defined and of class $C^1$, with

We conclude the following:

• Equation (75) in Proposition 3 for $g\equiv e$ there, together with Corollary 8 shows $\mathfrak {X}'=\mathfrak {X}_\psi$.

• Equation (75) in Proposition 3 for there, together with Corollary 8 and the previous point shows

  $$\begin{aligned} \mathfrak {X}-\mathfrak {X}_\psi (b')=\mathfrak {X}-\mathfrak {X}'(b')=\mathfrak {X}-\mathfrak {X}(a)=\mathfrak {X}_{\phi ,\psi }, \end{aligned}$$

  which proves the claim. $\square$

Corollary 9

Assume that G is a Banach Lie group, and let $\mathfrak {r}>0$ be as in A). Then, for $\phi \in C^0([a,b],\mathfrak {g})$ ($a<b$) with $\int _a^b \Vert \phi (s)\Vert \, \text {d} s< \mathfrak {r}$, we have

Proof

Apply Proposition 4 with $\psi =0$. $\square$

Example 8

(The BCDH Formula) Assume that we are in the situation of Proposition 3, with

$$\begin{aligned} {[}a,b]=[0,1],\qquad \phi =-\mathcal {C}_Y|_{[0,1]},\qquad g=\exp (-X),\qquad \mathfrak {X}(0)=-X \end{aligned}$$

(82)

for certain $X,Y\in \mathfrak {g}$ and $\mathfrak {X}\in C^1([0,1],\mathfrak {g})$. We obtain the well-known¹⁵ integral representation of the Baker–Campbell–Dynkin–Hausdorff formula:

$$\begin{aligned} \textstyle \exp (X)\cdot \exp (t\cdot Y)= \exp \big (X + \int _0^t \widetilde{\Psi }(\text {Ad} _{\exp (X)}\circ \text {Ad} _{\exp (s\cdot Y)})(Y)\,\text {d} s\big )\qquad \forall \, t\in [0,1]. \end{aligned}$$

(83)

Proof of Equation (83)

By (48), we have for each $t\in [0,1]$, hence

Then, (74) in Proposition 3 shows

for each $t\in [0,1]$, from which the claim follows. $\square$

For instance, the assumptions in (82) are fulfilled in the following situations:

• G is a Banach Lie group, with $\Vert X\Vert +\Vert Y\Vert <\mathfrak {r}$. Indeed, by A), then there exists $\mathfrak {X}\in C^1([0,1],\mathfrak {g})$ with $\mathfrak {X}(0)=-X$, such that (72) holds for $\phi$ and g as in (82).

• G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and $\{X,Y\}$ is a $\text {Nil} _{q}$-set for some $q\ge 2$. Indeed, we will reconsider this situation in Corollary 11 in Sect. 4.2. There, we construct some $\mathfrak {X}\in C^1([0,1],\mathfrak {g})$ with $\mathfrak {X}(0)=-X$, such that (72) holds for $\phi$ and g as in (82). $\square$

4.2 The product integral of nilpotent curves

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $\phi \in C^k([a,b],\mathfrak {g})$ for $a<b$ as well as $\psi \in C^k([a',b'],\mathfrak {g})$ for $a'<b'$ be given.

• If $\text {im} [\psi ]$ is a $\text {Nil} _{q}$-set for some $q\ge 2$, then for each $t\in [a',b']$ we set

  

  (84)

  The second line is due to (22) and Example 7.

• If $\text {im} [\phi ]\cup \text {im} [\psi ]$ is a $\text {Nil} _{q}$-set for some $q\ge 2$, then for each $t\in [a,b]$ we set

  

  (85)

  The second line is due to (22) and Example 7.

Notably, both maps take values in $\mathfrak {g}$ by Lemma 8 and are thus of class $C^1$ by (20). In this section, we combine Proposition 3 with the integral transformation introduced in Sect. 3.1 to prove the following statement.

Theorem 1

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $\phi \in C^k([a,b],\mathfrak {g})$, $\psi \in C^k([a',b'],\mathfrak {g})$ be given, such that $\text {N} :=\text {im} [\phi ]\cup \text {im} [\psi ]$ is a $\text {Nil} _{q}$-set for some $q\ge 2$. Then,

The proof of Theorem 1 is given in Sect. 4.2.2. We now first want to discuss some consequences.

4.2.1 Some consequences

Setting $\psi =0$ in Theorem 1, we immediately obtain the following statement.

Corollary 10

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $\phi \in C^k([a,b],\mathfrak {g})$ be given such that $\text {im} [\phi ]$ is a $\text {Nil} _{q}$-set for some $q\ge 2$. Then, we have

Example 9

Let G be abelian, as well as weakly $C^k$-regular for some $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$. Then, $\mathfrak {g}$ is a $\text {Nil} _{2}$-set, and Corollary 10 recovers formula (51) in Example 6. $\square$

Moreover, combining Theorem 1 with Example 8, we obtain the BCDH formula:

Corollary 11

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$. Let furthermore $X,Y\in \mathfrak {g}$ be given, such that $\{X,Y\}$ is a $\text {Nil} _{q}$-set for some $q\ge 2$. Then, we have

$$\begin{aligned} \textstyle \exp (X)\cdot \exp (t\cdot Y)= \exp \big (X + \int _0^t \widetilde{\Psi }\big (\text {Ad} _{\exp (X)}\circ \text {Ad} _{\exp (s\cdot Y)}\big )(Y)\,\text {d} s\big )\qquad \quad \forall \, t\in [0,1]. \end{aligned}$$

Proof

Set $\phi :=-\mathcal {C}_Y|_{[0,1]}$ as well as $\psi := -\mathcal {C}_X|_{[0,1]}$, and let $\mathfrak {X}_\psi ,\mathfrak {X}_{\phi ,\psi }$ be as in (84),(85). Then, $\mathfrak {X}:[0,1]\ni t\mapsto \mathfrak {X}_\psi (1) + \mathfrak {X}_{\phi ,\psi }(t) \in \mathfrak {g}$ is of class $C^1$, with $\mathfrak {X}(0)=\mathfrak {X}_\psi (1)=-X$, as well as

by Theorem 1. The claim is thus clear from Example 8. $\square$

Next, we define

$$\begin{aligned} \Gamma ^\infty _{\textrm{Nil}}&:=\{\phi \in \mathfrak {D}^\infty _{[0,1]}\,|\, \text {im} [\phi ] \,\text {is a}~ {\textrm{Nil}}_{q}\text {-}set\text { for some}\, q\ge 2\}\\ \text {and set}\qquad G^\infty _{\textrm{Nil}}&:={\text {evol} }_\infty (\Gamma ^\infty _{\textrm{Nil}}). \end{aligned}$$

We observe the following:

Lemma 14

If G is weakly $C^\infty$-regular, then $G^\infty _{\textrm{Nil}}=\text {im} [\exp ]$ holds.

Proof

Obviously, have $\mathfrak {i}(\text {dom} [\exp ])\subseteq \Gamma ^\infty _{\textrm{Nil}}$, hence $\text {im} [\exp ]\subseteq G^\infty _{\textrm{Nil}}$. Moreover, Corollary 10 shows $G^\infty _{\textrm{Nil}}\subseteq \text {im} [\exp ]$, which proves the claim. $\square$

Lemma 15

Assume that G is connected. Then, ${\text {evol} }_\infty$ is surjective.

Proof

Let $\mathcal {A}\subseteq G$ denote subgroup of all $g\in G$, such that there exist $\mu _1,\dots ,\mu _n\in C^\infty ([0,1],G)$ ($n\in \mathbb {N}$), with $g=\mu _n(1)\cdot {\dots }\cdot \mu _1(1)$ as well as $\mu _\ell (0)=e$ for $\ell =1,\dots ,n$. Set furthermore $\mathcal {O}:= \mathcal {U}\cap \textsf {inv} (\mathcal {U})$. Since G is connected, we have $G=\bigcup _{n\ge 1} \mathcal {O}^n$. Since $\mathcal {O}\subseteq \mathcal {U}\subseteq \mathcal {A}$ holds,¹⁶ we obtain $G=\mathcal {A}$. Let now $g\in G$ be fixed, and choose $\mu _1,\dots ,\mu _n\in C^\infty ([0,1],\mathfrak {g})$ with $\mu _1(0),\dots ,\mu _n(0)=e$ and $g=\mu _n(1)\cdot {\dots }\cdot \mu _1(1)$. We set for $\ell =1,\dots ,n$, and observe

Applying (a) inductively, we obtain (recall Remark 5)

which proves the claim. $\square$

Combining Lemma 14 with Lemma 15, we obtain the following proposition.

Proposition 5

Assume that G is weakly $C^\infty$-regular and connected. If $\Gamma ^\infty _{\textrm{Nil}}= C^\infty ([0,1],\mathfrak {g})$ holds, then $\exp :\mathfrak {g}\rightarrow G$ is surjective.

Proof

Lemma 14 shows $\text {im} [\exp ]=G^\infty _{\textrm{Nil}}={\text {evol} }_\infty (\Gamma ^\infty _{\textrm{Nil}})$, with ${\text {evol} }_\infty (\Gamma ^\infty _{\textrm{Nil}})={\text {evol} }_\infty (C^\infty ([0,1],\mathfrak {g}))$ by assumption. Since ${\text {evol} }_\infty$ is surjective by Lemma 15, the claim follows. $\square$

Proposition 5 immediately implies the following statement.

Corollary 12

Assume that G is weakly $C^\infty$-regular and connected. If $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is nilpotent, then $\exp$ is surjective.

Remark 11

• (1) The statement in Corollary 12 follows from Theorem IV.2.6 in [21] provided we replace “weakly $C^\infty$-regular” by “$\mathfrak {g}$ is Mackey complete and $\exp$ is smooth”.

• (2) Let $\mathfrak {g}$ be finite-dimensional. Then, the hypothesis $\Gamma ^\infty _{\textrm{Nil}}= C^\infty ([0,1],\mathfrak {g})$ in Proposition 5 is equivalent to nilpotency of $\mathfrak {g}$.

  In fact, the one implication is evident. Assume thus that $\Gamma ^\infty _{\textrm{Nil}}= C^\infty ([0,1],\mathfrak {g})$ holds, and let $\{X_0,\dots ,X_{n-1}\}\subseteq \mathfrak {g}$ be a base of $\mathfrak {g}$. Let furthermore $\rho :[0,1/n]\rightarrow [0,\infty )$ be a bump function, i.e., $\rho$ is smooth with

  $$\begin{aligned} \rho |_{(0,1/n)}>0\qquad \quad \text {as well as}\qquad \quad \rho ^{(\ell )}(0)=0=\rho ^{(\ell )}(1/n)\qquad \forall \,\ell \in \mathbb {N}. \end{aligned}$$

  We define $\phi :[0,1]\rightarrow \mathfrak {g}$ by $\phi (0):=0$ and

  $$\begin{aligned} \phi |_{(\ell /n,(\ell +1)/n]}:(\ell /n,(\ell +1)/n]\ni t \mapsto \rho (t-\ell /n)\cdot X_\ell \in \mathfrak {g}\qquad \quad \forall \, \ell =0,\dots ,n-1. \end{aligned}$$

  Then, $\phi$ is easily seen to be smooth (for details confer, e.g., Lemma 24 and Appendix B.2 in [8]). By assumption, there exists some $q\ge 2$, such that $\text {N} :=\text {im} [\phi ]$ is a $\text {Nil} _{q}$-set. Then, $\mathcal {V}_1(\text {N} )$ is a $\text {Nil} _{q}$-set by Remark 3, with $\{X_0,\dots ,X_{n-1}\}\subseteq \mathcal {V}_1(\text {N} )$, hence $\mathfrak {g}=\mathcal {V}_1(\text {N} )$. $\square$

We finally want to mention that Proposition 5 also provides an easy proof of the following well-known result in the finite-dimensional context.

Corollary 13

If G is compact and connected, then $\exp$ is surjective.

Proof

Let $T\subseteq G$ be a maximal torus with Lie algebra $\mathfrak {t}$. Since T is abelian, Corollary 12 shows that $\exp |_\mathfrak {t}:\mathfrak {t}\rightarrow T$ is surjective. Since G is covered by maximal tori (confer, e.g., Theorem 12.2.2.(iii) in [13]), the claim follows. $\square$

Remark 12

The proof of Corollary 13 can be simplified. Specifically, surjectivity of $\exp |_\mathfrak {t}:\mathfrak {t}\rightarrow T$ also follows from Lemma 15 and formula (51) (as T is abelian). Notably, in finite dimensions, (51) follows easily from smoothness of the exponential map:

Proof of (51) in the finite-dimensional case

Let G be abelian and finite-dimensional, and let $\phi \in C^0([a,b],\mathfrak {g})$ be given. Let $\psi \in C^0([a,b+\varepsilon ),\mathfrak {g})$ for $\varepsilon >0$ be a continuous extension of $\phi$. Then,

$$\begin{aligned} \textstyle \mu :[a,b+\varepsilon )\ni t \mapsto \exp \big (\!\int _a^t\psi (s)\,\text {d} s\big )\in G \end{aligned}$$

is of class $C^1$ as $\exp$ is smooth. Since G is abelian, we have

$$\begin{aligned} \textstyle (\mu (t+h)\cdot \mu (t)^{-1})=\exp \big (\int _t^{t+h}\psi (s) \,\text {d} s\big )\qquad \quad \forall \, t\in [a,b], \, 0<h<\varepsilon . \end{aligned}$$

Since $\text {d} _e\exp =\text {id} _\mathfrak {g}$ holds, we obtain , hence

by injectivity of . $\square$

4.2.2 The proof of Theorem 1

In this subsection, we prove Theorem 1. Our argumentation is based on the following (straightforward but technical) lemma that is proven in Sect. 4.2.3.

Lemma 16

Let G be weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $\phi \in C^k([a,b],\mathfrak {g})$ be given such that $\text {im} [\phi ]$ is a $\text {Nil} _{q}$-set for some $q\ge 2$. Then, the following assertions hold:

• We have $\mathfrak {T}^{q-1}(\phi |_{[a,\tau ]})\in C^\text {c} ([0,1],\mathfrak {g})$ for each $\tau \in [a,b]$, hence the map (recall (47))

  $$\begin{aligned} \mathfrak {X}:[a,b]\ni \tau \mapsto \mathfrak {i}^{-1}(\mathfrak {T}^q(\phi |_{[a,\tau ]}))\in \mathfrak {g}\end{aligned}$$

  is defined.

• We have $\mathfrak {X}\in C^1([a,b],\mathfrak {g})$, and $\text {im} [\mathfrak {X}]\cup \text {im} [{\dot{\mathfrak {X}}}]$ is a $\text {Nil} _{q}$-set.

In particular, iterated application of Proposition 1 yields

Proof

Confer Sect. 4.2.3. $\square$

Remark 13

Let $\psi \in C^0([x,y],\mathfrak {g})$ with $x<y$ be given. Let $\varrho :[x',y']\rightarrow [x,y]$ ($x'<y'$) be of class $C^1$, with $\varrho (x')=x$ and $\varrho (y')=y$. We set $\phi := {\dot{\varrho }}\cdot (\psi \circ \varrho )$ and obtain inductively from (21) that

$$\begin{aligned} \textstyle \int _{x'}^{y'}\varvec{\lambda }^+_{\ell ,\phi }[s](\phi (s))\, \text {d} s=\int _{x}^{y}\varvec{\lambda }^+_{\ell ,\psi }[s](\psi (s))\, \text {d} s \end{aligned}$$

holds for each $\ell \in \mathbb {N}$. $\square$

We are ready for the proof of Theorem 1.

Proof of Theorem 1

Let $\varrho _1:[0,1]\rightarrow [a',b']$ and $\varrho _2:[1,2]\rightarrow [a,b]$ be smooth, such that $\rho _{j}:={\dot{\varrho }}_j$ is a bump function for $j=1,2$, i.e., we have¹⁷

$$\begin{aligned} \rho _{1}|_{(0,1)},\rho _2|_{(1,2)}>0\qquad \quad \text {as well as}\qquad \quad \rho _1^{(\ell )}(0),\rho _1^{(\ell )}(1),\rho _2^{(\ell )}(1),\rho _2^{(\ell )}(2)=0\qquad \forall \, \ell \in \mathbb {N}. \end{aligned}$$

Define $\chi :[0,1]\rightarrow \mathfrak {g}$ by

$$\begin{aligned} \chi |_{[0,1]}:= \rho _1\cdot (\psi \circ \varrho _1)\qquad \quad \text {as well as}\qquad \quad \chi |_{(1,2]}:= (\rho _2\cdot (\phi \circ \varrho _2))|_{(1,2]}. \end{aligned}$$

Then, $\chi$ is easily seen to be of class $C^k$ (confer, e.g., Lemma 24 and Appendix B.2 in [8]), with $\text {im} [\chi ]\subseteq \mathcal {V}_1(\text {N} )\subseteq \overline{\mathcal {G}}_1(\text {N} )$ (a $\text {Nil} _{q}$-set by Remark 3). Then, for $\tau \in [0,1]$ and $t\in [1,2]$, we have

(86)

Now, Lemma 16 provides some $\tilde{\mathfrak {X}}\in C^1([0,2],\mathfrak {g})$ with , such that $\text {im} [\tilde{\mathfrak {X}}]\cup \text {im} [\dot{\tilde{\mathfrak {X}}}]$ is a $\text {Nil} _{q}$-set. In particular, we have

(87)

and obtain from Proposition 3 (specifically from (75) with $g\equiv e$ and $\mathfrak {X}\equiv \tilde{\mathfrak {X}}|_{[0,1]}$ there) and (68) in Corollary 8 (first step), as well as Remark 13 (second step) that

$$\begin{aligned} \begin{aligned} \textstyle \tilde{\mathfrak {X}}|_{[0,1]}(1)&\textstyle = \sum _{n=1}^{q-1} \frac{(-1)^{n-1}}{n}\cdot \int _{0}^1 \big (\sum _{\ell =1}^{q-2} \varvec{\lambda }^+_{\ell ,\chi }[s]\big )^{n-1} (\chi (s))\, \text {d} s\\&= \textstyle \sum _{n=1}^{q-1} \frac{(-1)^{n-1}}{n}\cdot \int _{a'}^{b'} \big (\sum _{\ell =1}^{q-2} \varvec{\lambda }^+_{\ell ,\psi }[s]\big )^{n-1} (\psi (s))\, \text {d} s\\&{\mathop {=}\limits ^{(84)}}\mathfrak {X}_\psi (b') \end{aligned} \end{aligned}$$

(88)

holds. Moreover, (75) in Proposition 3 applied with , $\phi \equiv \chi |_{[1,2]}$, $\mathfrak {X}\equiv \tilde{\mathfrak {X}}|_{[1,2]}$ there (first step) yields for $t\in [1,2]$ that

We obtain

From this, the claim follows, because $\varrho _2$ is strictly increasing on (1, 2), thus strictly increasing on [1, 2], hence bijective. $\square$

4.2.3 The proof of Lemma 16

Let $\text {N} \subseteq \mathfrak {g}$ and $\ell \ge 1$ be given:

We let $\mathcal {H}_\ell (\text {N} )$ denote the set of all maps $\alpha :[a,b]\times [0,1]\rightarrow \mathfrak {g}$ that admit the following properties:

(1):

We have $\text {im} [\alpha ]\subseteq \overline{\mathcal {G}}_{\ell +1}(\text {N} )$.

(2):

There exists a continuous map $L:[a,b]\times [0,1]\rightarrow \overline{\mathcal {G}}_1(\text {N} )\subseteq \mathfrak {g}$, as well as a map $\tilde{\alpha }:I\times [0,1]\rightarrow \mathfrak {g}$ for $I\subseteq \mathbb {R}$ an open interval with $[a,b]\subseteq I$, such that the following conditions hold:

(a):

$\tilde{\alpha }$ is continuous, with $\tilde{\alpha }|_{[a,b]\times [0,1]}=\alpha$.

(b):

To each $\tau \in [a,b]$, there exist $\delta _{\tau }>0$ as well as $\epsilon _\tau :(-\delta _\tau ,\delta _\tau )\times [0,1]\rightarrow \mathfrak {g}$ continuous with $\lim _{h\rightarrow 0}\textsf {p} _\infty (\epsilon _\tau (h,\cdot ))=0$ for each $\textsf {p} \in \textsf {Sem}(E)$, such that

$$\begin{aligned} \tilde{\alpha }(\tau +h,t)= \tilde{\alpha }(\tau ,t) + h\cdot L(\tau ,t) + h\cdot \epsilon _\tau (h,t)\qquad \quad (h,t)\in (-\delta _\tau ,\delta _\tau )\times [0,1]. \end{aligned}$$

In particular, $\Phi \equiv \tilde{\alpha }$ fulfills the assumptions in Theorem C for $(G,\cdot )\equiv (\mathfrak {g},+)$ there (recall Remark 8), i.e., we have

$$\begin{aligned} \textstyle \frac{\text {d} }{\text {d} h}\big |_{h=0}\int _0^1\alpha (\tau +h,s)\,\text {d} s= \int _0^1 L(\tau ,s)\,\text {d} s\qquad \quad \forall \, \tau \in [a,b]. \end{aligned}$$

(89)

We let $\mathcal {K}_{\ell }(\text {N} )$ denote the set of all (continuous) maps of the form

$$\begin{aligned} \textstyle \eta :[a,b]\times [0,1]\ni (\tau ,t)\mapsto \mathfrak {z}(\tau ) + \alpha (\tau ,t)\in \mathfrak {g}, \end{aligned}$$

for $\mathfrak {z}\in C^1([a,b],\mathfrak {g})$ with $\text {im} [\mathfrak {z}]\cup \text {im} [{\dot{\mathfrak {z}}}]\subseteq \overline{\mathcal {G}}_1(\text {N} )$, and $\alpha \in \mathcal {H}_\ell (\text {N} )$ (in particular, $\text {im} [\eta ]\subseteq \overline{\mathcal {G}}_1(\text {N} )$).

We recall Convention (50), and are ready for the proof of Lemma 16.

Proof of Lemma 16

Set $\text {N} :=\text {im} [\phi ]$. We prove by induction that

$$\begin{aligned} \mathcal {K}_{\ell }(\text {N} )\ni \eta _\ell :[a,b]\times [0,1]\ni (\tau ,t)\mapsto \mathfrak {T}^\ell \big (\phi |_{[a,\tau ]}\big )(t)\in \mathfrak {g}\qquad \quad \forall \, \ell =1,\dots ,q-1 \end{aligned}$$

(90)

holds. Evaluating this for $\ell =q-1$, the claim follows because:

• For $\alpha \in \mathcal {H}_{q-1}(\text {N} )$, we have $\text {im} [\alpha ]\subseteq \overline{\mathcal {G}}_q(\text {N} )=\{0\}$ by Condition 1) and Remark 3.1).

• $\overline{\mathcal {G}}_1(\text {N} )$ is a $\text {Nil} _{q}$-set by Remark 3.3).

To perform the induction, we fix $\varepsilon >0$ and $\tilde{\phi }\in C^k([a-\varepsilon ,b+\varepsilon ],\mathfrak {g})$ with $\tilde{\phi }|_{[a,b]}=\phi$. We observe that for each $\psi \in \mathfrak {D}_{[a,b]}$ with $\text {im} [\psi ]$ a $\text {Nil} _{q}$-set, and $\lambda ,t\in [a,b]$ we have (confer Example 7)

(91)

Induction Basis: Let $\ell =1$. By (91), for $\tau \in [a,b]$ and $t\in [0,1]$ we have

with

• $\mathfrak {z}\in C^1([a,b],\mathfrak {g})$ by (20), and $\text {im} [\mathfrak {z}]\cup \text {im} [{\dot{\mathfrak {z}}}]\subseteq \overline{\mathcal {G}}_1(\text {N} )$ by Remark 3.4).

• $\text {im} [\alpha ]\subseteq \overline{\mathcal {G}}_2(\text {N} )$ (hence Condition 1) for $\ell \equiv 1$ there) by Remark 3.4).

To verify Condition 2), for $\tau _1,\tau _2\in I:= (a-\varepsilon ,b+\varepsilon )$ and $t\in [0,1]$, we set

$$\begin{aligned} \tilde{\beta }(\tau _1,\tau _2,t):= \textstyle \sum _{p=1}^{q-2}\, t^p\cdot \int _{a-\varepsilon }^{\tau _1} \text {d} s \int _s^{\tau _2}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\tilde{\phi }(s_1)}\circ \dots \circ \text {ad} _{\tilde{\phi }(s_{p})})(\tilde{\phi }(s)), \end{aligned}$$

and define $\tilde{\alpha }(\tau ,t):= \tilde{\beta }(\tau ,\tau ,t) - \tilde{\beta }(a,\tau ,t)$ for $\tau \in I$ and $t\in [0,1]$. Then, $\tilde{\alpha }$ is continuous, with $\tilde{\alpha }|_{[a,b]\times [0,1]}=\alpha$, hence fulfills Condition 2).a). Moreover, for $\tau \in I$, $t\in [0,1]$, and $|h|>0$ suitably small, we have

$$\begin{aligned} \tilde{\alpha }(\tau +h,t)&\textstyle -\tilde{\alpha }(\tau ,t)\\&\textstyle =\sum _{p=1}^{q-2}\, t^p\cdot \int _{a}^{\tau } \text {d} s \int _\tau ^{\tau +h}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots }\\&\quad \textstyle \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\tilde{\phi }(s_1)}\circ \dots \circ \text {ad} _{\tilde{\phi }(s_{p})})(\tilde{\phi }(s))\\&\quad \,\textstyle +\sum _{p=1}^{q-2}\, t^p\cdot \int _{\tau }^{\tau +h} \text {d} s \int _s^{\tau +h}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \\&\quad \textstyle \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\tilde{\phi }(s_1)}\circ \dots \circ \text {ad} _{\tilde{\phi }(s_{p})})(\tilde{\phi }(s)),\\ \tilde{\alpha }(\tau ,t)-\tilde{\alpha }&\textstyle (\tau -h,t)\\&\textstyle =\sum _{p=1}^{q-2}\, t^p\cdot \int _{a}^{\tau -h} \text {d} s \int _{\tau -h}^{\tau }\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \\&\quad \textstyle \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\tilde{\phi }(s_1)}\circ \dots \circ \text {ad} _{\tilde{\phi }(s_{p})})(\tilde{\phi }(s))\\&\quad \,\textstyle +\sum _{p=1}^{q-2}\, t^p\cdot \int _{\tau -h}^{\tau } \text {d} s \int _s^{\tau }\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots }\\&\quad \textstyle \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\tilde{\phi }(s_1)}\circ \dots \circ \text {ad} _{\tilde{\phi }(s_{p})})(\tilde{\phi }(s)). \end{aligned}$$

We define $L:[a,b]\times [0,1]\rightarrow \overline{\mathcal {G}}_1(\text {N} )\subseteq \mathfrak {g}$ (recall Remark 3.4)) by

$$\begin{aligned} L(\tau , t):= \textstyle \sum _{p=1}^{q-2}\, t^p\cdot \int _{a}^{\tau } \text {d} s \int _s^{\tau }\text {d} s_2 \, {\dots } \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\phi (\tau )}\circ \text {ad} _{\phi (s_2)} \circ \dots \circ \text {ad} _{\phi (s_{p})})(\phi (s)) \end{aligned}$$

for all $\tau \in [a,b]$ and $t\in [0,1]$. Clearly, L is continuous; and it follows from (19) as well as continuity of the involved maps that

$$\begin{aligned} \textstyle \lim _{h\rightarrow 0} 1/|h|\cdot \textsf {p} _\infty \textstyle (\tilde{\alpha }(\tau +h,\cdot )-\tilde{\alpha }(\tau ,\cdot )- h\cdot L(\tau ,\cdot ))=0\qquad \quad \forall \, \textsf {p} \in \textsf {Sem}(E),\, \tau \in [a,b] \end{aligned}$$

holds, which verifies Condition 2).b). We thus have shown (90) for $\ell = 1$. In particular, this proves the claim for $q=2$.

Induction Step: Assume that (90) holds for some $1\le \ell \le q-1$ (with $q\ge 3$), hence we have

$$\begin{aligned} \textstyle \mathcal {K}_{\ell }(\text {N} )\ni \eta _\ell :[a,b]\times [0,1]\ni (\tau ,t)\mapsto \mathfrak {z}_\ell (\tau ) + \alpha _\ell (\tau ,t)\in \overline{\mathcal {G}}_1(\text {N} )\subseteq \mathfrak {g}\end{aligned}$$

(92)

for some $\mathfrak {z}_\ell \in C^1([a,b],\mathfrak {g})$ with $\text {im} [\mathfrak {z}_\ell ]\cup \text {im} [{\dot{\mathfrak {z}}}_\ell ]\subseteq \overline{\mathcal {G}}_1(\text {N} )$, as well as $\alpha _\ell \in \mathcal {H}_{\ell }(\text {N} )$. Let $I\subseteq \mathbb {R}$ be as in Condition 2) for $\alpha \equiv \alpha _\ell$ there. We can shrink I around [a, b] in such a way that $\mathfrak {z}_\ell$ extends to a $C^1$-curve $\tilde{\mathfrak {z}}_\ell \in C^1(I,\mathfrak {g})$, and set

$$\begin{aligned} \tilde{\eta }_\ell :I\times [0,1]\ni (\tau ,t)\mapsto \tilde{\mathfrak {z}}_\ell (\tau ) + \tilde{\alpha }_\ell (\tau ,t)\in \mathfrak {g}. \end{aligned}$$

(93)

Now, (91) for $\psi \equiv \eta _\ell (\tau ,\cdot )$ and $\lambda \equiv 1$ yields for $\tau \in [a,b]$ and $t\in [0,1]$ that

We obtain from (92) that $\mathfrak {z}_{\ell +1}=\mathfrak {z}_{\ell }+ \int _0^1 \alpha _\ell (\cdot ,s) \, \text {d} s$ holds, and conclude the following:

• Let $L:[a,b]\times [0,1]\rightarrow \overline{\mathcal {G}}_1(\text {N} )\subseteq \mathfrak {g}$ be as in Condition 2).b) for $\alpha \equiv \alpha _\ell$ there (induction hypothesis). Then, we have

  $$\begin{aligned} \textstyle {\dot{\mathfrak {z}}}_{\ell +1}(\tau ){\mathop {=}\limits ^{(89)}}{\dot{\mathfrak {z}}}_{\ell }(\tau ) + \int _0^1 L(\tau ,s)\, \text {d} s\in \overline{\mathcal {G}}_1(\text {N} ) \qquad \quad \forall \, \tau \in [a,b]. \end{aligned}$$

  Since L is continuous, this implies $\mathfrak {z}_{\ell +1}\in C^1([a,b],\mathfrak {g})$ with $\text {im} [\mathfrak {z}_{\ell +1}]\cup \text {im} [{\dot{\mathfrak {z}}}_{\ell +1}]\subseteq \overline{\mathcal {G}}_1(\text {N} )$.

• We have $\text {im} [\alpha _{\ell +1}]\subseteq \overline{\mathcal {G}}_{\ell +2}(\text {N} )$ by Remark 3.2), Remark 3.4), bilinearity of $\varvec{[}\cdot , \cdot \varvec{]}$, and

  • $(\text {ad} _{\mathfrak {z}_\ell (\tau )}\circ \dots \circ \text {ad} _{\mathfrak {z}_\ell (\tau )})(\mathfrak {z}_\ell (\tau ))=0$ for all $\tau \in [a,b]$,

  • $\text {im} [\mathfrak {z}_\ell ]\subseteq \overline{\mathcal {G}}_1(\text {N} )$,

  • $\text {im} [\alpha _\ell ]\subseteq \overline{\mathcal {G}}_{\ell +1}(\text {N} )$,

  which verifies Condition 1) (for $\alpha \equiv \alpha _{\ell +1}$ there).

It remains to verify Condition 2) (for $\alpha \equiv \alpha _{\ell +1}$ there). For this, we set (recall (93))

$$\begin{aligned} \tilde{\alpha }_{\ell +1}(\tau ,t):=&\textstyle \sum _{p=1}^{q-2}\, t^p\cdot \int _0^1 \text {d} s \int _s^{1}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \\&\textstyle \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\tilde{\eta }_\ell (\tau ,s_1)}\circ \dots \circ \text {ad} _{\tilde{\eta }_\ell (\tau ,s_{p})})(\tilde{\eta }_\ell (\tau ,s)) \end{aligned}$$

for $\tau \in I$ and $t\in [0,1]$. Clearly, $\tilde{\alpha }_{\ell +1}$ fulfills Condition 2).a) (for $\alpha \equiv \alpha _{\ell +1}$ there). Moreover, since $\mathfrak {z}_\ell$ is of class $C^1$ with $\text {im} [\mathfrak {z}]\cup \text {im} [{\dot{\mathfrak {z}}}]\subseteq \overline{\mathcal {G}}_1(\text {N} )$, by (93), and Condition 2).b) for $\alpha \equiv \alpha _\ell$ (induction hypothesis), there exists a continuous map $L:[a,b]\times [0,1]\rightarrow \overline{\mathcal {G}}_1(\text {N} )\subseteq \mathfrak {g}$, as well as to each fixed $\tau \in [a,b]$ some $\delta _{\tau }>0$ and $\epsilon _\tau :(-\delta _\tau ,\delta _\tau )\times [0,1]\rightarrow \mathfrak {g}$ continuous with $\lim _{h\rightarrow 0}\textsf {p} _\infty (\epsilon _\tau (h,\cdot ))=0$ for each $\textsf {p} \in \textsf {Sem}(E)$, such that

$$\begin{aligned} \tilde{\eta }_\ell (\tau +h,t)= \tilde{\eta }_\ell (\tau ,t) + h\cdot L(\tau ,t) + h\cdot \epsilon _\tau (h,t)\qquad \quad (h,t)\in (-\delta _\tau ,\delta _\tau )\times [0,1] \end{aligned}$$

(94)

holds. We define for $\tau \in I$ and $t\in [0,1]$

$$\begin{aligned} \tilde{L}(\tau ,t):=&\textstyle \sum _{p=1}^{q-2}\, t^p\cdot \int _0^1 \text {d} s \int _s^{1}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{p-1}}\text {d} s_p \sum _{k=1}^{p+1}\, \gamma _k(s_1,\dots ,s_p,s), \end{aligned}$$

where for $1\le p\le q-2$ and $0\le s_1,\dots ,s_p,s\le 1$, we set

$$\begin{aligned} \textstyle \gamma _{p+1}(s_1,\dots ,s_p,s):= (\text {ad} _{\tilde{\eta }_\ell (\tau ,s_1)}\circ \dots \circ \text {ad} _{\tilde{\eta }_\ell (\tau ,s_p)})(L(\tau ,s)) \end{aligned}$$

as well as for $1\le k\le p$

$$\begin{aligned} \textstyle \gamma _k(s_1,\dots ,s_p,s):= (\text {ad} _{\tilde{\eta }_\ell (\tau ,s_1)}\circ \dots&\textstyle \circ \text {ad} _{\tilde{\eta }_\ell (\tau ,s_{k-1})} \circ \text {ad} _{L_\tau (s_k)}\\&\textstyle \circ \text {ad} _{\tilde{\eta }_\ell (\tau ,s_{k+1})} \circ \dots \circ \text {ad} _{\tilde{\eta }_\ell (\tau ,s_p)})( \tilde{\eta }_\ell (\tau ,s)). \end{aligned}$$

It follows from continuity and bilinearity of $\varvec{[}\cdot , \cdot \varvec{]}$, (94), and (19) that

$$\begin{aligned} \textstyle \lim _{h\rightarrow 0} 1/|h|\cdot \textsf {p} _\infty (\tilde{\alpha }_{\ell +1}(\tau +h,\cdot ) - \tilde{\alpha }_{\ell +1}(\tau ,\cdot ) - h \cdot \tilde{L}(\tau ,\cdot ))=0\qquad \quad \forall \, \textsf {p} \in \textsf {Sem}(E) \end{aligned}$$

holds, which verifies Condition 2).b) for $\alpha \equiv \alpha _{\ell +1}$ there. $\square$

5 Asymptotic estimate Lie algebras

Throughout this section, $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})$ denotes a fixed sequentially complete and asymptotic estimate Lie algebra, i.e., the following conditions are fulfilled:

• $\mathfrak {q}\in \textsf {hlcVect}$ is sequentially complete.

• $\varvec{[}\cdot , \cdot \varvec{]}:\mathfrak {q}\times \mathfrak {q}\rightarrow \mathfrak {q}$ is bilinear, antisymmetric, continuous, and fulfills the Jacobi identity (25). Moreover, to each $\textsf {v} \in \textsf {Sem}(\mathfrak {q})$, there exist $\textsf {v} \le \textsf {w} \in \textsf {Sem}(\mathfrak {q})$ with

  $$\begin{aligned} \textsf {v} (\varvec{[}X_1,\varvec{[}X_2,\varvec{[}\dots ,\varvec{[}X_n,Y\varvec{]}{\dots }\varvec{]}\varvec{]}\varvec{]})\le \textsf {w} (X_1)\cdot {\dots }\cdot \textsf {w} (X_n)\cdot \textsf {w} (Y) \end{aligned}$$

  (95)

  for all $X_1,\dots ,X_n,Y\in \mathfrak {q}$ and $n\ge 1$.

The first part of this section (Sect. 5.1) is dedicated to a comprehensive analysis of the Lax equation

$$\begin{aligned} {\dot{\alpha }}=\varvec{[}\psi , \alpha \varvec{]}\qquad \text {with}\qquad \alpha (a)=X, \end{aligned}$$

for $X\in \mathfrak {q}$ and $\psi \in C^0([a,b],\mathfrak {q})$ ($a<b$) fixed, and $\alpha \in C^1([a,b],\mathfrak {q})$. We first show the existence and uniqueness of solutions and discuss their elementary properties (Sect. 5.1.1 and Sect. 5.1.2). More specifically, let ${\textbf {Ad}} ^\pm _\psi [X]$ be defined by the right hand side of (68). We show that the unique solution to the above equation is given by $\alpha ={\textbf {Ad}} _\psi ^+[X]$. We also show¹⁸

$$\begin{aligned} \textsf {Aut} (\mathfrak {q})\ni {\textbf {Ad}} _\psi ^\pm [t]:\mathfrak {q}\in X\mapsto {\textbf {Ad}} ^+_\psi [X](t)\in \mathfrak {q}\qquad \quad \forall \, t\in [a,b], \end{aligned}$$

with ${\textbf {Ad}} ^\pm _\psi [t]^{-1}={\textbf {Ad}} ^\mp _\psi [t]$ (both statements are proven in Proposition 6). This allows to define a group structure on $C^k([a,b],\mathfrak {q})$ for $k\in \mathbb {N}\cup \{\infty \}$ (in analogy to Remark 5 and Sect. 1.14 in [2]) by

$$\begin{aligned} \psi ^{-1}&:[a,b]\ni t \mapsto \qquad -{\textbf {Ad}} _\psi ^-[t](\psi (t))\in \mathfrak {q}\\ \phi \star \psi&:[a,b]\ni t\mapsto \phi (t) + {\textbf {Ad}} ^+_\phi [t](\psi (t))\in \mathfrak {q}\end{aligned}$$

for $\phi ,\psi \in C^k([a,b],\mathfrak {q})$. The group axioms are verified in Sect. 5.1.3.

Remark 14

(A Lie Group Construction) Let $k\in \mathbb {N}\cup \{\infty \}$ and $a<b$ be given; and set

$$\begin{aligned} G_k=E_k=\mathfrak {g}_k:=C^k([a,b],\mathfrak {q}),\qquad e_k:=0\in C^k([a,b],\mathfrak {q}),\qquad \Xi _k:=\text {id} _{C^k([a,b],\mathfrak {q})}. \end{aligned}$$

As already mentioned, we will show that the maps

(96)

are defined (confer Lemma 17), and that $(G_k,\textsf {m} _k,\textsf {inv} _k,e_k)$ is a group (confer Lemma 22). Expectably, the group operations (96) are smooth w.r.t. the $C^k$-topology; hence give rise to a Lie group $G_k$ with global chart $\Xi _k$, and sequentially complete Lie algebra $\mathfrak {g}_k$ (recall Lemma 3). It is furthermore to be expected that the corresponding Lie bracket $\varvec{[}\cdot , \cdot \varvec{]}_k:\mathfrak {g}_k\times \mathfrak {g}_k\rightarrow \mathfrak {g}_k$ is given by

$$\begin{aligned} \textstyle \varvec{[}\phi , \psi \varvec{]}_k(t) = \varvec{[}\int _a^t\phi (s)\,\text {d} s, \psi (t) \varvec{]} + \varvec{[}\phi (t), \int _a^t \psi (s)\,\text {d} s \varvec{]}\qquad \quad \forall \, t\in [a,b] \end{aligned}$$

for $\phi ,\psi \in \mathfrak {g}_k$, just as in Sect. 1.14 in [2] (confer Proposition 1.14.1 in [2]). Presumably, $(\mathfrak {g}_k,\varvec{[}\cdot , \cdot \varvec{]}_k)$ is asymptotic estimate, and $G_k$ is $C^0$-regular. The technical details will be worked out in a separate paper. This serves as a preparation for a possible extension of Lie’s third theorem to the infinite-dimensional asymptotic estimate case—just by performing the same (a similar) construction made in Sect. 1.14 in [2] to prove this theorem (Theorem 1.14.3 in [2]) in the finite-dimensional context. Notably, this construction had already been used in [1], to prove Lie’s third theorem in the context of Lie algebroids. $\square$

In the last part of Sect. 5.1, we use the proven statements to investigate the properties of the following integral transformation that mimics (49) in Sect. 3.1:

$$\begin{aligned} \textstyle \mathfrak {T}:C^0([a,b],\mathfrak {q})&\rightarrow C^\infty ([0,1],\mathfrak {q})\\ \phi&\textstyle \mapsto \big [ [0,1]\ni t \mapsto \int _a^b {\textbf {Ad}} ^+_{ t\cdot \phi |_{[s,b]}}[b](\phi (s)) \,\text {d} s\,\big ]. \end{aligned}$$

The investigations serve as a preparation for the discussions in Sect. 5.2. There, the above transformation is applied to the situation where $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})\equiv (\mathfrak {g},[\cdot ,\cdot ])$ equals the Lie algebra of a given Lie group G. Specifically, we prove the following regularity result (cf. Theorem I):

Theorem 2

Assume that $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is asymptotic estimate and sequentially complete. If G is weakly $C^\infty$-regular, then G is weakly $C^k$-regular for each $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$.

Remark 15

Note that in the situation of Theorem 2, G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$ if and only if G is $C^k$-semiregular, just because $\mathfrak {g}$ is assumed to be sequentially complete. $\square$

This result complements Theorem 2 in [9] that essentially states that $C^\infty$-regularity¹⁹ implies $C^k$-regularity for each $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, where for $k=0$ additionally sequentially completeness of the Lie group has to be assumed.²⁰ Hence, in a certain sense, the assumption in Theorem 2 in [9] that G is sequentially complete, had been replaced in Theorem 2 by the assumption that $\mathfrak {g}$ is sequentially complete. The key result proven in [9] (Theorem 1 in [9]) states that in the asymptotic estimate context, $C^\infty$-continuity of the product integral is equivalent to its $C^0$-continuity (local $\mu$-convexity of G), which makes the semiregularity results obtained in [8] applicable—Basically, in analogy to the definition of the Riemann integral as a limit over Riemann sums, in [8] the product integral is obtained as a limit over product integrals of piecewise integrable curves (under certain assumptions on the Lie group that are automatically fulfilled in the asymptotic estimate context). The proof of Theorem 2 works differently: For some given $\phi \in C^0([a,b],\mathfrak {g})$, we show that (this is defined by $C^\infty$-semiregularity of G, as well as $\text {im} [\mathfrak {T}]\subseteq C^\infty ([0,1],\mathfrak {g})$)

is of class $C^1$, with .

Remark 16

We will tacitly use throughout this section that sequentially completeness of $\mathfrak {q}$ implies (recall Remark 2)

$$\begin{aligned} \textstyle \int \phi (s)\,\text {d} s\in \mathfrak {q}\qquad \quad \forall \, \phi \in C^0([a,b],\mathfrak {q}),\, a<b. \end{aligned}$$

We will also use that $\varvec{[}\cdot , \cdot \varvec{]}$ is smooth (apply, e.g., the parts b), e) of Proposition A.1). $\square$

Theorem 2, together with Theorem 4 in [8] and Theorem 1 in [9], yields the following statement:

Corollary 14

If $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is asymptotic estimate and sequentially complete, then the following assertions are equivalent:

• (a) G is $C^k$-semiregular for some $k\in \mathbb {N}\cup \{\infty \}$, and ${\text {evol} }_\text {k}$ is of class $C^\ell$ for some $\ell \in \mathbb {N}\cup \{\infty \}$.

• (b) G is $C^0$-semiregular, and ${\text {evol} }_0$ is $C^0$-continuous ( G is locally $\mu$-convex ).

• (c) G is $C^k$-regular for all $k\in \mathbb {N}\cup \{\infty \}$.

Proof

The implication c) $\Rightarrow$ a) is evident. Moreover:

a) $\Rightarrow$ b): If the assertion b) holds, then

• G is $C^\infty$-semiregular, as $C^\infty ([0,1],\mathfrak {g})\subseteq C^k([0,1],\mathfrak {g})$ holds. Hence, G is weakly $C^\infty$-regular by Remark 15; thus, G is $C^0$-semiregular by Theorem 2.

• ${\text {evol} }_\text {k}$ is $C^k$-continuous, as of class $C^\ell$ (with $\ell \ge 0$). Hence, ${\text {evol} }_0$ is $C^0$-continuous (G is locally $\mu$-convex) by Theorem 1 in [9].

b) $\Rightarrow$ c): Assume that the assertion b) holds, and let $k\in \mathbb {N}\cup \{\infty \}$ be given. Then,

• G is $C^k$-semiregular, as $C^k([0,1],\mathfrak {g})\subseteq C^0([0,1],\mathfrak {g})$ holds.

• ${\text {evol} }_\text {k} ={\text {evol} }_0|_{C^k([0,1],\mathfrak {g})}$ is $C^k$-continuous, as even $C^0$-continuous.

Since $\mathfrak {g}$ is sequentially complete (hence, integral complete and Mackey complete), Theorem 4 in [8] shows that ${\text {evol} }_\text {k}$ is smooth. $\square$

5.1 The Lax equation

Let $a<b$ and $\psi \in \text {CP} ^0([a,b],\mathfrak {q})$ be given.²¹ Motivated by Sect. 3.3, for $X\in \mathfrak {q}$ we set

$$\begin{aligned} \varvec{\lambda }^\pm _{0,\psi }[X]:[a,b]\rightarrow \mathfrak {q},\qquad t\mapsto X, \end{aligned}$$

as well as (recall (23))

for each $\ell \ge 1$. We furthermore define (definedness is covered by Lemma 17)

$$\begin{aligned} \textstyle {\textbf {Ad}} ^\pm _\psi [X](t):= \sum _{\ell =0}^\infty \varvec{\lambda }^\pm _{\ell ,\psi }[X](t)\qquad \quad \forall \, X\in \mathfrak {q},\, t\in [a,b], \end{aligned}$$

and let

$$\begin{aligned} \varvec{\Lambda }_\psi ^\pm :[a,b]\times \mathfrak {q}\rightarrow \mathfrak {q},\qquad (t,X)\mapsto {\textbf {Ad}} _\psi ^\pm [X](t). \end{aligned}$$

To simplify the notations, we define

$$\begin{aligned} {\textbf {Ad}} _{\psi }^\pm [t]:\mathfrak {q}\ni X&\mapsto {\textbf {Ad}} _\psi ^\pm [X](t)\in \mathfrak {q}\qquad \quad \forall \, t\in [a,b]\\ {\textbf {Ad}} _\psi ^\pm&:= {\textbf {Ad}} _{\psi }^\pm [b]\\ \varvec{\lambda }_{\ell ,\psi }^\pm [t]:\mathfrak {q}\ni X&\mapsto \varvec{\lambda }_{\ell ,\psi }^\pm [X](t)\in \mathfrak {q}\qquad \quad \qquad \forall \, t\in [a,b],\,\, \ell \in \mathbb {N}\\ \varvec{\lambda }_{\ell ,\psi }^\pm&:= \varvec{\lambda }_{\ell ,\psi }^\pm [b]\qquad \quad \qquad \forall \, \ell \in \mathbb {N}. \end{aligned}$$

5.1.1 Elementary properties

In this section, we discuss the elementary properties of the objects defined in the beginning of Sect. 5.1. We start with the following lemma.

Lemma 17

Let $\psi \in C^0([a,b],\mathfrak {q})$ be given. Then, the following assertions hold:

(1):

The maps $\varvec{\Lambda }_\psi ^\pm$ are defined and of class $C^1$, and we have

$$\begin{aligned} \partial _1 \varvec{\Lambda }_\psi ^+(t,X)&\textstyle =\partial _t{\textbf {Ad}} _\psi ^+[X](t)=\varvec{[}\psi (t), {\textbf {Ad}} ^+_\psi [X](t) \varvec{]}\\ \partial _1 \varvec{\Lambda }^-_\psi (t,X)&\textstyle =\partial _t{\textbf {Ad}} _\psi ^-[X](t)=-{\textbf {Ad}} ^-_\psi [\varvec{[}\psi (t), X \varvec{]}](t) \end{aligned}$$

for all $t\in [a,b]$ and $X\in \mathfrak {q}$.

(2):

$\varvec{\Lambda }^\pm _\psi (t,\cdot )={\textbf {Ad}} _\psi ^\pm [t]$ is linear and continuous for each $t\in [a,b]$.

(3):

If $\psi \in C^k([a,b],\mathfrak {q})$ holds for $k\in \mathbb {N}\cup \{\infty \}$, then $\varvec{\Lambda }_\psi ^\pm$ is of class $C^{k+1}$.

Proof

The proof is straightforward but technical. It is provided in Appendix A.7. $\square$

Remark 17

The substitution formula (21) yields the following:

• Let $a<b$, $a'<b'$, $\psi \in C^0([a,b],\mathfrak {q})$, as well as $\varrho :[a',b']\rightarrow [a,b]$ be of class $C^1$ with $\varrho (a')=a$ and $\varrho (b')=b$. We set $\phi := {\dot{\varrho }}\cdot (\psi \circ \varrho )$ and obtain inductively from (21) that

  $$\begin{aligned} \textstyle \varvec{\lambda }^\pm _{\ell ,\phi }[t]=\varvec{\lambda }^\pm _{\ell ,\psi }[\varrho (t)]\qquad \quad \forall \, t\in [a',b'] \end{aligned}$$

  holds for all $\ell \in \mathbb {N}$. Hence, we have ${\textbf {Ad}} ^\pm _{{\dot{\varrho }}\,\cdot \, (\psi \,\circ \,\varrho )}[\cdot ]={\textbf {Ad}} ^\pm _{\psi }[\varrho (\cdot )]$. In particular, if additionally ${\dot{\varrho }}|_{(a,b)}>0$ holds, then we have

  $$\begin{aligned} {\textbf {Ad}} ^\pm _{({\dot{\varrho }}\,\cdot \, (\psi \,\circ \,\varrho ))|_{[x',y']}}[\cdot ]={\textbf {Ad}} ^\pm _{\psi |_{[\varrho (x'),\varrho (y')]}}[\varrho (\cdot )]\qquad \quad \forall \, a'\le x'<y\le b'. \end{aligned}$$

  (97)

• The previous point applied to affine transformations yields²²

  $$\begin{aligned} {\textbf {Ad}} ^\pm _\chi [b]={\textbf {Ad}} ^\pm _{\chi (\cdot -x)}[b+x] \end{aligned}$$

  for $a<b$, $\chi \in C^0([a,b],\mathfrak {q})$, and $x\in \mathbb {R}$.$\square$

Corollary 15

For $\psi \in C^0([a,b],\mathfrak {q})$ and $t\in [a,b]$, we have ${\textbf {Ad}} _\psi ^-[t]\circ {\textbf {Ad}} _\psi ^+[t]=\text {id} _\mathfrak {q}$.

Proof

Let $Z\in \mathfrak {q}$ be given, and define

$$\begin{aligned} \alpha _Z:[a,b]\ni t\mapsto \varvec{\Lambda }_\psi ^-(t,\varvec{\Lambda }_\psi ^+(t,Z))=({\textbf {Ad}} ^-_\psi [t]\circ {\textbf {Ad}} ^+_\psi [t])(Z)\in \mathfrak {q}. \end{aligned}$$

Then, Lemma 17 shows $\alpha _Z\in C^1([a,b],\mathfrak {q})$; and the parts b), b), e) of Proposition A.1 yield

$$\begin{aligned} {\dot{\alpha }}_Z(t)&\textstyle = -\varvec{\Lambda }_\psi ^-(t,\varvec{[}\psi (t), \varvec{\Lambda }_\psi ^+(t,Z) \varvec{]}) + \varvec{\Lambda }_\psi ^-(t,\varvec{[}\psi (t), \varvec{\Lambda }_\psi ^+(t,Z) \varvec{]})=0\qquad \quad \forall \, t\in [a,b], \end{aligned}$$

hence $\alpha _Z(t)=Z+ \int _a^t {\dot{\alpha }}_Z(s) \,\text {d} s=Z$ for each $t\in [a,b]$ by (17). $\square$

Lemma 18

For $\psi \in \text {CP} ^0([a,b],\mathfrak {q})$ and $a< c< b$, we have ${\textbf {Ad}} ^+_\psi ={\textbf {Ad}} ^+_{\psi |_{[c,b]}}\circ {\textbf {Ad}} ^+_{\psi |_{[a,c]}}$.

Proof

Let $X\in \mathfrak {q}$ be fixed. We first prove by induction that

$$\begin{aligned} \varvec{\lambda }^+_{\ell ,\psi }[X](s)\textstyle =\sum _{m=0}^\ell \varvec{\lambda }^+_{m,\psi |_{[c,b]}}[\varvec{\lambda }^+_{\ell -m,\psi |_{[a,c]}}[X](c)](s)\qquad \quad \forall \, c<s\le b \end{aligned}$$

(98)

holds for each $\ell \in \mathbb {N}$. It is clear from the definitions that

$$\begin{aligned} \varvec{\lambda }^+_{0,\psi }[X](s)=X= \varvec{\lambda }^+_{0,\psi |_{[c,b]}}[\varvec{\lambda }^+_{0,\psi |_{[a,c]}}[X](c)](s) \qquad \quad \forall \, c<s\le b. \end{aligned}$$

We thus can assume that (98) holds for some $\ell \in \mathbb {N}$, and obtain for $c<s\le b$ that

holds, so that (98) follows by induction. Then, for $n\ge 1$ we have

$$\begin{aligned} \Delta _n:=&\textstyle \sum _{\ell =0}^{2 n}\varvec{\lambda }^+_{\ell ,\psi }(X) - \sum _{p=0}^{n}\varvec{\lambda }^+_{p,\psi |_{[c,b]}}( \sum _{q=0}^{n}\varvec{\lambda }^+_{q,\psi |_{[a,c]}}(X)) \end{aligned}$$

(99)

$$\begin{aligned} =&\textstyle \sum _{\ell =0}^{2 n}\sum _{m=0}^\ell \varvec{\lambda }^+_{m,\psi |_{[c,b]}}(\varvec{\lambda }^+_{\ell -m,\psi |_{[a,c]}}(X)) - \sum _{p,q=0}^{n}\varvec{\lambda }^+_{p,\psi |_{[c,b]}}(\varvec{\lambda }^+_{q,\psi |_{[a,c]}}(X)). \end{aligned}$$

(100)

For $\textsf {v} \le \textsf {w}$ as in (95), we obtain from (100) and the binomial theorem that

$$\begin{aligned} \textstyle \textsf {v} (\Delta _n)&\textstyle \le \sum _{\ell =n+1}^{2 n}\sum _{m=0}^\ell \textsf {v} (\varvec{\lambda }^+_{m,\psi |_{[c,b]}}(\varvec{\lambda }^+_{\ell -m,\psi |_{[a,c]}}(X)))\\&\textstyle \le \textsf {w} (X)\cdot \sum _{\ell =n+1}^{\infty } (\sum _{m=0}^\ell \frac{1}{m!\cdot (\ell -m)!}\cdot (b-c)^m\cdot (c-a)^{\ell -m})\cdot \textsf {w} _\infty (\psi )^\ell \\&\textstyle = \textsf {w} (X)\cdot \sum _{\ell =n+1}^{\infty } \frac{1}{\ell !}\cdot (b-a)^\ell \cdot \textsf {w} _\infty (\psi )^\ell \end{aligned}$$

holds, which yields

It thus remains to show $\lim _n \beta _n={\textbf {Ad}} ^+_{\psi |_{[c,b]}}({\textbf {Ad}} ^+_{\psi |_{[a,c]}}(X))$. For this, we observe that

holds for each $n\in \mathbb {N}$; and choose $\textsf {w} \le \textsf {m}$ as in (95) for $\textsf {v} \equiv \textsf {w}$ and $\textsf {w} \equiv \textsf {m}$ there. We obtain

$$\begin{aligned} \textstyle \textsf {v} (\textrm{T}_1(n))&\textstyle \le \text {e} ^{(b-c)\cdot \textsf {w} _\infty (\psi )}\cdot \textsf {w} (\sum _{q=n+1}^{\infty }\varvec{\lambda }^+_{q,\psi |_{[a,c]}}(X))\\&\textstyle \le \text {e} ^{(b-c)\cdot \textsf {w} _\infty (\psi )}\cdot \sum _{q=n+1}^{\infty } \frac{1}{q!}\cdot (c-a)^q\cdot \textsf {m} _\infty (\psi )^p, \end{aligned}$$

hence $\lim _n \textrm{T}_1(n)=0$. We furthermore obtain

$$\begin{aligned} \textstyle \textsf {v} (\textrm{T}_2(n))&\textstyle \le (\sum _{p=n+1}^{\infty } \frac{1}{p!}\cdot (b-c)^q\cdot \textsf {w} _\infty (\psi )^p) \cdot \textsf {w} ({\textbf {Ad}} ^+_{\psi |_{[a,c]}}(X))\\&\textstyle \le (\sum _{p=n+1}^{\infty }\frac{1}{p!}\cdot (b-c)^p\cdot \textsf {w} _\infty (\psi )^p )\cdot \text {e} ^{(c-a)\cdot \textsf {m} _\infty (\psi )} \cdot \textsf {m} (X), \end{aligned}$$

hence $\lim _n \textrm{T}_2(n)=0$. It follows that $\lim _n\widetilde{\Delta }_n=0$ holds, which proves the claim. $\square$

Remark 18

Lemma 18, together with the second point in Remark 17 shows

$$\begin{aligned} {\textbf {Ad}} ^+_\psi [b]={\textbf {Ad}} ^+_{\psi (\cdot -x)}[b+x] \end{aligned}$$

for $a<b$, $\psi \in \text {CP} ^0([a,b],\mathfrak {q})$, and $x\in \mathbb {R}$. $\square$

Lemma 19

Let $\psi \in \text {CP} ^0([a,b],\mathfrak {q})$ and $\{\psi _n\}_{n\in \mathbb {N}}\subseteq \text {CP} ^0([a,b],\mathfrak {q})$ be given, with $\{\psi _n\}_{n\in \mathbb {N}}\rightarrow \psi$ w.r.t. the $C^0$-topology. Then, for each $X\in \mathfrak {q}$, we have $\{{\textbf {Ad}} ^\pm _{\psi _n}[X]\}_{n\in \mathbb {N}} \rightarrow {\textbf {Ad}} _\psi ^\pm [X]$ w.r.t. the $C^0$-topology.

Proof

For each $\ell \ge 1$ and $X_1,\dots ,X_\ell ,Y_1,\dots ,Y_\ell ,X\in \mathfrak {q}$, we have

$$\begin{aligned} \begin{aligned} (\text {ad} _{X_1}\circ {\dots } \circ ~&\text {ad} _{X_\ell })(X)\textstyle -(\text {ad} _{Y_1}\circ {\dots }\circ \text {ad} _{Y_\ell })(X)\\&\textstyle = \sum _{p=1}^\ell \, (\text {ad} _{Y_1}\circ \dots \circ \text {ad} _{Y_{p-1}}\circ \text {ad} _{X_p-Y_p}\circ \text {ad} _{X_{p+1}}\circ \dots \circ \text {ad} _{X_\ell })(X). \end{aligned} \end{aligned}$$

(101)

Let $\textsf {v} \le \textsf {w}$ be as in (95). For $N\in \mathbb {N}$ suitably large, we have $\textsf {w} _\infty (\psi _n)\le 2\cdot \textsf {w} _\infty (\psi )$ for each $n\ge N$. We obtain for $n\ge N$ from (101) (second step) that

$$\begin{aligned}&\textsf {v} ({\textbf {Ad}} ^\pm _{\psi _n}[X](t)-{\textbf {Ad}} ^\pm _\psi [X](t))\\&\quad \textstyle \le \sum _{\ell =0}^\infty \textsf {v} (\varvec{\lambda }^\pm _{\ell ,\psi _n}[X](t)-\varvec{\lambda }^\pm _{\ell ,\psi }[X](t))\\&\quad \textstyle \le (b-a)\cdot \textsf {w} _\infty (\psi _n-\psi ) \cdot \textsf {w} (X)\cdot \sum _{\ell =1}^\infty \frac{(b-a)^{\ell -1}}{(\ell -1)!}\cdot 2^{\ell -1}\cdot \textsf {w} _\infty (\psi )^{\ell -1}\\&\quad \textstyle = (b-a)\cdot \textsf {w} _\infty (\psi _n-\psi ) \cdot \textsf {w} (X)\cdot \text {e} ^{2\cdot (b-a)\cdot \textsf {w} _\infty (\psi )} \end{aligned}$$

holds for each $t\in [a,b]$, which proves the claim. $\square$

5.1.2 Uniqueness of the solution

For $\psi \in C^0([a,b],\mathfrak {q})$ with $a<b$, we define ${\mathfrak {inv}(\psi )}\in C^0([a,b],\mathfrak {q})$ as in Example 4 by

$$\begin{aligned} {\mathfrak {inv}(\psi )}:[a,b]\ni t\mapsto -\psi (a+b-t)\in \mathfrak {q}. \end{aligned}$$

Notably, we have ${\mathfrak {inv}({\mathfrak {inv}(\psi )})}=\psi$. In this section, we prove the following proposition.²³

Proposition 6

Let $\psi \in C^0([a,b],\mathfrak {q})$ be given. Then, the following assertions hold:

• (1) We have ${\textbf {Ad}} ^\pm _\psi [t]\in \textsf {Aut} (\mathfrak {q})$ for each $t\in [a,b]$, with ${\textbf {Ad}} ^\pm _\psi [t]^{-1}={\textbf {Ad}} ^\mp _\psi [t]={\textbf {Ad}} ^\pm _{{\mathfrak {inv}(\psi |_{[a,t]})}}$.

• (2) Let $X\in \mathfrak {q}$ be given. Then, ${\textbf {Ad}} ^+_\psi [X]$ is the unique solution $\alpha \in C^1([a,b],\mathfrak {q})$ to the differential equation (Lax equation)

  $$\begin{aligned} {\dot{\alpha }}=\varvec{[}\psi , \alpha \varvec{]}\qquad \,\,\text {with initial condition}\qquad \,\, \alpha (a)=X. \end{aligned}$$

We obtain the following corollaries.

Corollary 16

For each $\psi \in C^0([a,b],\mathfrak {q})$, we have

$$\begin{aligned} {\textbf {Ad}} ^\pm _\psi [\varvec{[}X, Y \varvec{]}]=\varvec{[}{\textbf {Ad}} ^\pm _\psi [X], {\textbf {Ad}} ^\pm _\psi [Y] \varvec{]}\qquad \quad \forall \, X,Y\in \mathfrak {q}. \end{aligned}$$

Proof

Let $X,Y\in \mathfrak {g}$ and $\psi \in C^0([a,b],\mathfrak {q})$ be given. Lemma 17 shows

$$\begin{aligned} C^1([a,b],\mathfrak {q})\ni \alpha :[a,b]\ni t\mapsto \varvec{[}{\textbf {Ad}} ^+_\psi [X](t), {\textbf {Ad}} ^+_\psi [Y](t) \varvec{]}\in \mathfrak {q}, \end{aligned}$$

with $\alpha (a)=\varvec{[}X, Y \varvec{]}$. The parts b), e) of Proposition A.1 (first step), and the Jacobi identity (25) (second step) show that

$$\begin{aligned} {\dot{\alpha }}(t)&\textstyle =\varvec{[}\varvec{[}\psi (t), {\textbf {Ad}} ^+_\psi [X](t) \varvec{]}, {\textbf {Ad}} ^+_\psi [Y](t) \varvec{]} + \varvec{[}{\textbf {Ad}} ^+_\psi [X](t), \varvec{[}\psi (t), {\textbf {Ad}} ^+_\psi [Y](t) \varvec{]} \varvec{]}\\&\textstyle = \varvec{[}\psi (t), \varvec{[}{\textbf {Ad}} ^+_\psi [X](t), {\textbf {Ad}} ^+_\psi [Y](t) \varvec{]} \varvec{]}\\&\textstyle =\varvec{[}\psi (t), \alpha (t) \varvec{]} \end{aligned}$$

holds for each $t\in [a,b]$. Proposition 6.2) yields $\alpha ={\textbf {Ad}} ^+_\psi [\varvec{[}X, Y \varvec{]}]$, which proves the one part of the statement. Then, Proposition 6.1) shows

$$\begin{aligned} {\textbf {Ad}} ^-_\psi [\varvec{[}X, Y \varvec{]}](t)&= {\textbf {Ad}} ^-_\psi [t](\varvec{[}({\textbf {Ad}} ^+_\psi [t] \circ {\textbf {Ad}} ^-_\psi [t])(X), ({\textbf {Ad}} ^+_\psi [t]\circ {\textbf {Ad}} ^-_\psi [t])(Y) \varvec{]})\\&= ({\textbf {Ad}} ^-_\psi [t]\circ {\textbf {Ad}} ^+_\psi [t])(\varvec{[}{\textbf {Ad}} ^-_\psi [X](t), {\textbf {Ad}} ^-_\psi [Y](t) \varvec{]})\\&=\varvec{[}{\textbf {Ad}} ^-_\psi [X](t), {\textbf {Ad}} ^-_\psi [Y](t) \varvec{]} \end{aligned}$$

for each $t\in [a,b]$ and $X,Y\in \mathfrak {q}$, which establishes the claim. $\square$

Corollary 17

For $\psi \in C^0([a,b],\mathfrak {q})$ and $\varrho :[a',b']\rightarrow [a,b]$ of class $C^1$, we have

$$\begin{aligned} \textstyle {\textbf {Ad}} ^+_{\psi }[\varrho (t)]={\textbf {Ad}} ^+_{{\dot{\varrho }}\,\cdot \,(\psi \,\circ \,\varrho )}[t]\circ {\textbf {Ad}} _{\psi }^+[\varrho (a')]\qquad \quad \forall \, t\in [a',b']. \end{aligned}$$

Proof

For $Z\in \mathfrak {q}$, define

$$\begin{aligned} \alpha _Z(t):={\textbf {Ad}} ^+_{\psi }[\varrho (t)]({\textbf {Ad}} _\psi ^-[\varrho (a')](Z))\qquad \quad \forall \, t\in [a',b']. \end{aligned}$$

Proposition 6 shows $\alpha _Z\in C^1([a',b'],\mathfrak {q})$, with $\alpha _Z(a')=Z$ and (additionally apply b) in Proposition A.1)

$$\begin{aligned} {\dot{\alpha }}_Z(t)= \varvec{[}{\dot{\varrho }}(t)\cdot (\psi \circ \varrho )(t), \alpha _Z(t) \varvec{]} \qquad \quad \forall \, t\in [a',b']. \end{aligned}$$

The uniqueness statement in Proposition 6 (second identity) yields

$$\begin{aligned} {\textbf {Ad}} ^+_{\psi }[\varrho (t)]({\textbf {Ad}} _\psi ^-[\varrho (a')](Z))=\alpha _Z(t)= {\textbf {Ad}} ^+_{{\dot{\varrho }}\,\cdot \,(\psi \,\circ \,\varrho )}[t](Z)\qquad \quad \forall \, t\in [a,b],\, Z\in \mathfrak {q}. \end{aligned}$$

Given $X\in \mathfrak {q}$, we set $Z:={\textbf {Ad}} _\psi ^+[\varrho (a')](X)$ and obtain from Proposition 6.1) that

$$\begin{aligned} \textstyle {\textbf {Ad}} ^+_{\psi }[\varrho (t)](X)={\textbf {Ad}} ^+_{{\dot{\varrho }}\,\cdot \,(\psi \,\circ \,\varrho )}[t]( {\textbf {Ad}} _{\psi }^+[\varrho (a')](X))\qquad \quad \forall \, t\in [a',b'] \end{aligned}$$

holds, from which the claim is clear. $\square$

Remark 19

Let $\mathcal {L}:\mathfrak {q}\rightarrow \mathfrak {q}$ be linear and continuous, with

$$\begin{aligned} \mathcal {L}(\varvec{[}X, Y \varvec{]})=\varvec{[}\mathcal {L}(X), \mathcal {L}(Y) \varvec{]}\qquad \quad \forall \, X,Y\in \mathfrak {q}. \end{aligned}$$

(102)

It follows by iterated application of (102) and (22) that

$$\begin{aligned} \mathcal {L}\circ {\textbf {Ad}} ^\pm _\psi [t] = {\textbf {Ad}} ^\pm _{\mathcal {L}\,\circ \, \psi }[t]\circ \mathcal {L}\qquad \quad \forall \, a<b,\, \psi \in C^0([a,b],\mathfrak {q}),\,t\in [a,b] \end{aligned}$$

(103)

holds. This equality alternatively follows from Proposition 6:

Proof of Equation (103)

Given $\psi \in C^0([a,b],\mathfrak {q})$ and $Z\in \mathfrak {q}$, define $\alpha :=\mathcal {L}\circ {\textbf {Ad}} ^+_{\psi }[Z]\in C^1([a,b],\mathfrak {q})$. Then, we have $\alpha (a)=\mathcal {L}(Z)$, with

$$\begin{aligned} {\dot{\alpha }}=\mathcal {L}(\varvec{[}\psi , {\textbf {Ad}} ^+_\psi [Z] \varvec{]})=\varvec{[}\mathcal {L}\circ \psi , \alpha \varvec{]}. \end{aligned}$$

Proposition 6.2) shows $\alpha ={\textbf {Ad}} ^+_{\mathcal {L}\,\circ \, \psi }[\mathcal {L}(Z)]$, which establishes the “$+$”-case in (103). Together with Proposition 6.1), we obtain

$$\begin{aligned} \mathcal {L}\circ {\textbf {Ad}} ^-_{\psi }[t](Z)&=\mathcal {L}\circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\psi )}|_{[a,t]}}(Z)\\&={\textbf {Ad}} ^+_{\mathcal {L}\, \circ \,{\mathfrak {inv}(\psi |_{[a,t]})}}(\mathcal {L}(Z))\\&={\textbf {Ad}} ^+_{{\mathfrak {inv}(\mathcal {L}\,\circ \,\psi |_{[a,t]})}}(\mathcal {L}(Z))\\&={\textbf {Ad}} ^-_{{\mathfrak {inv}({\mathfrak {inv}(\mathcal {L}\,\circ \,\psi |_{[a,t]})})}}(\mathcal {L}(Z))\\&={\textbf {Ad}} ^-_{\mathcal {L}\,\circ \,\psi |_{[a,t]}}(\mathcal {L}(Z)) \end{aligned}$$

for $a<t\le b$ and $\psi \in C^0([a,b],\mathfrak {q})$, which establishes the “−”-case in (103). $\square$

For the proof of Proposition 6, we shall need the following facts and definitions:

• For $\phi \equiv \{\phi [p]\}_{0\le p\le n-1}\in \text {CP} ^0([a,b],\mathfrak {q})$, we define ${\mathfrak {inv}(\phi )}\in \text {CP} ^0([a,b],\mathfrak {q})$ by

  $$\begin{aligned} {\mathfrak {inv}(\phi )}:= \{{\mathfrak {inv}(\phi [(n-1)-p])}\}_{0\le p\le n-1}. \end{aligned}$$

• Let $\phi \equiv \{\phi [p]\}_{0\le p\le n-1}\in \text {CP} ^0([a,b],\mathfrak {q})$ ($a<b$, $n\ge 1$) and $\phi '\equiv \{\phi '[q]\}_{0\le q\le n'-1}\in \text {CP} ^0([a',b'],\mathfrak {q})$ ($a'<b'$, $n'\ge 1$) be given. We set

  $$\begin{aligned} \psi [\ell ]:= {\left\{ \begin{array}{ll} \phi [\ell ] &{}\text {for}\quad 0\le \ell \le n-1\\ \phi '[\ell -n] &{}\text {for}\quad n\le \ell \le n+n'-1, \end{array}\right. } \end{aligned}$$

  and define $\phi \triangleright \phi '\in \text {CP} ^0([a,b+(b'-a')],\mathfrak {q})$ by

  $$\begin{aligned} \phi \triangleright \phi ':= \{\psi [\ell ]\}_{0\le \ell \le n+n'-1}. \end{aligned}$$

  Lemma 18 together with Remark 18 shows

  $$\begin{aligned} {\textbf {Ad}} ^+_{\phi \,\triangleright \, \phi '}\textstyle ={\textbf {Ad}} ^+_{\phi \,\triangleright \, \phi '}|_{[b,b + (b'-a')]}\circ {\textbf {Ad}} ^+_{\phi \,\triangleright \, \phi '}|_{[a,b]} = {\textbf {Ad}} ^+_{\phi '}\circ {\textbf {Ad}} ^+_{\phi }. \end{aligned}$$

  (104)

• We have ${\mathfrak {inv}({\mathfrak {inv}(\psi )})}=\psi$ for each $\psi \in C^0([a,b],\mathfrak {q})$.

• For $a<b$ and $Z\in \mathfrak {q}$, we have (observe ${\mathfrak {inv}(\mathcal {C}_Z|_{[a,b]})}=-\mathcal {C}_Z|_{[a,b]}$)

  $$\begin{aligned} \textstyle {\textbf {Ad}} ^+_{\mathcal {C}_Z|_{[a,b]}}[b]\textstyle =\sum _{\ell =0}^\infty \, \frac{(b-a)^\ell }{\ell !}\cdot \text {ad} _{Z}^\ell ={\textbf {Ad}} ^-_{{\mathfrak {inv}(\mathcal {C}_Z|_{[a,b]})}}[b]. \end{aligned}$$

  (105)

  Corollary 15 (second step) yields

  $$\begin{aligned} {\textbf {Ad}} ^+_{\mathcal {C}_Z|_{[a,b]}}[b]\circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\mathcal {C}_Z|_{[a,b]})}}[b]{\mathop {=}\limits ^{(105)}}{\textbf {Ad}} ^-_{{\mathfrak {inv}(\mathcal {C}_Z|_{[a,b]})}}[b]\circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\mathcal {C}_Z|_{[a,b]})}}[b]=\text {id} _\mathfrak {q}. \end{aligned}$$

  (106)

We obtain the following statement.

Lemma 20

For $\psi \in C^0([a,b],\mathfrak {q})$, we have ${\textbf {Ad}} ^+_\psi \circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\psi )}}=\text {id} _\mathfrak {q}$.

Proof

For $n\ge 1$ and $p=0,\dots ,n$, we define $t_{n,p}:=a+ \frac{p}{n}\cdot (b-a)$ and $Z_{n,p}:=\psi (t_{n,p})$. We set

$$\begin{aligned} \chi _{n,p}:=\mathcal {C}_{Z_{n,p}}|_{[t_{n,p},t_{n,p+1}]} \qquad \quad \forall \, n\ge 1,\, p=0,\dots ,n-1. \end{aligned}$$

For $n\ge 1$, we define $\psi ^\pm _n\equiv \{\psi _n^\pm [p]\}_{0\le p\le n-1}\in \text {CP} ^0([a,b],\mathfrak {q})$ by

$$\begin{aligned} \textstyle \psi _n^+[p] := \chi _{n,p}\qquad \text {as well as}\qquad \psi _n^-[p] := {\mathfrak {inv}(\chi _{n,(n-1) -p})}\qquad \text {for}\qquad p=0,\dots ,n-1. \end{aligned}$$

(hence $\psi _n^-={\mathfrak {inv}(\psi _n^+)}$. Obviously, we have $\{\psi ^+_n\}_{n\ge 1}\rightarrow \psi$ as well as $\{\psi ^-_n\}_{n\ge 1}\rightarrow {\mathfrak {inv}(\psi )}$ w.r.t. the $C^0$-topology. Moreover, for $n\ge 1$, we have by Lemma 18 and (106) that

$$\begin{aligned} {\textbf {Ad}} ^+_{\psi ^+_n}\circ {\textbf {Ad}} ^+_{\psi ^-_n}={\textbf {Ad}} ^+_{\chi _{n,0}}\circ \dots \circ ({\textbf {Ad}} ^+_{\chi _{n,n-1}} \circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\chi _{n,n-1})}} )\circ {\dots }\circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\chi _{n,0})}}=\text {id} _\mathfrak {q}\end{aligned}$$

(107)

holds. We define $\phi :={\mathfrak {inv}(\psi )}\triangleright \psi$, as well as $\phi _n:=\psi ^-_n\triangleright \psi ^+_n$ for each $n\ge 1$. Then, $\{\phi _n\}_{n\ge 1}\rightarrow \phi$ converges w.r.t. the $C^0$-topology; and we obtain from Lemma 19 (second step) that

$$\begin{aligned} \textstyle ({\textbf {Ad}} ^+_{\psi }\circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\psi )}})(X){\mathop {=}\limits ^{(104)}} {\textbf {Ad}} ^+_{\phi }(X)=\lim _n {\textbf {Ad}} ^+_{\phi _n}(X){\mathop {=}\limits ^{(104)}} \lim _n ({\textbf {Ad}} ^+_{\psi _n^+}\circ {\textbf {Ad}} ^+_{\psi _n^-})(X){\mathop {=}\limits ^{(107)}}X \end{aligned}$$

holds for each $X\in \mathfrak {q}$, which proves the claim. $\square$

We are ready for the proof of Proposition 6.

Proof of Proposition 6

• (1) The claim is clear for $t=a$. Then, it suffices to prove the statement for $t=b$, as then we can apply it to the restriction $\psi |_{[a,t]}$ for each $a<t<b$. By Lemma 20 (first step), and Corollary 15 (third step), we have

  $$\begin{aligned} {\textbf {Ad}} _\psi ^-[b]&={\textbf {Ad}} _\psi ^-[b]\circ ({\textbf {Ad}} _\psi ^+[b]\circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\psi )}}[b])\\&=({\textbf {Ad}} _\psi ^-[b]\circ {\textbf {Ad}} _\psi ^+[b])\circ {\textbf {Ad}} ^+_{{\mathfrak {inv}(\psi )}}[b] ={\textbf {Ad}} ^+_{{\mathfrak {inv}(\psi )}}[b]. \end{aligned}$$

  Applying ${\textbf {Ad}} _\psi ^+$ from the left, Lemma 20 yields (Corollary 15 in the second equality)

  $$\begin{aligned} {\textbf {Ad}} _\psi ^+[b]\circ {\textbf {Ad}} _\psi ^-[b]=\text {id} _\mathfrak {q}={\textbf {Ad}} _\psi ^-[b]\circ {\textbf {Ad}} _\psi ^+[b]. \end{aligned}$$

  Together, we have shown the following properties:

• ${\textbf {Ad}} ^\pm _{\psi }[b]\in \textsf {Aut} (\mathfrak {q})$   with   ${\textbf {Ad}} ^\pm _{\psi }[b]^{-1}={\textbf {Ad}} ^\mp _{\psi }[b]$,

• ${\textbf {Ad}} _\psi ^-[b]={\textbf {Ad}} ^+_{{\mathfrak {inv}(\psi )}}[b]$.

Applying the third property to $\widetilde{\psi }:={\mathfrak {inv}(\psi )}$ (second step), we obtain from ${\mathfrak {inv}({\mathfrak {inv}(\psi )})}=\psi$ (third step) that

$$\begin{aligned} {\textbf {Ad}} ^-_{{\mathfrak {inv}(\psi )}}[b]={\textbf {Ad}} ^-_{\widetilde{\psi }}[b]={\textbf {Ad}} _{{\mathfrak {inv}(\widetilde{\psi })}}^+[b]={\textbf {Ad}} _\psi ^+[b] \end{aligned}$$

holds, which completes the proof.

• (2) By Lemma 17, for $\alpha \equiv {\textbf {Ad}} ^+_\psi [X]$, we have $\alpha \in C^1([a,b],\mathfrak {q})$ with $\alpha (a)=X$ and ${\dot{\alpha }}=\varvec{[}\psi , \alpha \varvec{]}$. This shows the solution property. For uniqueness, let $\alpha \in C^1([a,b],\mathfrak {q})$ be given, such that $\alpha (a)=X$ and ${\dot{\alpha }}=\varvec{[}\psi , \alpha \varvec{]}$ holds. By Lemma 17, we have $\beta :=\varvec{\Lambda }_\psi ^-(\cdot ,\alpha (\cdot ))\in C^1([a,b],\mathfrak {q})$ with (use Part e) of Proposition A.1 on the left side, as well as (20) on the right side)

  $$\begin{aligned}&{\dot{\beta }}=-\varvec{\Lambda }_\psi ^-(\cdot ,\varvec{[}\psi (\cdot ), \alpha (\cdot ) \varvec{]})+ \varvec{\Lambda }_\psi ^-(\cdot ,\varvec{[}\psi (\cdot ), \alpha (\cdot ) \varvec{]})=0\\&\quad \textstyle \Longrightarrow \qquad \,\,\, \beta =\beta (a)+ \int _a^\bullet {\dot{\beta }}(s) \,\text {d} s=X. \end{aligned}$$

  Applying Part 1), we obtain

  $$\begin{aligned} {\textbf {Ad}} ^+_\psi [X](t)={\textbf {Ad}} _\psi ^+[t](\beta (t))=({\textbf {Ad}} _\psi ^+[t]\circ {\textbf {Ad}} _\psi ^-[t])(\alpha (t))=\alpha (t)\qquad \quad \forall \, t\in [a,b], \end{aligned}$$

  which proves the claim.$\square$

5.1.3 A group structure

For $\phi ,\psi \in C^0([a,b],\mathfrak {q})$ with $a<b$, we define

$$\begin{aligned} C^0([a,b],\mathfrak {q})\ni \psi ^{-1}:= - \varvec{\Lambda }_\psi ^-(\cdot ,\psi (\cdot ))&:[a,b]\ni t \mapsto - {\textbf {Ad}} _\psi ^-[t](\psi (t))\in \mathfrak {q}\\ C^0([a,b],\mathfrak {q})\ni \phi \star \psi := \phi (\cdot ) + \varvec{\Lambda }^+_\phi (\cdot ,\psi (\cdot ))&:[a,b]\ni t\mapsto \phi (t) + {\textbf {Ad}} ^+_\phi [t](\psi (t))\in \mathfrak {q}. \end{aligned}$$

We observe the following.

Lemma 21

Let $\phi ,\psi \in C^0([a,b],\mathfrak {q})$ be fixed. Then, the following assertions hold:

• (1) We have ${\textbf {Ad}} ^+_{\psi ^{-1}}[t]={\textbf {Ad}} _\psi ^-[t]$  for each $t\in [a,b]$.

• (2) We have ${\textbf {Ad}} ^+_{\phi \,\star \,\psi }[t]={\textbf {Ad}} ^+_\phi [t]\circ {\textbf {Ad}} ^+_\psi [t]$  for each $t\in [a,b]$.

Proof

• (1) Let $X\in \mathfrak {q}$ be given, and set $\alpha _X:= {\textbf {Ad}} _\psi ^-[X]\in C^1([a,b],\mathfrak {q})$. We have $\alpha _X(a)=X$; and Lemma 17 shows $\alpha _X:= {\textbf {Ad}} _\psi ^-[X]\in C^1([a,b],\mathfrak {q})$ with (for the second step use Corollary 16)

  $$\begin{aligned} {\dot{\alpha }}_X(t)\textstyle = -{\textbf {Ad}} _\psi ^-[t](\varvec{[}\psi (t), X \varvec{]})=\varvec{[}-{\textbf {Ad}} _\psi ^-[t](\psi (t)), {\textbf {Ad}} _\psi ^-[X](t) \varvec{]}=\varvec{[}\psi ^{-1}(t), \alpha _X(t) \varvec{]} \end{aligned}$$

  for each $t\in [a,b]$. Then, Proposition 6.2) gives $\alpha _X={\textbf {Ad}} ^+_{\psi ^{-1}}[X]$.

• (2) For $X\in \mathfrak {q}$, we define

  $$\begin{aligned} \alpha _X:[a,b]\ni t\mapsto ({\textbf {Ad}} ^+_\phi [t]\circ {\textbf {Ad}} ^+_\psi [t])(X) = \varvec{\Lambda }^+_\phi (t,{\textbf {Ad}} ^+_\psi [X](t))\in \mathfrak {q}. \end{aligned}$$

  We have $\alpha _X(a)=X$; and Lemma 17 shows $\alpha _X\in C^1([a,b],\mathfrak {q})$ with (for the first step additionally apply the parts b) and e) of Proposition A.1; and for the second step apply Corollary 16)

  $$\begin{aligned} {\dot{\alpha }}_X(t)&\textstyle = \varvec{[}\phi (t), \alpha _X(t) \varvec{]}+ {\textbf {Ad}} ^+_\phi [t](\varvec{[}\psi (t), {\textbf {Ad}} ^+_\psi [t](X) \varvec{]})\\&\textstyle = \varvec{[}\phi (t), \alpha _X(t) \varvec{]} + \varvec{[}{\textbf {Ad}} ^+_{\phi }[t](\psi (t)), \alpha _X(t) \varvec{]} \\&\textstyle = \varvec{[}(\phi \star \psi )(t), \alpha _X(t) \varvec{]}. \end{aligned}$$

  Proposition 6.2) yields $\alpha _X={\textbf {Ad}} ^+_{\phi \,\star \,\psi }[X]$, which proves the claim. $\square$

Lemma 22

Given $\phi ,\psi ,\chi \in C^0([a,b],\mathfrak {q})$, then we have

$$\begin{aligned}&0\star \psi =\psi =\psi \star 0,\\&\psi ^{-1}\star \psi =0=\psi \star \psi ^{-1},\\&(\phi \star \psi )\star \chi =\phi \star (\psi \star \chi ). \end{aligned}$$

Proof

We have ${\textbf {Ad}} ^+_0[t]=\text {id} _\mathfrak {q}$ and ${\textbf {Ad}} ^+_\psi [t](0)=0$ for each $t\in [a,b]$; hence,

$$\begin{aligned} (0\star \psi )(t)= 0 + {\textbf {Ad}} ^+_0[t](\psi (t))= \psi (t)= \psi (t) + {\textbf {Ad}} ^+_\psi [t](0)= (\psi \star 0)(t)\qquad \quad \forall \, t\in [a,b]. \end{aligned}$$

Next, Proposition 6.1) yields

$$\begin{aligned} (\psi \star \psi ^{-1})(t)&= \psi (t) - ({\textbf {Ad}} ^+_\psi [t]\circ {\textbf {Ad}} ^{-}_\psi [t])(\psi (t))= 0\qquad \quad \forall \, t\in [a,b]. \end{aligned}$$

Moreover, Lemma 21.1) shows

$$\begin{aligned} (\psi ^{-1}\star \psi )(t)&= \psi ^{-1}(t) + {\textbf {Ad}} ^+_{\psi ^{-1}}[t](\psi (t))= \psi ^{-1}(t) + {\textbf {Ad}} ^-_{\psi }[t](\psi (t))\\&= \psi ^{-1}(t)-\psi ^{-1}(t)=0 \end{aligned}$$

for each $t\in [a,b]$. Finally, we obtain from Lemma 21.2) that

$$\begin{aligned} ((\phi \star \psi )\star \chi )(t)&= (\phi \star \psi )(t) + {\textbf {Ad}} ^+_{\phi \,\star \, \psi }[t](\chi (t))\\&=\phi (t) + {\textbf {Ad}} ^+_\phi [t](\psi (t)) + ({\textbf {Ad}} ^+_{\phi }[t]\circ {\textbf {Ad}} ^+_\psi [t])(\chi (t))\\&= \phi (t) + {\textbf {Ad}} ^+_\phi [t]((\psi \star \chi )(t))\\&= (\phi \star (\psi \star \chi ))(t) \end{aligned}$$

holds for each $t\in [a,b]$, which establishes the proof. $\square$

Corollary 18

For $\phi ,\psi \in C^0([a,b],\mathfrak {q})$ and $t\in [a,b]$, we have

$$\begin{aligned} {\textbf {Ad}} ^+_{\phi +\psi }[t]&={\textbf {Ad}} ^+_{\phi }[t]\circ {\textbf {Ad}} ^+_{{\textbf {Ad}} ^-_{\phi }[\cdot ](\psi (\cdot ))}[t]. \end{aligned}$$

Proof

By Lemma 22, we have

$$\begin{aligned} (\phi ^{-1}\star (\phi + \psi ))(t)&= \phi ^{-1}(t) + {\textbf {Ad}} ^+_{\phi ^{-1}}[t](\phi (t)) + {\textbf {Ad}} ^+_{\phi ^{-1}}[t](\psi (t))\\&=(\phi ^{-1}\star \phi )(t) + {\textbf {Ad}} ^+_{\phi ^{-1}}[t](\phi (t))\\&={\textbf {Ad}} ^+_{\phi ^{-1}}[t](\phi (t)) \end{aligned}$$

for each $t\in [a,b]$. Lemma 21.2) (first step) and Lemma 21.1) (third step) yield

$$\begin{aligned} {\textbf {Ad}} ^+_{\phi ^{-1}}[t]\circ {\textbf {Ad}} ^+_{\phi + \psi }[t]={\textbf {Ad}} ^+_{\phi ^{-1}\,\star \, (\phi + \psi )}[t]={\textbf {Ad}} ^+_{{\textbf {Ad}} ^+_{\phi ^{-1}}[\cdot ](\psi (\cdot ))}[t]={\textbf {Ad}} ^+_{{\textbf {Ad}} ^-_{\phi }[\cdot ](\psi (\cdot ))}[t] \end{aligned}$$

for each $t\in [a,b]$. Applying ${\textbf {Ad}} ^+_\phi [t]$ to both hand sides, the claim follows from Lemma 21.2), Lemma 22, as well as ${\textbf {Ad}} ^+_0[t]=\text {id} _\mathfrak {q}$ for each $t\in [a,b]$. $\square$

5.1.4 An integral transformation

In analogy to (49), for $a<b$ and $\phi \in C^0([a,b],\mathfrak {q})$, we set

$$\begin{aligned} \textstyle \mathfrak {T}(\phi ):[0,1]\ni t\mapsto \int _a^b {\textbf {Ad}} ^+_{t\cdot \phi |_{[s,b]}}(\phi (s)) \,\text {d} s\in \mathfrak {q}\end{aligned}$$

(108)

as well as $\mathfrak {T}(\phi |_{[a,a]}):[0,1]\ni t \mapsto 0\in \mathfrak {q}$. Observe that (108) is defined as the integrand is continuous; because, by Lemma 18 and Proposition 6 we have

$$\begin{aligned} {\textbf {Ad}} ^+_{t\cdot \phi |_{[s,b]}}(\phi (s))={\textbf {Ad}} ^+_{t\cdot \phi }[b]({\textbf {Ad}} ^-_{t\cdot \phi }[s](\phi (s)))\qquad \quad \forall \, s\in [a,b]. \end{aligned}$$

Continuity of the integrand also implies the following statement.

Remark 20

Let $a<b$ and $\psi \in C^0([a,b],\mathfrak {q})$. Let $\varrho :[a',b']\rightarrow [a,b]$ ($a'<b'$) be of class $C^1$, with ${\dot{\varrho }}|_{(a',b')}>0$ as well as $\varrho (a')=a$ and $\varrho (b')=b$. Then, (97) in Remark 17 (second step) shows

$$\begin{aligned} \textstyle \mathfrak {T}({\dot{\varrho }}\cdot (\psi \circ \varrho ))(t)&\textstyle =\int _{a'}^{b'} {\textbf {Ad}} ^+_{ ({\dot{\varrho }}\,\cdot \, ((t\cdot \psi )\,\circ \,\varrho ))|_{[s,b']}}({\dot{\varrho }}(s)\cdot (\psi \circ \varrho )(s)) \,\text {d} s\\&\textstyle =\int _{a'}^{b'} {\dot{\varrho }}(s)\cdot {\textbf {Ad}} ^+_{t\cdot \psi |_{[\varrho (s),b]}}( \psi (\varrho (s))) \,\text {d} s\\&\textstyle =\int _{a}^{b} {\textbf {Ad}} ^+_{t\cdot \psi |_{[s,b]}}( \psi (s)) \,\text {d} s\\&=\mathfrak {T}(\psi ) \end{aligned}$$

where we have applied the substitution formula (21) in the third step. $\square$

In this section, we investigate the elementary properties of this map, and provide some statements that we shall need in Sect. 5.2. We start with some technical properties:

Lemma 23

Let $a<b$, $\psi \in C^0([a,b],\mathfrak {q})$, and $X\in \mathfrak {q}$ be given. Then, $\eta _{\psi ,X}:\mathbb {R}\ni t \mapsto {\textbf {Ad}} ^+_{t\cdot \psi }[b](X)\in \mathfrak {q}$ is smooth, with

$$\begin{aligned} \textstyle \eta _{\psi ,X}^{(\ell )}:\mathbb {R}\ni t\mapsto \sum _{p=\ell }^\infty \frac{p!}{(p - \ell )!}\cdot t^{p-\ell }\cdot \varvec{\lambda }^+_{p,\psi }[b](X)\qquad \quad \forall \, n\in \mathbb {N}. \end{aligned}$$

Proof

It follows from the definitions that $\eta _{\psi ,X}(t)=\sum _{p=0}^\infty t^p\cdot \varvec{\lambda }^+_{p,\psi }[b](X)$ holds for each $t\in \mathbb {R}$. Now, for $\textsf {v} \le \textsf {w}$ as in (95), we have

$$\begin{aligned} \textstyle \textsf {v} (\varvec{\lambda }^+_{p,\psi }[b](X))\le \textsf {w} (X)\cdot \frac{(b-a)^p}{p!}\cdot \textsf {w} _\infty (\psi )^p\qquad \quad \forall \, p\in \mathbb {N}. \end{aligned}$$

Consequently, $\textstyle \gamma [\textsf {q} ]:\mathbb {R}\ni t\mapsto \sum _{p=0}^\infty t^p\cdot \textsf {q} (\varvec{\lambda }^+_{p,\psi }[b](X))\in [0,\infty )$ is defined for each $\textsf {q} \in \textsf {Sem}(\mathfrak {q})$, so that the claim is clear from Corollary 2. $\square$

Next, let $a<b$ and $\psi \in C^0([a,b],\mathfrak {q})$ be given:

• For $a\le s< z\le b$, we define $\mathcal {Y}_0(\psi ,s,z):=\psi (s)$, as well as

  $$\begin{aligned} \textstyle \mathcal {Y}_p(\psi ,s,z):= \int _s^{z}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{p-1}}\text {d} s_p \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{p})})(\psi (s))\qquad \quad \forall \, p \ge 1. \end{aligned}$$

  For each $t\in [0,1]$ and $a<z\le b$, we have

  $$\begin{aligned} \mathfrak {T}(\psi |_{[a,z]})(t)= \textstyle \int _a^z \sum _{p=0}^\infty \varvec{\lambda }^+_{p,\, t\cdot \psi |_{[s,z]}}(\psi (s))\, \text {d} s=\textstyle \int _a^z \sum _{p=0}^\infty t^p \cdot \mathcal {Y}_p(\psi ,s,z)\, \text {d} s. \end{aligned}$$

  (109)

• We define

  $$\begin{aligned} \textstyle \mathcal {T}_p(\psi ,z,t):= \textstyle t^p\cdot \int _a^z \mathcal {Y}_p(\psi ,s,z)\, \text {d} s \qquad \quad \forall \, p\in \mathbb {N},\,\, t\in [0,1],\,\, z\in [a,b]. \end{aligned}$$

  (110)

We have the following statement.

Lemma 24

For each $a<b$, we have $\mathfrak {T}:C^0([a,b],\mathfrak {q})\rightarrow C^\infty ([0,1],\mathfrak {q})$. Moreover,

$$\begin{aligned} \textstyle \mathfrak {T}(\psi |_{[a,z]})= \sum _{p=0}^\infty \mathcal {T}_p(\psi ,z,\cdot ) \end{aligned}$$

converges w.r.t. the $C^\infty$-topology for each $\psi \in C^0([a,b],\mathfrak {q})$ and $a<z\le b$; and we have

$$\begin{aligned} \sup \{a\le z\le b\,|\, \textsf {q} _\infty ^\text {s} (\mathfrak {T}(\psi |_{[a,z]}))\}<\infty \qquad \quad \forall \, \psi \in C^0([a,b],\mathfrak {q}),\, \textsf {q} \in \textsf {Sem}(\mathfrak {q}),\,\text {s} \in \mathbb {N}. \end{aligned}$$

(111)

Proof

Let $\psi \in C^0([a,b],\mathfrak {q})$ be given, and let $\textsf {v} \le \textsf {w}$ be as in (95). Then, we have

$$\begin{aligned} \textstyle \textsf {v} _\infty ( \mathcal {Y}_p(\psi ,\cdot ,z))\le \textsf {w} _\infty (\psi )\cdot \frac{(b-a)^p}{p!}\cdot \textsf {w} _\infty (\psi )^p\qquad \quad \forall \, p\in \mathbb {N},\, a< z\le b. \end{aligned}$$

It follows that $\textstyle \{\sum _{p=0}^\ell t^p\cdot \mathcal {Y}_p(\psi ,\cdot ,z)\}_{\ell \in \mathbb {N}}\rightarrow \sum _{p=0}^\infty t^p\cdot \mathcal {Y}_p(\psi ,\cdot ,z)$ converges uniformly on [a, b], for each fixed $t\in [0,1]$ and $a< z\le b$. Since the Riemann integral is $C^0$-continuous, we obtain

Now, for each $p\in \mathbb {N}$ and $a<z\le b$, we have

$$\begin{aligned} \textstyle \textsf {v} (X_p(z))\le (b-a)\cdot \textsf {w} _\infty (\psi )\cdot \frac{(b-a)^p}{p!}\cdot \textsf {w} _\infty (\psi )^p. \end{aligned}$$

Consequently, $\textstyle \gamma [\textsf {q} ]:\mathbb {R}\ni t\mapsto \sum _{p=0}^\infty t^p\cdot \textsf {q} (X_p(z))\in [0,\infty )$ is defined for each $\textsf {q} \in \textsf {Sem}(\mathfrak {q})$ and $a<z\le b$, so that the claim is clear from Corollary 2. $\square$

Lemma 25

For $\psi \in C^0([a,b],\mathfrak {q})$, we have

$$\begin{aligned} {\textbf {Ad}} ^+_{t\cdot \psi }[b]={\textbf {Ad}} ^+_{\mathfrak {T}(\psi )}[t]\qquad \quad \forall \,t\in [0,1]. \end{aligned}$$

Proof

Let $X\in \mathfrak {q}$ be given. Lemma 23 yields the following:

• We have $C^\infty (\mathbb {R},\mathfrak {q})\ni \eta _{\psi ,X}:\mathbb {R}\ni t \mapsto {\textbf {Ad}} ^+_{t\cdot \psi }[b](X)\in \mathfrak {q}$.

• By Lemma 17, we have

  $$\begin{aligned} \chi _t:={\textbf {Ad}} ^-_{t\cdot \psi }[\cdot ](\psi (\cdot ))\in C^0([a,b],\mathfrak {q})\qquad \quad \forall \, t\in \mathbb {R}. \end{aligned}$$

  Lemma 23 shows that $\eta _{\chi _t,X}:\mathbb {R}\ni h\mapsto {\textbf {Ad}} ^+_{h\cdot \chi _t}[b](X)\in \mathfrak {q}$ is smooth for each $t\in \mathbb {R}$, with

  $$\begin{aligned} {\dot{\eta }}_{\chi _t,X}(0)\textstyle = \varvec{\lambda }^+_{1,\chi _t}[b](X){\mathop {=}\limits ^{(22)}}\varvec{[}\int _a^b \chi _t(s)\,\text {d} s, X \varvec{]}. \end{aligned}$$

  (112)

By Lemma 18 (second step) and Proposition 6.1) (third step), we have

$$\begin{aligned}&{\textbf {Ad}} ^+_{t\cdot \psi }[b]\circ {\textbf {Ad}} ^-_{t\cdot \psi }[s]=({\textbf {Ad}} ^+_{t\cdot \psi |_{[s,b]}}\circ {\textbf {Ad}} ^+_{t\cdot \psi |_{[a,s]}})\circ {\textbf {Ad}} ^-_{t\cdot \psi |_{[a,s]}}={\textbf {Ad}} ^+_{t\cdot \psi |_{[s,b]}}\\&\qquad \quad \forall \, t\in \mathbb {R},\,\, a\le s\le b. \end{aligned}$$

This yields (second step), together with Corollary 16 and (112) (first step) that

$$\begin{aligned} \begin{aligned} {\textbf {Ad}} ^+_{t\cdot \psi }[b]({\dot{\eta }}_{\chi _t,X}(0))&\textstyle = \varvec{[}\int _a^b{\textbf {Ad}} ^+_{t\cdot \psi }[b](\chi _t(s))\,\text {d} s, {\textbf {Ad}} ^+_{t\cdot \psi }[b](X) \varvec{]}) \\&\textstyle = \varvec{[}\int _a^b {\textbf {Ad}} ^+_{t\cdot \psi |_{[s,b]}}[b](\psi (s))\,\text {d} s, {\textbf {Ad}} ^+_{t\cdot \psi }[b](X) \varvec{]}) \\&\textstyle = \varvec{[}\mathfrak {T}(\psi )(t), \eta _{\psi ,X}(t) \varvec{]}) \end{aligned} \end{aligned}$$

(113)

holds for each $t\in [0,1]$. Now, Corollary 18 (second step) shows

$$\begin{aligned} \eta _{\psi ,X}(t+h)\textstyle&={\textbf {Ad}} ^+_{t\cdot \psi + h\cdot \psi }[b](X)\\&\textstyle ={\textbf {Ad}} ^+_{t\cdot \psi }[b]\big ( {\textbf {Ad}} ^+_{h\cdot {\textbf {Ad}} ^-_{t\cdot \psi }[\cdot ](\psi (\cdot ))}[b](X)\big )\\&={\textbf {Ad}} ^+_{t\cdot \psi }[b](\eta _{\chi _t,X}(h)) \end{aligned}$$

for $t,h\in \mathbb {R}$. Hence, (113), together with Lemma 17 and Part b) of Proposition A.1 yields

$$\begin{aligned} {\dot{\eta }}_{\psi ,X}(t)= \varvec{[}\mathfrak {T}(\psi )(t), \eta _{\psi ,X} \varvec{]})\qquad \quad \forall \, t\in [0,1], \end{aligned}$$

with $\eta _{\psi ,X}(0)=X$. Then, Proposition 6.2) ( second step) shows

$$\begin{aligned} {\textbf {Ad}} ^+_{t\cdot \psi }[b](X)=\eta _{\psi ,X}(t)={\textbf {Ad}} ^+_{\mathfrak {T}(\psi )}[t](X)\qquad \quad \forall \, t\in [0,1]. \end{aligned}$$

Since $X\in \mathfrak {q}$ was arbitrary, the claim follows. $\square$

We obtain the following statement.

Lemma 26

Let $\psi \in C^0([a,b+\varepsilon ],\mathfrak {q})$ with $a< b$ and $\varepsilon >0$ be given. Then, we have

$$\begin{aligned} \mathfrak {T}(\psi |_{[a,z+h]})=\mathfrak {T}(\psi |_{[z,z+h]})\star \mathfrak {T}(\psi |_{[a,z]})\qquad \quad \forall \, a< z\le b,\,\, 0< h\le \varepsilon . \end{aligned}$$

Proof

Let $a< z\le b$ and $0< h\le \varepsilon$ be fixed. For $t\in [0,1]$, we have

Now, Lemma 18 and (22) (first step), as well as Lemma 25 (third step) yield

$$\begin{aligned} \alpha (t)&\textstyle = {\textbf {Ad}} ^+_{\, t\cdot \psi |_{[z,z+h]}}[\int _a^{z} {\textbf {Ad}} ^+_{\, t\cdot \psi |_{[s,z]}}[\psi (s)] \,\text {d} s]\\&\textstyle = {\textbf {Ad}} ^+_{\, t\cdot \psi |_{[z,z+h]}}[z+h](\mathfrak {T}(\psi |_{[a,z]})(t))\\&\textstyle ={\textbf {Ad}} ^+_{\mathfrak {T}(\psi |_{[z,z+h]})}[t](\mathfrak {T}(\psi |_{[a,z]})(t)) \end{aligned}$$

for each $t\in [0,1]$. We obtain

$$\begin{aligned} \mathfrak {T}(\psi |_{[a,z+h]})(t)&= \mathfrak {T}(\psi |_{[z,z+h]}) + {\textbf {Ad}} ^+_{\mathfrak {T}(\psi |_{[z,z+h]})}[t](\mathfrak {T}(\psi |_{[a,z]})(t))\\&= \big (\mathfrak {T}(\psi |_{[z,z+h]})\star \mathfrak {T}(\psi |_{[a,z]})\big )(t) \end{aligned}$$

for each $t\in [0,1]$, which proves the claim. $\square$

We obtain the following statement (in analogy to Remark 9).

Proposition 7

Let $a<c<b$, $\psi \in C^k([a,b],\mathfrak {q})$, and $\varrho :[a',b']\rightarrow [a,b]$ ($a'<b'$) be of class $C^1$ with ${\dot{\varrho }}|_{(a',b')}>0$, $\varrho (a')=a$, $\varrho (b')=b$. Then, we have

$$\begin{aligned} \mathfrak {T}(\psi )^{-1}&=\mathfrak {T}({\mathfrak {inv}(\psi )}) \end{aligned}$$

(114)

$$\begin{aligned} \mathfrak {T}(\psi )&=\mathfrak {T}(\psi |_{[c,b]})\star \mathfrak {T}(\psi |_{[a,c]}) \end{aligned}$$

(115)

$$\begin{aligned} \mathfrak {T}(\psi )&=\mathfrak {T}({\dot{\varrho }}\cdot (\psi \circ \varrho )). \end{aligned}$$

(116)

Proof

Lemma 26 shows (115), and Remark 20 shows (116). To prove (114), let $\varrho _1:[a,b]\rightarrow [a,b]$ with $\varrho _1(a)=a$, $\varrho _1(b)=b$, as well as $\varrho _2:[b,2b+a]\rightarrow [a,b]$ with $\varrho _2(b)=a$, $\varrho _2(2b+a)=b$ both be of class $C^1$, such that ${\dot{\varrho }}_1|_{(a,b)}>0$ as well as ${\dot{\varrho }}_2|_{(b,2b+a)}>0$ holds with

$$\begin{aligned} {\dot{\varrho }}_1(a),\,{\dot{\varrho }}_1(b),\, {\dot{\varrho }}_2(b),\,{\dot{\varrho }}_2(2b+a)=0. \end{aligned}$$

We set $\chi _1:={\dot{\varrho }}_1\cdot (\psi \circ \varrho _1)\in C^0([a,b],\mathfrak {q})$ as well as $\chi _2:={\dot{\varrho }}_2\cdot ({\mathfrak {inv}(\psi )}\circ \varrho _2)\in C^0([b,2b+a],\mathfrak {q})$, and define $\phi \in C^0([a,2b+a],\mathfrak {q})$ by

$$\begin{aligned} \phi (t):= {\left\{ \begin{array}{ll} \chi _1(t) &{}\text {for}\quad t\in [a,b]\\ \chi _2(t) &{}\text {for}\quad t\in (b,2b+a] \end{array}\right. } \end{aligned}$$

for each $t\in [a,2b+a]$. Then, we have

$$\begin{aligned} \mathfrak {T}(\psi )\star \mathfrak {T}({\mathfrak {inv}(\psi )}){\mathop {=}\limits ^{(116)}}\mathfrak {T}(\chi _1) \star \mathfrak {T}(\chi _2){\mathop {=}\limits ^{(115)}} \mathfrak {T}(\phi ). \end{aligned}$$

(117)

Let now $t\in [0,1]$ and $X\in \mathfrak {q}$ be given (and observe $t\cdot {\mathfrak {inv}(\psi )}={\mathfrak {inv}(t\cdot \psi )}$).

• For $s\in [a,b]$, we have by Lemma 18 (first step), (97) in Remark 17 (third step), as well as Lemma 18 and Proposition 6.1) (fourth step) that

  $$\begin{aligned} \begin{aligned} {\textbf {Ad}} ^+_{t\cdot \phi |_{[s,2b+a]}}(X)&={\textbf {Ad}} ^+_{t\cdot \phi |_{[b,2b+a]}}\big ({\textbf {Ad}} ^+_{t\cdot \phi |_{[s,b]}}(X)\big )\\&={\textbf {Ad}} ^+_{t\cdot \chi _2}\big ({\textbf {Ad}} ^+_{t\cdot \chi _1|_{[s,b]}}(X)\big )\\&={\textbf {Ad}} ^+_{ {\mathfrak {inv}(t\cdot \psi )}}\big ({\textbf {Ad}} ^+_{t\cdot \psi |_{[\varrho _1(s),b]}}(X)\big )\\&={\textbf {Ad}} ^-_{t\cdot \psi |_{[a,\varrho _1(s)]}}(X). \end{aligned} \end{aligned}$$

  (118)

• For $s\in [b,2b+a]$, we define

  $$\begin{aligned} \chi :[\varrho _2(s),b]\ni z \mapsto {\mathfrak {inv}(t\cdot \psi |_{[a,a+b-\varrho _2(s)]})}(z-(\varrho _2(s)-a)) \in \mathfrak {q}. \end{aligned}$$

  We observe the following:

• (i) We have $\chi ={\mathfrak {inv}(t\cdot \psi )}|_{[\varrho _2(s),b]}$ as

  $$\begin{aligned} \chi (z)&=t\cdot \psi (2a+b-\varrho _2(s)- (z-(\varrho _2(s)-a)))=t\cdot \psi (a+b-z)\\&={\mathfrak {inv}(t\cdot \psi )}|_{[\varrho _2(s),b]}(z) \end{aligned}$$

  holds for each $z\in [\varrho _2(s),b]$.

• (ii) By the second point in Remark 17 (second step), we have

  $$\begin{aligned} {\textbf {Ad}} ^+_{\chi }&={\textbf {Ad}} ^+_{\chi }[b]\\&= {\textbf {Ad}} ^+_{\chi (\cdot -(a-\varrho _2(s)))}[b+(a-\varrho _2(s))]\\&={\textbf {Ad}} ^+_{{\mathfrak {inv}(t\cdot \psi |_{[a,a+b-\varrho _2(s)]})}}[b+(a-\varrho _2(s))]\\&={\textbf {Ad}} ^+_{{\mathfrak {inv}(t\cdot \psi |_{[a,a+b-\varrho _2(s)]})}}. \end{aligned}$$

We obtain from (97) in Remark 17 (second step), Point i) (third step), Point ii) (fourth step), as well as Proposition 6.1) (fifth step) that

$$\begin{aligned} \begin{aligned} {\textbf {Ad}} ^+_{t\cdot \phi |_{[s,2b+a]}}(X)&={\textbf {Ad}} ^+_{t\cdot \chi _2|_{[s,2b+a]}}(X)\\&={\textbf {Ad}} ^+_{{\mathfrak {inv}(t\cdot \psi )}|_{[\varrho _2(s),b]}}(X)\\&={\textbf {Ad}} ^+_{\chi }(X)\\&={\textbf {Ad}} ^+_{{\mathfrak {inv}(t\cdot \psi |_{[a,a+b-\varrho _2(s)]})}}(X)\\&={\textbf {Ad}} ^-_{t\cdot \psi |_{[a,a+b-\varrho _2(s)]}}(X). \end{aligned} \end{aligned}$$

(119)

We obtain from (118) and (119) (third step), (21) (fourth step), as well as the second point in Remark 2 (sixth step) that

$$\begin{aligned} \mathfrak {T}(\phi )(t)&\textstyle = \int _a^{2b + a} {\textbf {Ad}} ^+_{t\cdot \phi |_{[s,2b+a]}}(\phi (s)) \,\text {d} s\\&\textstyle = \int _a^{b} {\textbf {Ad}} ^+_{t\cdot \phi |_{[s,2b+a]}}(\phi (s)) \,\text {d} s + \int _b^{2b + a} {\textbf {Ad}} ^+_{t\cdot \phi |_{[s,2b+a]}}(\phi (s)) \,\text {d} s\\&\textstyle = \int _a^{b} {\dot{\varrho }}_1(s)\cdot {\textbf {Ad}} ^-_{t\cdot \psi |_{[a,\varrho _1(s)]}}(\psi (\varrho _1(s))) \,\text {d} s \\&\quad \textstyle + \int _b^{2b + a}{\dot{\varrho }}_2(s)\cdot {\textbf {Ad}} ^-_{t\cdot \psi |_{[a, a+b-\varrho _2(s)]}}({\mathfrak {inv}(\psi )}(\varrho _2(s))) \,\text {d} s\\&\textstyle = \int _a^{b} {\textbf {Ad}} ^-_{t\cdot \psi |_{[a,s]}}(\psi (s)) \,\text {d} s + \int _a^{b}{\textbf {Ad}} ^-_{t\cdot \psi |_{[a,a+b-s]}}({\mathfrak {inv}(\psi )}(s)) \,\text {d} s\\&\textstyle = \int _a^{b} {\textbf {Ad}} ^-_{t\cdot \psi |_{[a,s]}}(\psi (s)) \,\text {d} s - \int _a^{b}{\textbf {Ad}} ^-_{t\cdot \psi |_{[a,a+b-s]}}(\psi (a+b-s)) \,\text {d} s\\&=0 \end{aligned}$$

holds for each $t\in [0,1]$. Together with (117), this proves (116). $\square$

Moreover, applying Lemma 26 to the situation where $(\mathfrak {q},\varvec{[}\cdot , \cdot \varvec{]})\equiv (\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is an asymptotic estimate and sequentially complete Lie algebra that is inherited by a Lie group G, we obtain the following statement.

Corollary 19

Let G be weakly $C^\infty$-regular, with sequentially complete and asymptotic estimate Lie algebra $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$. Then, for $\psi \in C^0([a,b+\varepsilon ])$ with $a< b$ and $\varepsilon >0$, we have

Proof

Let $a< z\le b$ and $0< h\le \varepsilon$ be fixed. Corollary 8 shows²⁴

We thus obtain from (a) (first step) as well as Lemma 26 (last step) that

holds, which proves the claim. $\square$

5.2 Weak regularity

In this section, we prove Theorem 2. In the following, let thus G be weakly $C^\infty$-regular, with asymptotic estimate and sequentially complete Lie algebra $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$.

Let $a\le b$, $0<\varepsilon <1$, and $\psi \in C^0([a,b+\varepsilon ],\mathfrak {g})$ be given.

• We define $\Phi _\psi :(a-1,b+\varepsilon )\times [0,1]\rightarrow \mathfrak {g}$ by

  $$\begin{aligned} \begin{aligned} \Phi _\psi (z,t) := {\left\{ \begin{array}{ll} (z-a)\cdot \psi (a) &{}\text {for}\quad z\in (a-1,a)\\ \mathfrak {T}(\psi |_{[a,z]})(t) &{}\text {for}\quad z\in [a,b+\varepsilon ) \end{array}\right. } \end{aligned} \end{aligned}$$

  (120)

  for each $t\in [0,1]$. Lemma 24 yields the following:

  • For $z\in (a-1,b+\varepsilon )$, we have $\Phi _\psi (z,\cdot )\in C^\infty ([0,1],\mathfrak {g})$.

  • For $z\in [a,b+\varepsilon )$, we have

    $$\begin{aligned} \textstyle \Phi _\psi (z,t)=\int _a^z {\textbf {Ad}} ^+_{\, t\cdot \psi |_{[s,z]}}[\psi (s)] \,\text {d} s =\sum _{\ell =0}^\infty \mathcal {T}_\ell (\psi ,z,t)\qquad \quad \forall \, t\in [0,1], \end{aligned}$$

    where $\Phi _\psi (z,\cdot ) =\sum _{\ell =0}^\infty \mathcal {T}_\ell (\psi ,z,\cdot )$ converges w.r.t. the $C^\infty$-topology.

  • For $\textsf {p} \in \textsf {Sem}(E)$ and $\text {s} \in \mathbb {N}$, we have

    $$\begin{aligned} \sup \{a-1\le z\le b+\varepsilon \,|\, \textsf {p} _\infty ^\text {s} (\Phi _\psi (z,\cdot ))\}<\infty . \end{aligned}$$

    (121)

• For each $t\in [0,1]$, we define the map

  

  (122)

  We observe the following:

  • We have

    

    (123)

  • We have

    $$\begin{aligned} \Phi _\psi (a,\cdot )=\mathfrak {T}(\psi |_{[a,a]})(\cdot )=0 \qquad \text {hence}\qquad \mu _{t,\psi }(a)=e\qquad \text {for each}\qquad t\in [0,1]. \end{aligned}$$

    (124)

We obtain the following lemma.

Lemma 27

Let $a\le b$, $0<\varepsilon <1$, and $\psi \in C^0([a,b+\varepsilon ],\mathfrak {g})$ be given. We define $\varvec{\psi }\in C^0((a-1,b+\varepsilon ),\mathfrak {g})$ by

$$\begin{aligned} \varvec{\psi }(z):= {\left\{ \begin{array}{ll} \psi (a) &{}\text {for}\quad z\in (a-1,a)\\ \psi (z) &{}\text {for}\quad z\in [a,b+\varepsilon ), \end{array}\right. } \end{aligned}$$

and set

$$\begin{aligned} \mathcal {D}(z,t):=\varvec{\psi }(z)+t\cdot \varvec{[}\varvec{\psi }(z), \Phi _\psi (z,t) \varvec{]}\qquad \quad \forall \, z\in (a-1,b+\varepsilon ),\, t\in [0,1]. \end{aligned}$$

Then, the following assertions hold:

• (1) For $z\in (a-1,b+\varepsilon )$, $\textsf {p} \in \textsf {Sem}(E)$, and $\text {s} \in \mathbb {N}$, we have

  $$\begin{aligned} \textstyle \lim _{ h\rightarrow 0}\frac{1}{|h|}\cdot \textsf {p} ^\text {s} _\infty \big (\Phi _\psi (z+h,\cdot )-\Phi _\psi (z,\cdot )- h\cdot \mathcal {D}(z,\cdot )\big )=0. \end{aligned}$$

  (125)

• (2) The following assertions hold:

  • (a) We have $\partial _1\Phi _\psi (z,\cdot )=\mathcal {D}(z,\cdot )\in C^\infty ([0,1],\mathfrak {g})$ for each $z\in (a-1,b+\varepsilon )$.

  • (b) To each $\textsf {p} \in \textsf {Sem}(E)$, $\text {s} \preceq k$, and $z\in (a-1,b+\varepsilon )$, there exists $L_{\textsf {p} ,\text {s} }\ge 0$ as well as $I_{\textsf {p} ,\text {s} }\subseteq (a-1,b+\varepsilon )$ open with $z\in I_{\textsf {p} ,\text {s} }$, such that

    $$\begin{aligned} \textstyle \textsf {p} ^\text {s} _\infty (\Phi (z+h,\cdot )-\Phi (z,\cdot ))\le |h|\cdot L_{\textsf {p} ,\text {s} }\qquad \quad \forall \, h\in \mathbb {R}_{\ne 0}\,\text { with }\, z+h\in I_{\textsf {p} ,\text {s} }. \end{aligned}$$

  • (c) $\Phi _\psi$ is continuous.

• (3) We have $\mu _{t,\psi }\in C^1((a-1,b+\varepsilon ),G)$ for each $t\in [0,1]$, with

  $$\begin{aligned}&\textstyle {\dot{\mu }}_{t,\psi }(z)\textstyle =\text {d} _e\text {L} _{\mu _{t,\psi }(z)}\big (\int _0^t \text {Ad} _{\mu _{s,\psi }(z)^{-1}}\big (\varvec{\psi }(z)+s\cdot \varvec{[}\varvec{\psi }(z), \Phi _\psi (z,s) \varvec{]}\big )\,\text {d} s\,\big )\\&\qquad \forall \,z\in (a-1,b+\varepsilon ). \end{aligned}$$

  In particular, for each $t\in [0,1]$ and $z\in [a,b+\varepsilon )$, we have

  $$\begin{aligned} \textstyle {\dot{\mu }}_{t,\psi }(z)\textstyle =\text {d} _e\text {L} _{\mu _{t,\psi }(z)}\big (\int _0^t \text {Ad} _{\mu _{s,\psi }(z)^{-1}}(\psi (z)+s\cdot \varvec{[}\psi (z), \mathfrak {T}(\psi |_{[a,z]})(s) \varvec{]})\,\text {d} s\,\big ). \end{aligned}$$

  (126)

Proof

• (1) For $t\in [0,1]$, $z\in (a-1,a]$ and $h\ne 0$ with $z+h\in (a-1,a]$, we have

  $$\begin{aligned} \Phi _\psi (z+h,t)\,&-\Phi _\psi (z,t)- h\cdot \mathcal {D}(z,t)\\&\textstyle =(z+h-a)\cdot \psi (a)- (z-a)\cdot \psi (a) \\&\quad -h\cdot (\psi (a) + t\cdot \varvec{[}\psi (a), (z-a)\cdot \psi (a) \varvec{]})\\&=0. \end{aligned}$$

  Hence, (125) holds for each $z\in (a-1,a)$, and we have

  $$\begin{aligned}&\textstyle \lim _{0> h\rightarrow 0}\frac{1}{|h|}\cdot \textsf {p} ^\text {s} _\infty \big (\Phi _\psi (a+h,\cdot )-\Phi _\psi (a,\cdot )- h\cdot \mathcal {D}(a,\cdot )\big )=0 \nonumber \\&\qquad \quad \forall \, \textsf {p} \in \textsf {Sem}(E),\, \text {s} \in \mathbb {N}. \end{aligned}$$

  (127)

  For the remaining cases, we need the following observations:

• For $t\in [0,1]$ and $z\in [a,b+\varepsilon )$, we define $L_0(z,t):=\psi (z)$ as well as (recall (110))

  $$\begin{aligned}&\textstyle L_\ell (z,t):=\,\textstyle t\cdot \varvec{[}\psi (z), \mathcal {T}_{\ell -1}(\psi ,z,t) \varvec{]}\\&\quad \textstyle =t\cdot \varvec{[}\psi (z), t^{\ell -1}\cdot \int _a^z \mathcal {Y}_{\ell -1}(\psi ,s,z)\, \text {d} s \varvec{]} \\&\quad \textstyle =t^\ell \cdot \int _a^{z}\text {d} s \int _s^{z}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (z)}\circ \text {ad} _{\psi (s_2)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s)) \end{aligned}$$

  for $\ell \ge 1$. Since $(\mathfrak {g},\varvec{[}\cdot , \cdot \varvec{]})$ is asymptotic estimate and sequentially complete, Corollary 2 shows that the map

  $$\begin{aligned} \textstyle L:[a,b+\varepsilon )\times [0,1]\rightarrow \mathfrak {g},\qquad (z,t)\mapsto \sum _{\ell =0}^\infty L_\ell (z,t) \end{aligned}$$

  (128)

  is defined, such that $C^\infty ([0,1],\mathfrak {g})\ni L(z,\cdot )=\sum _{\ell =0}^\infty L_\ell (z,t)$ converges w.r.t. the $C^\infty$-topology for each $z\in [a,b+\varepsilon )$.

• Let $t\in [0,1]$, $z\in [a,b+\varepsilon )$, and $h>0$ be fixed.

  • If $z+h<b$ holds, we have $\mathcal {T}_0(\psi ,z+h,t)-\mathcal {T}_0(\psi ,z,t)=\int _z^{z+h} \psi (s)\, \text {d} s$, as well as for $\ell \ge 1$

    $$\begin{aligned}&\mathcal {T}_\ell (\psi ,z+h,t) -\mathcal {T}_\ell (\psi ,z,t)\\&\quad \textstyle =t^\ell \cdot \int _a^{z} \text {d} s \textstyle \int _z^{z+h}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\qquad \textstyle + t^\ell \cdot \int _z^{z+h}\text {d} s \textstyle \int _s^{z+h}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\quad \textstyle = t^\ell \cdot \int _a^{z} \text {d} s \textstyle \int _z^{z+h}\text {d} s_1\int _s^{z}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\qquad \textstyle +t^\ell \cdot \int _a^{z} \text {d} s \textstyle \int _z^{z+h}\text {d} s_1\int _z^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\qquad \textstyle + t^\ell \cdot \int _z^{z+h}\text {d} s \textstyle \int _s^{z+h}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s)). \end{aligned}$$

    (In the second step, we have split the third integral in the first summand at z.)

  • If $a<z-h$ holds, we have $\mathcal {T}_0(\psi ,z,t)-\mathcal {T}_0(\psi ,z-h,t)=\int _{z-h}^{z} \psi (s)\, \text {d} s$, as well as for $\ell \ge 1$

    $$\begin{aligned}&\mathcal {T}_\ell (\psi ,z,t)-\mathcal {T}_\ell (\psi ,z-h,t)\\&\quad \textstyle = t^\ell \cdot \int _a^{z-h} \text {d} s \textstyle \int _{z-h}^{z}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\qquad \textstyle + t^\ell \cdot \int _{z-h}^{z}\text {d} s \textstyle \int _s^{z}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\quad \textstyle =t^\ell \cdot \int _a^{z-h} \text {d} s \textstyle \int _{z-h}^{z}\text {d} s_1\int _s^{z-h}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\qquad \textstyle + t^\ell \cdot \int _a^{z-h} \text {d} s \textstyle \int _{z-h}^{z}\text {d} s_1\int _{z-h}^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s))\\&\qquad \textstyle + t^\ell \cdot \int _{z-h}^{z}\text {d} s \textstyle \int _s^{z}\text {d} s_1\int _s^{s_1}\text {d} s_2 \, {\dots } \int _s^{s_{\ell -1}}\text {d} s_\ell \, (\text {ad} _{\psi (s_1)}\circ \dots \circ \text {ad} _{\psi (s_{\ell })})(\psi (s)). \end{aligned}$$

    (In the second step, we have split the third integral in the first summand at $z-h$.)

Let $\textsf {v} \le \textsf {w}$ be as in (95), and set

$$\begin{aligned} \textstyle \Delta (\textsf {w} ,z,h):=\textsf {w} _\infty \big (\psi (z)-\psi |_{[z-|h|,z+|h|]\cap [a,b+\varepsilon ]}\big )\qquad \quad \forall \,z\in [a,b+\varepsilon ),\, h\ne 0. \end{aligned}$$

Let $z\in [a,b+\varepsilon )$ be fixed, and set

$$\begin{aligned} D:=\{x-z\,|\, x\in [a,b+\varepsilon )\}. \end{aligned}$$

For $\text {s} \in \mathbb {N}$, $\ell \ge 1$, and $h\in D_z$, we obtain

$$\begin{aligned} \textsf {v} ^\text {s} _\infty (\mathcal {T}_0(\psi ,z+h,\cdot )&\textstyle -\mathcal {T}_0(\psi ,z,\cdot )-h\cdot L_0(z,\cdot ))\\&\textstyle = \textsf {v} \big (\!\int _z^{z+h} \psi (s)\,\text {d} s-\psi (z)\big )\\&\textstyle \le |h|\cdot \Delta (\textsf {w} ,z,h),\\ \textstyle \textsf {v} ^\text {s} _\infty (\mathcal {T}_\ell (\psi ,z+h,\cdot )&-\mathcal {T}_\ell (\psi ,z,\cdot )-h\cdot L_\ell (z,\cdot ))\\&\textstyle \le \text {s} !\cdot |h|\cdot \Delta (\textsf {w} ,z,h)\cdot (b+\varepsilon -a)\cdot \textsf {w} _\infty (\psi ) \cdot \frac{(b+\varepsilon -a)^{\ell -1}}{(\ell -1)!}\cdot \textsf {w} _\infty (\psi )^{\ell -1}\\&\textstyle \quad \, + \text {s} !\cdot (b+\varepsilon -a)\cdot \textsf {w} _\infty (\psi ) \cdot |h|^2\cdot \textsf {w} _\infty (\psi )^2\cdot \frac{(b+\varepsilon -a)^{\ell -2}}{(\ell -2)!}\cdot \textsf {w} _\infty (\psi )^{\ell -2} \\&\textstyle \quad \, + \text {s} !\cdot |h|^2\cdot \textsf {w} _\infty (\psi )^2\cdot \frac{(b+\varepsilon -a)^{\ell -1}}{(\ell -1)!}\cdot \textsf {w} _\infty (\psi )^{\ell -1}. \end{aligned}$$

Now, the maps (with $h\in D$)

$$\begin{aligned} \textstyle \Phi _\psi (z+h,\cdot )&=\textstyle \sum _{\ell =0}^\infty \mathcal {T}_\ell (\psi ,z+h,\cdot ),\quad \textstyle \Phi _\psi (z,\cdot )=\sum _{\ell =0}^\infty \mathcal {T}_\ell (\psi ,z,\cdot ),\\&\qquad \qquad \quad \qquad L(z,\cdot )=\textstyle \sum _{\ell =0}^\infty L_\ell (z,\cdot ) \end{aligned}$$

converge w.r.t. the $C^\infty$-topology; and continuity of $\psi$ implies $\lim _{h\rightarrow 0}\Delta (\textsf {w} ,z,h)=0$. The triangle inequality thus yields

$$\begin{aligned} \textstyle \lim _{D\ni h\rightarrow 0}\frac{1}{|h|}\cdot \textsf {v} ^\text {s} _\infty (\Phi _\psi (z+h,\cdot )-\Phi _\psi (z,\cdot )- h\cdot L(z,\cdot ))=0. \end{aligned}$$

Together with (127) this proves the claim, because for $z\in [a,b+\varepsilon )$ and $t\in [0,1]$, we have

$$\begin{aligned} \textstyle L(z,t){\mathop {=}\limits ^{(128)}}\psi (z) + t\cdot \sum _{\ell =1}^\infty \varvec{[}\psi (z), \mathcal {T}_{\ell -1}(\psi ,z,t) \varvec{]}=\psi (z) + t\cdot \varvec{[}\psi (z), \Phi _\psi (z,t) \varvec{]}=\mathcal {D}(z,t) \end{aligned}$$

by continuity of $\varvec{[}\cdot , \cdot \varvec{]}$.

• (2) Point (a) is clear from Part 1). For the points (b) and (c), we observe the following:

• For $\textsf {p} \in \textsf {Sem}(E)$, $\text {s} \in \mathbb {N}$, $z\in (a-1,b+\varepsilon )$, and $h\in \mathbb {R}$ with $z+h\in (a-1,b+\varepsilon )$, we have

  $$\begin{aligned} \begin{aligned}&\textsf {p} ^\text {s} _\infty (\Phi _\psi (z+h,\cdot )-\Phi _\psi (z,\cdot ))\\&\quad \le \textsf {p} ^\text {s} _\infty (\Phi _\psi (z+h,\cdot )-\Phi _\psi (z,\cdot )-h\cdot \mathcal {D}(z,\cdot )) + |h|\cdot \textsf {p} ^\text {s} _\infty ( \mathcal {D}(z,\cdot )). \end{aligned} \end{aligned}$$

  (129)

• Since $\varvec{[}\cdot , \cdot \varvec{]}$ is continuous and bilinear, there exists $\textsf {q} \in \textsf {Sem}(E)$ with

  $$\begin{aligned} \textsf {p} (\varvec{[}X, Y \varvec{]})\le \textsf {q} (X)\cdot \textsf {q} (Y) \qquad \quad \forall \, X,Y\in \mathfrak {g}. \end{aligned}$$

  Consequently, (121) implies

  $$\begin{aligned} \sup \{a-1\le z\le b+\varepsilon \,|\, \textsf {p} _\infty ^\text {s} (\mathcal {D}(z,\cdot ))\}<\infty \qquad \quad \forall \, \textsf {p} \in \textsf {Sem}(E). \end{aligned}$$

  (130)

Point (b) is now clear from Part 1), (129), and (130). For point (c), we set $\text {s} :=0$. Then, by Part 1) and (130), both summands in (129) tend to zero if h tends to zero. Point (c) now follows from the triangle inequality, as well as continuity of $\Phi _\psi (z,\cdot )$ for each $z\in (a-1,b+\varepsilon )$.

• (3) Let $z\in (a-1,b+\varepsilon )$ and $\mathbb {R}_{\ne 0}\supseteq \{h_n\}_{n\in \mathbb {N}}\rightarrow 0$ be given. Then, Part 2).(b) implies that holds (recall (14)). Theorem B and Lemma A yield

  

  It follows that the map

  

  (131)

  is continuous. In particular, $\mu _{t,\psi }$ is continuous for each $t\in [0,1]$.

  Let now $t\in [0,1]$ be fixed. The points (a) and (b) in Part 2) show that the map $\Phi _\psi |_{(a-1,b+\varepsilon )\times [0,t]}$ fulfills the assumptions in Theorem C. We obtain (apply Part 2).(a) in the third step)

  

  for each $z\in (a-1,b+\varepsilon )$. It follows from smoothness of the group operations, continuity of (131), and continuity of $\Phi _\psi$ (by Part 2).(c)) that ${\dot{\mu }}_{t,\psi }$ is continuous. This shows that $\mu _{t,\psi }$ is of class $C^1$, which proves the claim. $\square$

We are ready for the proof of Theorem 2.

Proof of Theorem 2

Since $\mathfrak {g}$ is sequentially complete, $\mathfrak {g}$ is both Mackey complete and integral complete. To prove the claim, it thus suffices to show that G is $C^0$-semiregular. For this, let $\phi \in C^0([0,1],\mathfrak {g})$ be given, and fix a continuous extension $\psi :[0,1+\varepsilon ]\rightarrow \mathfrak {g}$ for some $\varepsilon >0$. We define $\mu _{1,\psi }:(-1,1+\varepsilon )\rightarrow G$ as in (122) (for $a\equiv 0$ and $b\equiv 1$ there). Lemma 27.2) shows that $\mu _{1,\psi }$ is of class $C^1$, so that it remains to verify . For this, let $z\in [0,1]$ be fixed. Corollary 19 shows ($\Phi _\psi$ is defined as in (120))

(132)

for each $h\in [0,\varepsilon )$. Set $\widehat{\psi }:=\psi |_{[z,z+\varepsilon ]}$. Then,

holds by definition, so that we have

$$\begin{aligned} \mu _{1,\psi }(z+h)\cdot \mu _{1,\psi }(z)^{-1}{\mathop {=}\limits ^{(132)}} \mu _{1,\widehat{\psi }}(z+h)\qquad \quad \forall \, h\in [0,\varepsilon ). \end{aligned}$$

(133)

Lemma 27.2) shows $\mu _{1,\widehat{\psi }}\in C^1((z-1,z+\varepsilon ),G)$, and Equation (126) in Lemma 27.2) together with both sides of (124) (first step) yields

$$\begin{aligned} {\dot{\mu }}_{1,\widehat{\psi }}(z)=\widehat{\psi }(z)=\psi (z)=\phi (z). \end{aligned}$$

Then, (133) shows , which proves the claim. $\square$

Appendix

1.1 Bastiani’s Differential Calculus

In this appendix, we recall Bastiani’s differential calculus, confer also [3, 7, 19,20,21,22]. Let $E,F\in \textsf {hlcVect}$ be given. A map $f:U\rightarrow F$, with $U\subseteq E$ open, is said to be differentiable if

$$\begin{aligned} \textstyle (D_v f)(x):=\lim _{t\rightarrow 0}1/t\cdot (f(x+t\cdot v)-f(x))\in F \end{aligned}$$

exists for each $x\in U$ and $v\in E$. The map f is said to be k-times differentiable for $k\ge 1$ if

$$\begin{aligned} D_{v_k,\dots ,v_1}f:= D_{v_k}(D_{v_{k-1}}( {\dots } (D_{v_1}(f))\dots )):U\rightarrow F \end{aligned}$$

is defined for all $v_1,\dots ,v_k\in E$. Implicitly, this means that f is p-times differentiable for each $1\le p\le k$, and we set

$$\begin{aligned} \text {d} ^p_xf(v_1,\dots ,v_p)\equiv \text {d} ^p f(x,v_1,\dots ,v_p):=D_{v_p,\dots ,v_1}f(x)\qquad \quad \forall \, x\in U,\,v_1,\dots ,v_p\in E \end{aligned}$$

for $p=1,\dots ,k$. We furthermore define $\text {d} f:= \text {d} ^1 f$, as well as $\text {d} _x f:= \text {d} ^1_x f$ for each $x\in U$. The map $f:U\rightarrow F$ is said to be

• of class $C^0$ if it is continuous. In this case, we define $\text {d} ^0 f:= f$.

• of class $C^k$ for $k\ge 1$ if it is k-times differentiable, such that

  $$\begin{aligned} \text {d} ^p f:U\times E^p\rightarrow F,\qquad (x,v_1,\dots ,v_p)\mapsto D_{v_p,\dots ,v_1}f(x) \end{aligned}$$

  is continuous for $p=0,\dots ,k$. In this case, $\text {d} ^p_x f$ is symmetric and p-multilinear for each $x\in U$ and $p=1,\dots ,k$, cf. [3].

• of class $C^\infty$ if it is of class $C^k$ for each $k\in \mathbb {N}$.

Remark

Assume that E, F are normed spaces. Let $U\subseteq E$ be non-empty open, as well as $k\in \mathbb {N}\cup \{\infty \}$. Let $\mathcal {F}C^k(U,F)$ denote the set of all k-times Fréchet differentiable maps $U\rightarrow F$. Then, $C^{k+1}(U,F)\subseteq \mathcal {F}C^k(U,F)\subseteq C^k(U,F)$ holds [20, 26], hence $C^\infty (U,F)=\mathcal {F}C^\infty (U,F)$. $\square$

We have the following differentiation rules [3].

Proposition A.1

• (a) A map $f:E\supseteq U\rightarrow F$ is of class $C^k$ for $k\ge 1$ if and only if $\text {d} f$ is of class $C^{k-1}$ when considered as a map $E \times E \supseteq U\times E \rightarrow F$.

• (b) Let $f:E\rightarrow F$ be linear and continuous. Then, f is smooth, with $\text {d} ^1_xf=f$ for each $x\in E$, as well as $\text {d} ^pf=0$ for each $p\ge 2$.

• (c) Let $F_1,\dots ,F_m$ be Hausdorff locally convex vector spaces, and $f_q:E\supseteq U\rightarrow F_q$ be of class $C^k$ for $k\ge 1$ and $q=1,\dots ,m$. Then,

  $$\begin{aligned} f:=f_1\times {\dots }\times f_m :U\rightarrow F_1\times {\dots }\times F_m,\qquad x\mapsto (f_1(x),\dots ,f_m(x)) \end{aligned}$$

  if of class $C^k$, with $\text {d} ^p f=\text {d} ^pf_1{\times } \dots \times \text {d} ^pf_m$ for $p=1,\dots ,k$.

• (d) Let $F,\bar{F},\bar{\bar{F}}\in \textsf {hlcVect}$, $1\le k\le \infty$, as well as $f:F\supseteq U\rightarrow \bar{U}\subseteq \bar{F}$ and $\bar{f}:\bar{F}\supseteq \bar{U}\rightarrow \bar{\bar{U}}\subseteq \bar{\bar{F}}$ be of class $C^k$. Then, $\bar{f}\circ f:U\rightarrow \bar{\bar{F}}$ is of class $C^k$, with

  $$\begin{aligned} \text {d} _x(\bar{f}\circ f)=\text {d} _{f(x)}\bar{f}\circ \text {d} _x f\qquad \quad \forall \, x\in U. \end{aligned}$$

• (e) Let $F_1,\dots ,F_m,E\in \textsf {hlcVect}$, and $f:F_1\times {\dots } \times F_m\supseteq U\rightarrow E$ be of class $C^0$. Then, f is of class $C^1$ if and only if for $p=1,\dots ,m$, the partial derivatives

  $$\begin{aligned}&\partial _p f :U\times F_p\ni ((x_1,\dots ,x_m),v_p)\mapsto \lim _{t\rightarrow 0} 1/t\\&\quad \cdot (f(x_1,\dots , x_p+t\cdot v_p,\dots ,x_m)-f(x_1,\dots ,x_m)) \end{aligned}$$

  exist in E and are continuous. In this case, we have

  $$\begin{aligned} \textstyle \text {d} f((x_1,\dots ,x_m),v_1,\dots ,v_m)&\textstyle =\sum _{p=1}^m\partial _p f((x_1,\dots ,x_m),v_p)\\ \big (\!&\textstyle = \sum _{p=1}^m ~~\text {d} f((x_1,\dots ,x_m),(0,\dots ,0, v_p,0,\dots ,0))\,\big ) \end{aligned}$$

  for each $(x_1,\dots ,x_m)\in U$, and $v_p\in F_p$ for $p=1,\dots ,m$.

1.2 Proof of Equation (26)

The claim is clear for $m=1$ and $n\ge 1$. Assume now that (26) holds for some $m\ge 1$, as well as each $n\ge 1$. For $X\in \mathcal {V}_1(S)$, $\alpha \in \mathcal {V}_m(S)$, and $\beta \in \mathcal {V}_n(S)$ with $n\ge 1$, we obtain from (25) and antysimmetry of $\varvec{[}\cdot , \cdot \varvec{]}$ that

$$\begin{aligned} \varvec{[}\varvec{[}X, \alpha \varvec{]}, \beta \varvec{]}=\varvec{[}X, \varvec{[}\alpha , \beta \varvec{]} \varvec{]}-\varvec{[}\alpha , \varvec{[}X, \beta \varvec{]} \varvec{]}\in \mathcal {V}(S)_{(m+1)+n} \end{aligned}$$

holds. Since $\varvec{[}\cdot , \cdot \varvec{]}$ is bilinear, this implies $\varvec{\langle } \varvec{[}\mathcal {V}_{m+1}(S), \mathcal {V}_n(S) \varvec{]}\varvec{\rangle } \subseteq \mathcal {V}_{(m+1)+n}(S)$ for each $n\ge 1$, so that the claim follows inductively.

1.3 Proof of Lemma 5

Lemma 5

Let $V\subseteq F$ be open with $0\in V$. Let furthermore $\Psi :V\times E\rightarrow E$ be smooth with $\Psi (0,\cdot )=\text {id} _{E}$, such that $\Psi (x,\cdot )$ is linear for each $x \in V$. Then, to each $\textsf {p} \in \textsf {Sem}(E)$, there exist $\textsf {q} \in \textsf {Sem}(F)$ and $\textsf {w} \in \textsf {Sem}(E)$ with

$$\begin{aligned} \textsf {p} (\Psi (x,Y)-Y)\le \textsf {q} (x)\cdot \textsf {w} (Y)\qquad \quad \forall \, x\in \text {B} _{\textsf {q} ,1}\subseteq V,\, Y\in E. \end{aligned}$$

Proof of Lemma 5

We consider the continuous map

$$\begin{aligned} \Phi :V \times E \times F \rightarrow E,\qquad (x,Y,Z)\mapsto \partial _1 \Psi ((x,Y),Z) \end{aligned}$$

that is linear in the last two arguments. Then, for $\textsf {p} \in \textsf {Sem}(E)$ fixed, by Lemma 4 there exist $\textsf {q} \in \textsf {Sem}(F)$ and $\textsf {p} \le \textsf {w} \in \textsf {Sem}(E)$ with

$$\begin{aligned} (\textsf {p} \circ \Phi )((x,Y),Z)&\le \textsf {w} (Y)\cdot \textsf {q} (Z)\qquad \quad \forall \, x\in \text {B} _{\textsf {q} ,1}\subseteq V,\, Y\in E,\, Z\in F. \end{aligned}$$

(134)

For $x\in \text {B} _{\textsf {q} ,1}$, we let $\gamma :[0,1]\ni t\mapsto t\cdot x \in \text {B} _{\textsf {q} ,1}$, and define

$$\begin{aligned} C^\infty ([0,1],E)\ni \alpha _Y:[0,1]\ni t\mapsto \Psi (\gamma (t),Y)\qquad \quad \forall \, Y\in E. \end{aligned}$$

Then, the parts b) and e) of Proposition A.1 yield ${\dot{\alpha }}_Y=\Phi ((\gamma ,Y),{\dot{\gamma }})=\Phi ((\gamma ,Y),x)$, and we obtain

$$\begin{aligned} \textsf {p} (\Psi (x,Y)-Y)&=\textsf {p} (\alpha _Y(1)-\alpha _Y(0)) \textstyle {\mathop {\le }\limits ^{(18)}} \int _0^1 \textsf {p} ({\dot{\alpha }}_Y(s)) \,\text {d} s\\&\textstyle \le \int _0^1 \textsf {p} (\Phi ((\gamma (s),Y),x)) \,\text {d} s \textstyle \le \textsf {p} _\infty (\Phi ((\gamma ,Y),x)) \textstyle {\mathop {\le }\limits ^{(134)}} \textsf {w} (Y)\cdot \textsf {q} (x) \end{aligned}$$

for all $x\in \text {B} _{\textsf {q} ,1}$ and $Y\in E$. $\square$

1.4 Appendix to Sect. 2.3

For combinatorial reasons, there exist $c[p]_0,\dots ,c[p]_{p^2}\in \mathbb {K}$ with

$$\begin{aligned} f_p(g_p(z))&\textstyle =(f_p\diamond g_p)(z)=\sum _{n=0}^{p^2} c[p]_{n}\cdot z^n\\ f_p(g_p(\mathfrak {n}))&\textstyle =(f_p\diamond g_p)(\mathfrak {n})=\sum _{n=0}^q c[p]_{n}\cdot \mathfrak {n}^n\\ f_p(g_p(\mathfrak {a}))&\textstyle =(f_p\diamond g_p)(\mathfrak {a})=\sum _{n=0}^{p^2} c[p]_{n}\cdot \mathfrak {a}^n. \end{aligned}$$

Set $c[p]_n:=0$ for $n\ge p^2+1$. Since $\{f_p\circ g_p\}_{p\in \mathbb {N}}\rightarrow f\circ g$ converges compactly on $U_{S}(0)$, the Weierstrass convergence theorem yields

$$\begin{aligned} \textstyle \lim _{p\rightarrow \infty } c[p]_{n}= c_n\qquad \quad \forall \, n\in \mathbb {N}. \end{aligned}$$

(135)

Proof of Equation (31)

Since $b_0=0$ holds, we have $c[q+\ell ]_{n}=c[q]_{n}$ for $n=0,\dots ,q$ and $\ell \in \mathbb {N}$, hence $c[q]_n=c_n$ for $n=0,\dots ,q$ by (135). We obtain

$$\begin{aligned} \textstyle f(g(\mathfrak {n}))&~\textstyle =\sum _{n=0}^q a_n\cdot (\sum _{\ell =1}^q b_\ell \cdot \mathfrak {n}^\ell )^n=f_{q}(g_{q}(\mathfrak {n}))=\sum _{n=0}^q c[q]_n\cdot \mathfrak {n}^n\\&~\textstyle =\sum _{n=0}^q c_n\cdot \mathfrak {n}^n=(f\diamond g)(\mathfrak {n}) \end{aligned}$$

which proves (31). $\square$

Proof of Equation (32)

Fix $\Vert \mathfrak {a}\Vert _\mathcal {A}< s <S$, and set $M:=|f|_{|g|_s}$. Then, $|(f_p\circ g_p)(z)|\le M$ holds for all $p\in \mathbb {N}$ and $z\in \mathbb {C}$ with $|z|\le s$. Cauchy’s estimate yields $\textstyle |c[p]_n|\le \frac{M}{s^n}$ for all $p,n\in \mathbb {N}$, hence

$$\begin{aligned} \textstyle \sum _{n=N+1}^\infty |c[p]_n|\cdot (\Vert \mathfrak {a}\Vert _\mathcal {A})^n= \sum _{n=N+1}^\infty |c[p]_n|\cdot s^n\cdot \frac{(\Vert \mathfrak {a}\Vert _\mathcal {A})^n}{s^n}\le M\cdot \sum _{n=N+1}^\infty \frac{(\Vert \mathfrak {a}\Vert _\mathcal {A})^n}{s^n} \end{aligned}$$

for all $N,p\in \mathbb {N}$. Given $\varepsilon >0$, there thus exists $N_\varepsilon \in \mathbb {N}$ with

$$\begin{aligned} \textstyle \Vert \sum _{n=N_\varepsilon +1}^\infty c_n\cdot \mathfrak {a}^n\Vert _\mathcal {A}<\frac{\varepsilon }{4}\qquad \quad \text {as well as}\qquad \quad \Vert \sum _{n=N_\varepsilon +1}^\infty c[p]_n\cdot \mathfrak {a}^n\Vert _\mathcal {A}<\frac{\varepsilon }{4} \qquad \forall \, p\in \mathbb {N}. \end{aligned}$$

(136)

By (135), there exists $m\in \mathbb {N}$ with

$$\begin{aligned} \textstyle |c[p]_n-c_n|<\frac{\varepsilon }{4(N_\varepsilon +1)\cdot \max (\Vert \mathfrak {a}^q\Vert _\mathcal {A}\,|\, 0\le q\le N_\varepsilon )}\qquad \quad \forall \, 0\le n\le N_\varepsilon ,\, p\ge m. \end{aligned}$$

(137)

Increasing m if necessary, we can additionally assume²⁵

$$\begin{aligned} \textstyle \Vert f(g(\mathfrak {a}))-f_p(g_p(\mathfrak {a}))\Vert _\mathcal {A}<\frac{\varepsilon }{4}\qquad \quad \forall \, p\ge m. \end{aligned}$$

Set $p:= m$. Then, we have

with

$$\begin{aligned} \textrm{A}&\textstyle \le \Vert \sum _{n=0}^{N_\varepsilon } c[p]_n\cdot \mathfrak {a}^n - \sum _{n=0}^{N_\varepsilon } c_n\cdot \mathfrak {a}^n \Vert _\mathcal {A} \textstyle + \Vert \sum _{n=N_\varepsilon +1}^\infty c[p]_n\cdot \mathfrak {a}^n - \sum _{n=N_\varepsilon +1}^\infty c_n\cdot \mathfrak {a}^n\Vert _\mathcal {A}\\&\textstyle \le \sum _{n=0}^{N_\varepsilon } |c[p]_n-c_n|\cdot \Vert \mathfrak {a}^n\Vert _\mathcal {A}\textstyle + \Vert \sum _{n=N_\varepsilon +1}^\infty c[p]_n\cdot \mathfrak {a}^n\Vert _\mathcal {A} + \Vert \sum _{n=N_\varepsilon +1}^\infty c_n\cdot \mathfrak {a}^n\Vert _\mathcal {A} \\&\textstyle \le \frac{3}{4}\cdot \varepsilon \end{aligned}$$

by (136) and (137).$\square$

1.5 Proof of Lemma 9

Lemma 9

Assume that G is weakly $C^k$-regular for $k\in \mathbb {N}\cup \{\text {lip} ,\infty \}$, and let $\phi \in C^k([a,b])$ ($a<b$) be given. Then, is continuous.

Proof

Let $z\in [a,b]$, $t\in \mathbb {R}$, and $U\subseteq G$ be an open neighborhood of e. We fix $W\subseteq G$ open with $e\in W$ and $W\cdot W\subseteq U$, and observe that

is continuous. Since [a, b] is compact, there exists some open $V\subseteq G$ with $e\in V=V^{-1}$, such that

Theorem B together with Lemma 15 in [11] implies that there exists $\delta >0$ with

For $a\le x\le b$ and $|h|<\delta$, we obtain

Since $\kappa (t,\cdot )$ is continuous, we can shrink $\delta >0$ such that for each $x\in (z-\delta ,z+\delta ) \cap [a,b]$, we have . We obtain

for $|h|<\delta$ and $x\in (z-\delta ,z+\delta ) \cap [a,b]$. $\square$

1.6 Proof of Equation (64)

Proof of Equation (64)

By definition, we have for certain $\mu \in C^1((a-\varepsilon ,b+\varepsilon ),G)$ and $\varepsilon >0$. Replacing $\mu$ by $\mu \cdot \mu (a)^{-1}$ if necessary, we can assume that $\mu (a)=e$ holds, hence

Then, $\alpha ^\pm := \text {Ad} _{\mu ^\pm }(X)\in C^1((a-\varepsilon ,b+\varepsilon ))$ holds by Lemma 8, with $\alpha ^\pm |_{[a,b]}={\textbf {Ad}} ^\pm _\psi [X]$. This shows ${\textbf {Ad}} _\psi ^\pm [X]\in C^1([a,b],\mathfrak {g})$, with ${\textbf {Ad}} ^\pm _\psi [X](a)=\text {Ad} _e(X)=X$. Let now $t\in [a,b]$ and $0<h<\varepsilon$ be given. We observe

and obtain from (d) that

Differentiating both hand sides and then applying (9), we obtain (64). $\square$

1.7 The Proof of Lemma 17

Lemma 17

Let $\psi \in C^0([a,b],\mathfrak {q})$ be given. Then, the following assertions hold:

• (1) The maps $\varvec{\Lambda }_\psi ^\pm$ are defined and of class $C^1$, and we have

  $$\begin{aligned} \partial _1 \varvec{\Lambda }_\psi ^+(t,X)&\textstyle =\partial _t{\textbf {Ad}} _\psi ^+[X](t)=\varvec{[}\psi (t), {\textbf {Ad}} ^+_\psi [X](t) \varvec{]}\\ \partial _1 \varvec{\Lambda }^-_\psi (t,X)&\textstyle =\partial _t{\textbf {Ad}} _\psi ^-[X](t)=-{\textbf {Ad}} ^-_\psi [\varvec{[}\psi (t), X \varvec{]}](t) \end{aligned}$$

  for all $t\in [a,b]$ and $X\in \mathfrak {q}$.

• (2) $\varvec{\Lambda }^\pm _\psi (t,\cdot )={\textbf {Ad}} _\psi ^\pm [t]$ is linear and continuous for each $t\in [a,b]$.

• (3) If $\psi \in C^k([a,b],\mathfrak {q})$ holds for $k\in \mathbb {N}\cup \{\infty \}$, then $\varvec{\Lambda }_\psi ^\pm$ is of class $C^{k+1}$.

Proof

Let $X\in \mathfrak {q}$ be fixed. Then, $\varvec{\lambda }^\pm _{\ell ,\psi }[X]$ is of class $C^1$ for each $\ell \in \mathbb {N}$ by (20); with $\partial _t\varvec{\lambda }^\pm _{0,\psi }[X]=0$ as well as

$$\begin{aligned} \partial _t\varvec{\lambda }^+_{\ell ,\psi }[X](t)= \varvec{[}\psi (t), \varvec{\lambda }^+_{{\ell -1},\psi }[X](t) \varvec{]}\quad \text {and}\quad \partial _t\varvec{\lambda }^-_{\ell ,\psi }[X](t)= -\varvec{\lambda }^-_{{\ell -1},\psi }[\varvec{[}\psi (t), X \varvec{]}](t) \end{aligned}$$

for each $\ell \ge 1$ and $t\in [0,1]$. For $\textsf {v} \le \textsf {w}$ as in (95), we have

$$\begin{aligned} \textstyle \sum _{\ell =0}^\infty \textsf {v} _\infty (\varvec{\lambda }^\pm _{\ell ,\psi }[X])\textstyle \le \textsf {w} (X)\cdot \text {e} ^{(b-a)\cdot \textsf {w} _\infty (\psi )}<\infty . \end{aligned}$$

(138)

Corollary 1 thus implies $\varvec{\Lambda }^\pm (\cdot ,X)={\textbf {Ad}} _\psi ^\pm [X]\in C^1([a,b],\mathfrak {q})$, with

$$\begin{aligned} \textstyle \partial _t{\textbf {Ad}} _\psi ^+[X](t)&\textstyle =\sum _{\ell =0}^\infty \varvec{[}\psi (t), \varvec{\lambda }_\psi ^+[X](t) \varvec{]}=\varvec{[}\psi (t), {\textbf {Ad}} ^+_\psi [X](t) \varvec{]}\\ \textstyle \partial _t{\textbf {Ad}} _\psi ^-[X](t)&\textstyle =-\sum _{\ell =0}^\infty \varvec{\lambda }_\psi ^-[\varvec{[}\psi (t), X \varvec{]}](t)=-{\textbf {Ad}} ^-_\psi [\varvec{[}\psi (t), X \varvec{]}](t) \end{aligned}$$

for each $t\in [a,b]$, which establishes Part 1). Now, it is clear from the definitions that $\varvec{\Lambda }^\pm _\psi (t,\cdot )$ is linear for each $t\in [a,b]$; and we furthermore observe that

$$\begin{aligned} \begin{aligned}&\textstyle \textsf {v} \big (\sum _{\ell =0}^\infty \varvec{\lambda }^\pm _{\ell ,\psi }[X](t')\textstyle -\sum _{\ell =0}^\infty \varvec{\lambda }^\pm _{\ell ,\psi }[X](t)\big )\\&\quad \textstyle \le |t-t'| \cdot \textsf {w} (X)\cdot \textsf {w} _\infty (\psi )\cdot \sum _{\ell =1}^\infty \frac{(b-a)^{\ell -1}}{(\ell -1)!}\cdot \textsf {w} _\infty (\psi )^{\ell -1}\\&\quad \textstyle = |t-t'| \cdot \textsf {w} (X)\cdot \textsf {w} _\infty (\psi )\cdot \text {e} ^{(b-a)\cdot \textsf {w} _\infty (\psi )} \end{aligned} \end{aligned}$$

(139)

holds for all $t,t'\in [a,b]$ and $X\in \mathfrak {q}$. We obtain from (138) and (139) that

$$\begin{aligned} \textsf {v} (\varvec{\Lambda }_\psi ^\pm (t',X')-\varvec{\Lambda }_\psi ^\pm (t,X))&\textstyle \le \textsf {v} (\varvec{\Lambda }_\psi ^\pm (t',X')-\varvec{\Lambda }_\psi ^\pm (t',X))+\textsf {v} (\varvec{\Lambda }_\psi ^\pm (t',X)-\varvec{\Lambda }_\psi ^\pm (t,X))\\&\le \textstyle (\textsf {w} (X'-X) + |t-t'| \cdot \textsf {w} (X)\cdot \textsf {w} _\infty (\psi ) )\cdot \text {e} ^{(b-a)\cdot \textsf {w} _\infty (\psi )} \end{aligned}$$

for all $t,t'\in [a,b]$ and $X,X'\in \mathfrak {q}$. This shows that $\varvec{\Lambda }_\psi ^\pm$ is continuous, which establishes Part 2). We furthermore conclude that the partial derivatives

$$\begin{aligned} \partial _1\varvec{\Lambda }_\psi ^+((t,X),\lambda )\,&= \lambda \cdot \varvec{[}\psi (t), \varvec{\Lambda }_\psi ^+(t,X) \varvec{]}\qquad \quad \,\forall \, t\in [a,b],\,\lambda \in \mathbb {R},\, X\in \mathfrak {q}\\ \partial _1\varvec{\Lambda }_\psi ^-((t,X),\lambda )\,&= -\lambda \cdot \varvec{\Lambda }_\psi ^-(t,\varvec{[}\psi (t), X \varvec{]})\qquad \quad \forall \, t\in [a,b],\,\lambda \in \mathbb {R},\, X\in \mathfrak {q}\\ \partial _2\varvec{\Lambda }_\psi ^\pm ((t,X),Y)&= \varvec{\Lambda }_\psi ^\pm (t,Y) \qquad \qquad \qquad \qquad ~~~~\,\,\forall \, t\in [a,b],\, X,Y\in \mathfrak {q}\end{aligned}$$

are continuous. Part e) of Proposition A.1 thus shows that the maps $\varvec{\Lambda }_\psi ^\pm$ are of class $C^1$, with

$$\begin{aligned} \begin{aligned} \text {d} \varvec{\Lambda }_\psi ^+((t,X),(\lambda ,Y))&= \lambda \cdot \varvec{[}\psi (t), \varvec{\Lambda }_\psi ^+(t,X) \varvec{]} + \varvec{\Lambda }_\psi ^+(t,Y)\\ \text {d} \varvec{\Lambda }_\psi ^-((t,X),(\lambda ,Y))&= -\lambda \cdot \varvec{\Lambda }_\psi ^-(t,\varvec{[}\psi (t), X \varvec{]}) + \varvec{\Lambda }_\psi ^-(t,Y). \end{aligned} \end{aligned}$$

(140)

Finally, assume that $\psi \in C^k([a,b],\mathfrak {q})$ holds for $k \ge 1$. Then, (140) together with the parts a), c), d) of Proposition A.1 imply that given $1\le \ell \le k$, then $\varvec{\Lambda }_\psi ^\pm$ is of class $C^{\ell +1}$ if $\varvec{\Lambda }_\psi ^\pm$ is of class $C^\ell$, which establishes Part 3). It thus follows by induction that $\varvec{\Lambda }_\psi ^\pm$ is of class $C^{k+1}$. $\square$