Linear Combinations of Projections in Type III Factors

Abstract

It is shown that every self-adjoint operator in a $\sigma$-finite von Neumann factor of type III can be written as a real linear combination of 3 projections. This is in stark contrast to factors of type $I_\infty ,\, II_1$ and $II_\infty$, where there are always operators that require at least 4 projections in such a representation.

1 Introduction

There are many questions concerned with linear combinations of projections in operator algebras that are an area of active research. For example, we can ask which elements in a $C^*$-algebra are linear combinations of projections. For some classes of operator algebras, each of their elements can be represented in such a way. This is obvious, for example, for algebras of n-by-n matrices. This is also true for the algebra B(H) with infinite-dimensional H,  as shown by Fillmore in 1967 (see [1]). The authors’ paper [2] from 1992 shows that this is also true, among others, for all von Neumann factors. The $C^*$-algebra case was considered in [3, 4].

There are several variants of this basic question, concerned mainly with writing positive operators as either linear combinations of projections with positive coefficients or as (strong) sums of finite or infinite families of projections (see [5] for a survey). In particular, this latter variant has now a complete solution in case of von Neumann factors (see [6,7,8]).

Whenever an algebra is a (complex) linear span of its projections, the following question presents itself: What is the smallest number of projections such that each self-adjoint element of such an algebra can be written as a linear combination of this specific number of projections? Some positive results in this direction have already been given in [1, 9,10,11,12,13] and [2]. An important next step was a construction of self-adjoint operators that are not linear combinations of 3 projections—for factors of type $I_n$ with n infinite or $n\ge 76$ by [13], and for factors of type II in our recent publication [14]. An interested reader can find the history of the problem and many bibliographic references to earlier works in [14].

In [14] it was shown that any self-adjoint operator in an infinite factor can be written as a linear combination of 4 projections, but in each semifinite factor (including $II_1$ ones) there is a self-adjoint operator that is not a linear combination of 3 projections. This paper shows that the situation in $\sigma$-finite type III factors is diametrally different—each self-adjoint operator is a linear combination of 3 projections. The proof of the fact requires the development of a completely new technique, and some of results obtained on the way seem to be of independent interest.

The general form of all real linear combinations of two projections was described by r15 in [15], and also in our previous paper [14]. In particular, in any infinitely dimensional von Neumann algebra there is a self-adjoint operator that is not a real linear combination of two projections.

Most of the time, our notation is standard. The set ${\mathbb {N}}$ of natural numbers starts from 1. For $\mu \in {\mathbb {R}},$ $\mu ^+=\max \{0,\mu \}$ and $\mu ^-=\max \{0,-\mu \}$. For $\mu ,\nu \in {\mathbb {R}},$ the open interval $(\mu ,\nu )$ denotes the empty set $\emptyset$ whenever $\mu \ge \nu$.

For a normed space ${\mathcal {D}}$ and for $a\in {\mathcal {D}},\, \varepsilon >0$, we denote by ${\mathbf {B}}(a; \varepsilon )$ the closed ball with center a and radius $\epsilon$.

${\mathcal {M}}$ is an arbitrary von Neumann algebra acting in a complex Hilbert space H. For any subset ${\mathcal {A}}$ of ${\mathcal {M}}$, ${\mathcal {A}}_h$ denotes the self-adjoint (hermitian) part of ${\mathcal {A}}$. Whenever we treat ${\mathcal {M}}_h$ or its subspaces as normed spaces, we always mean the norm inherited from the usual $C^*$-norm on ${\mathcal {M}}$. The group of all automorphisms of ${\mathcal {M}}$ is denoted by ${{\,\mathrm{Aut}\,}}({\mathcal {M}})$ and the subgroup of inner automorphisms of ${\mathcal {M}}$ by ${{\,\mathrm{Inn}\,}}({\mathcal {M}})$. We always endow the spaces with the usual operator norm topology.

The unit of ${\mathcal {M}}$ is denoted by $\mathbb {1}$ or $\mathbb {1}_{\mathcal {M}}$. We agree on $a^0=\mathbb {1}$ for all $a\in {\mathcal {M}}$. For $a\in {\mathcal {M}}_h$, we denote by ${{\,\mathrm{supp}\,}}a$ the support projection of a, that is the smallest projection in ${\mathcal {M}}$ such that $a=ea=ae$. For hermitian a, b we write $a\perp b$ if ${{\,\mathrm{supp}\,}}a\perp {{\,\mathrm{supp}\,}}b$. Note that in this case $\Vert a+b\Vert =\max (\Vert a\Vert ,\Vert b\Vert )$. Note that infinite sums of mutually orthogonal operators $a_i\in {\mathcal {M}}_h$ with uniformly bounded norms always exist in strong operator topology. For $a\in {\mathcal {M}}_h,$ we write $a^+$ for $(|a|+a)/2$ and $a^-$ for $(|a|-a)/2$.

The lattice of (orthogonal) projections in ${\mathcal {M}}$ is denoted by ${{\,\mathrm{Proj}\,}}{\mathcal {M}}$. For $p,q\in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ we write $p \sim q$ (resp., $p \simeq q$) if there exists a partial isometry $u\in {\mathcal {M}}$ such that $p = u^*u,\,q = uu^*$ (resp., a unitary u such that $q=upu^*$). For $b, c \in {\mathcal {M}}_h$, $b \sim c$ (resp. $b \simeq c$) means that $c = ubu^*$ for a partial isometry u satisfying $uu^* = {{\,\mathrm{supp}\,}}c$, $u^*u = {{\,\mathrm{supp}\,}}b$ (resp., a unitary u such that $c=ubu^*$). For a projection e from ${\mathcal {M}}$, the reduced von Neumann algebra ${\mathcal {M}}_e$ consists of restrictions to e(H) of elements of $e{\mathcal {M}}e$.

The spectrum of $a\in {\mathcal {M}}_h$ is denoted by ${{\,\mathrm{sp}\,}}a$. The set of Borel subsets of ${\mathbb {R}}$ is written as ${\mathcal {B}}({\mathbb {R}})$. For $Z \in {\mathcal {B}}({\mathbb {R}}),$ we denote by $e_a(Z)$ the spectral projection of a corresponding to Z,  and by $e_a^Z$ the spectral projection of a corresponding to $Z \setminus \{0\}$. Note that $e_a^Z\le {{\,\mathrm{supp}\,}}a$ for all $Z \in {\mathcal {B}}({\mathbb {R}})$. The operator $a_{|Z}:=ae_a(Z)$ will be called the spectral window of a corresponding to the set Z or a restriction of a to Z. We shall write ${\mathcal {M}}_Z$ for the set $\{ a \in {\mathcal {M}}_h:a = a_{|Z} \}$ for $Z \in {\mathcal {B}}({\mathbb {R}})$. One notion we introduce deserves a separate definition:

Definition 1

Let $\mu \in [-1,1)$. We say that an operator $a \in {\mathcal {M}}_h$ is of type $\mu$, if $\max {{\,\mathrm{sp}\,}}a= 1$ and $\inf ({{\,\mathrm{sp}\,}}a\setminus \{0\}) = \mu$.

The content of the paper is as follows. In Sect. 2 we present the final result. We construct there a linear combination of three projections p, q, r,  equal to the given operator $a\in {\mathcal {M}}_h,$ for a $\sigma$-finite factor ${\mathcal {M}}$ of type III. The construction is different from analogous representations in other algebras ${\mathcal {M}}$ but is relatively simple. However, it uses a convoluted Lemma D from Sect. 2. We will obtain this lemma by a certain generalization of the construction of two projections in generic position. We will tentatively explain this at the end of Sect. 2.

The proof of Lemma D is contained in Sect. 5, it is based on the results of Sects. 3 and 4. The essential motivation leading to the results of Sects. 3 and 4 can be stated as follows. We have mappings $g_X(\cdot ),$ acting between subsets of the space ${\mathcal {Y}}={\mathcal {M}}_h$ (with operator norm). For a given $b\in {\mathcal {M}}_h$ we are looking for a solution $a\in {\mathcal {M}}_h$ of the relation

$$\begin{aligned} g_X(a)^+\sim b^+. \end{aligned}$$

(1)

In this sense we are looking for the inverse of an equivalent point.

We assume that X belongs to a neighborhood of 0 in a normed space ${\mathcal {N}}$ and describes a perturbation of the function $g_0(\cdot )$. Further, we assume that the mapping $(X,a)\mapsto g_X(a)$ is (locally) Lipschitzian and that there exist solutions a of the relation $g_0(a)^+\sim b^+$. We give additional sufficient conditions for a solution a of relation (1) to exist. In Sect. 3 we put forward a general scheme using any Banach space ${\mathcal {Y}}$ (the analogues of the relation “$\sim$” are certain isometries), cf. Theorem B.

In Sect. 4 we show and recall (for the convenience of the reader) some properties of projections in any von Neumann algebra. In particular, we describe how to estimate the perturbation of the spectral projection of a hermitian operator corresponding to its positive part by the perturbation of the operator itself (Lemma 23). The principal result, Theorem C, will then be an automatic conclusion of Theorem B.

The conclusion of Theorem C is similar to that of the aforementioned Lemma D, but the assumptions are slightly stronger. Applying Theorem C to the components of a suitable decomposition of the operator a yields Lemma D in full generality. This construction is described in Sect. 5.

In Sect. 6 we note that the proof of Theorem B reduces to the study of a special case of equation $f_X(a)=b,$ with unknown a sought in a Banach space ${\mathfrak {X}}$. The point b belongs to a unitary space ${\mathcal {Y}},$ while we interpret X as a perturbation (belonging to a unitary space ${\mathcal {N}}$). The examples and remarks in Sect. 6.1 are of a more general (and elementary) nature. The reasoning carried out in Sect. 6.2 rather non-trivially leads to a general scheme for studying spectral windows of an operator D. That is, we study the behaviour of spectral windows $D_{|Z}$ of D for an arbitrary Borel set Z under small perturbations of the operator (Theorem 37 with the operator D of the form $g_X(c))$. One application of this theorem is, in particular, a representation of an operator $a\in {\mathcal {M}}_h$ as a linear combination of 4 projections when ${\mathcal {M}}$ is a factor of type $II_1$ (Claim 41).

The entire final Sect. 6.3 is a brief summary of current problems concerning linear combinations of projections.

2 Main Result

MainTheorem

If ${\mathcal {M}}$ is a $\sigma$-finite factor of type III and $a \in {\mathcal {M}}_h$, then $a= \alpha p+ \alpha q+ \gamma r$ for some $p,q,r \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and $\alpha , \gamma \in {\mathbb {R}}$.

In this section, we shall describe a principal path leading to the result. The proof will be based on a technical Lemma D. Sections 3, 4 and 5 will be devoted to establishing the lemma. We start with the following theorem.

TheoremA

If ${\mathcal {M}}$ is a $\sigma$-finite factor of type III and a is an operator of type $\mu$ for some $\mu \in [-1,1)$, then $a+(1- \mu ) r=p+q$ for some $p,q,r \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$.

The Main Theorem is a direct consequence of Theorem A, by the following lemma:

Lemma 2

If ${\mathcal {M}}$ is an arbitrary von Neumann algebra and $a \in {\mathcal {M}}_h$ is not a multiple of a projection, then $\alpha a$ is of type $\mu$ for some $\alpha \ne 0, \mu \in [-1,1)$.

Proof

For $\beta =\inf ({{\,\mathrm{sp}\,}}a \setminus \{0\}),\, \gamma = \sup ({{\,\mathrm{sp}\,}}a \setminus \{0\})$ we have $\beta <\gamma$. It is enough to take $\alpha := 1/\beta ,\,\mu :=\gamma /\beta$ for $\beta < -\gamma$, and $\alpha := 1 / \gamma , \,\mu :=\beta /\gamma$ for $\beta \ge -\gamma$.    $\square$

The proof of Theorem A requires a decomposition of an operator of type $\mu$ into a series of mutually orthogonal operators of the same type. This is possible in any algebra without minimal projections, and is the content of a corollary to the following lemma:

Lemma 3

Let ${\mathcal {M}}$ be a von Neumann algebra without minimal projections. If a is of type $\mu ,\, \mu \in [-1,1)$, then a is a sum $a = \sum _{i \ge 0} a_i$ (in strong topology) of mutually orthogonal operators $a_i\in {\mathcal {M}}_h$ with $a_i \in {\mathcal {M}}_{[\mu , 1]}$ for $i \ge 0$ and $e_{a_i}^{[\mu , \mu + \varepsilon _i)}, e_{a_i}^{(1 - \varepsilon _i, 1]} \ne 0$ for $i \in {\mathbb {N}}$, for some numbers $\varepsilon _1> \varepsilon _2 > \ldots , \varepsilon _i \rightarrow 0$.

Proof

If $e_a^{\{\mu \}} \ne 0$, then we write $e_a^{\{\mu \}}$ as a sum $\Sigma _{i\in {\mathbb {N}}} p_i$ of mutually orthogonal, non-zero $p_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and put $b_i := \mu p_i$. If $e_a^{\{\mu \}} = 0$, then we find by induction positive numbers $\frac{1 - \mu }{2}> \delta _1> \delta _2 > \ldots , \delta _i \rightarrow 0$, satisfying $e_a^{[\mu , \mu + \delta _i]} \ne 0$ for $i \in {\mathbb {N}}$ and $e_a^{(\mu + \delta _i, \mu + \delta _{i-1}]} \ne 0$ for $i \ge 2,$ and put $b_i := ae_a^{(\mu + \delta _{i+1}, \mu + \delta _i]}$, for $i \in {\mathbb {N}}$. Operators $b_i$ are then mutually orthogonal, $b_i a = a b_i,\, 0 \ne b_i \in {\mathcal {M}}_{[\mu , \mu + \delta _i]}$ and ${{\,\mathrm{supp}\,}}b_i \le e_a^{[\mu , \frac{\mu + 1}{2})}$ for $i \in {\mathbb {N}}$. Similarly, we will find mutually orthogonal operators $c_i, \,ac_i = c_ia,\, 0 \ne c_i \in {\mathcal {M}}_{[1 - \gamma _i,1]}$ for some $\frac{1 - \mu }{2}> \gamma _1> \gamma _2 > \ldots ,$ $\gamma _i \rightarrow 0$ and ${{\,\mathrm{supp}\,}}c_i \le e_a^{(\frac{\mu + 1}{2}, 1]}$ for $i \in {\mathbb {N}}$. It is enough to put $a_i: = b_i + c_i,\,\varepsilon _i := \max (\delta _i, \gamma _i)$ for $i \in {\mathbb {N}}$, $a_0 := a - \sum _{i \in {\mathbb {N}}} a_i$. $\square$

Corollary 4

Let ${\mathcal {M}}$ be a von Neumann algebra without minimal projections. If a is of type $\mu ,\, \mu \in [-1,1)$, then $a = \sum _{j \in {\mathbb {N}}} a_j$ for some mutually orthogonal operators $a_j$ of type $\mu$.

Proof

We find a decomposition $a = \sum _{i \ge 0} a'_i$ as in Lemma 3. Then we put $a_1 := a'_0 + \sum _{i \in {\mathbb {N}}, 2\not \mid i} a'_i$, $a_j := \sum _{i \in {\mathbb {N}}, 2^{j-1} \mid i, 2^j \not \mid i} a'_i$ for $j = 2, 3, \ldots$. $\square$

Here is the statement of Lemma D, as formulated in Sect. 5 (cf. explanations at the end of this section):

LemmaD

Let ${\mathcal {M}}$ be a $\sigma$-finite factor of type III. If a is of type $\mu ,\, \mu \in [-1,1)$ and $b \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2- \mu ^+)}$ with $b \perp a$, then there is a decomposition $a = a' + a'',\, a' \perp a''$, in which $a'' \in {\mathcal {M}}_{[\mu , 1] }$ and there are $p,q,r \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and $c \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2-\mu ^+)}$ satisfying

$$\begin{aligned} a' + (1 - \mu ) r + b = p + q + c, \end{aligned}$$

with $c \perp p \vee q,\,p \vee q\le {{\,\mathrm{supp}\,}}(a'+b)$ and $r\vee {{\,\mathrm{supp}\,}}c \le {{\,\mathrm{supp}\,}}a'$.

Additionally, $p \wedge q = p \wedge q^{\perp } = p^{\perp } \wedge q = 0$.

For the proof of Theorem A, we need to get rid of the operator $a''$ from the decomposition of a in Lemma D. The following lemma is instrumental to this aim:

Lemma 5

Let ${\mathcal {M}}$ be an arbitrary von Neumann algebra. If an operator $a \in {\mathcal {M}}_{[\mu , 1]}$ for some $\mu \in [-1,1)$, then there are $r(a), s(a) \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $c(a) \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 1+\mu ^-)}$ (so in particular $c(a) \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2-\mu ^+)}$) satisfying

$$\begin{aligned} a + (1 - \mu ) r(a) = c(a) + s(a). \end{aligned}$$

Moreover, $r(a)\vee s(a) \vee {{\,\mathrm{supp}\,}}c(a) \le {{\,\mathrm{supp}\,}}a$ and $c(a) \perp s(a)$. $\square$

Proof

Note that

$$\begin{aligned} a + (1 - \mu ) e_a^{\{\mu \} \cup (\mu , 0)}&= a e_a^{\{\mu \}\cup (\mu ,0)\cup (\mu ^+, 1)\cup \{1\}} + (1 - \mu ) e_a^{\{\mu \}\cup (\mu , 0)}\\&= c(a) + s(a). \end{aligned}$$

Here, $c(a) := ae_a^{(\mu ^+, 1)} + (a + (1 - \mu ) \mathbb {1}_{{\mathcal {M}}}) e_a^{(\mu , 0)}$, $s(a) := ae_a^{\{\mu , 1\}}+(1-\mu )e_a^{\{\mu \}}=e_a^{\{\mu ,1\}}$. Moreover, $c(a) \perp s(a)$ and

$$\begin{aligned} c(a) \in {\mathcal {M}}_{(\mu ^+, 1) \cup [(1 - \mu ) + (\mu , 0)]} = {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 1+ \mu ^-)}. \end{aligned}$$

Putting $r(a):= e_a^{\{\mu \} \cup (\mu , 0)}$ ends the proof. $\square$

Lemma 5 leads to the following strengthening of Lemma D we need to prove Theorem A.

Corollary 6

Let ${\mathcal {M}}$ be a $\sigma$-finite factor of type III. If a is of type $\mu ,\, \mu \in [-1,1)$ and $b \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2- \mu ^+)}$ with $b \perp a$, then there are $p,q,r \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and $c \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2-\mu ^+)}$ satisfying

$$\begin{aligned} a + (1 - \mu ) r + b = p + q + c, \end{aligned}$$

with $c \perp p \vee q,\,p \vee q\le {{\,\mathrm{supp}\,}}(a+b)$ and $r\vee {{\,\mathrm{supp}\,}}c \le {{\,\mathrm{supp}\,}}a$.

Proof

Let, in accordance with Lemma D,

$$\begin{aligned}&a = a' + a'',\, a' \perp a'',\, a' + (1 - \mu ) r' + b = p' + q +c',\\&r' \le {{\,\mathrm{supp}\,}}a',\,c' \perp p' \vee q ,\, p' \vee q\le {{\,\mathrm{supp}\,}}(a' + b),\,{{\,\mathrm{supp}\,}}c'\le {{\,\mathrm{supp}\,}}a' \end{aligned}$$

for some $p', q, r' \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $c' \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2 - \mu ^+)}$ and $a'' \in {\mathcal {M}}_{[\mu , 1]}$. We apply Lemma 5 to the operator $a''$ and get $r(a'')$, $s(a'') \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $c(a'') \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2 - \mu ^+)}$. We may now define $r := r' + r(a''),$ $p := p' + s(a'')$ and $c := c' + c(a'')$. They are sums of mutually orthogonal terms, and the result follows easily from $a'' \perp a' + b$. $\square$

Proof of Theorem A

As in Corollary 4, we take a decomposition $a = \sum _{i \in {\mathbb {N}}} a_i$ into mutually orthogonal terms of type $\mu$. We put $b_0 := 0$. If for $i \in {\mathbb {N}}$ one has an operator $b_{i-1} \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2 - \mu ^+)}$, satisfying additionally ${{\,\mathrm{supp}\,}}b_{i-1} \le {{\,\mathrm{supp}\,}}a_{i-1}$ in the case $i \ge 2$, then, following Corollary 6, we define $r_i, p_i, q_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}},\,c_i \in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2 - \mu ^+)}$ satisfying $a_i + (1 - \mu ) r_i + b_{i-1} = p_i + q_i + c_i,\, r_i\vee {{\,\mathrm{supp}\,}}c_i \le {{\,\mathrm{supp}\,}}a_i,\, c_i \perp p_i \vee q_i$, $p_i \vee q_i \le {{\,\mathrm{supp}\,}}(a_i + b_{i-1})$. Then we put $b_i := c_i$, getting in particular ${{\,\mathrm{supp}\,}}b_i \le {{\,\mathrm{supp}\,}}a_i, b_i\perp a_{i+1}$ and $b_i\in {\mathcal {M}}_{(\mu ^+, 1) \cup (1, 2 - \mu ^+)}$.

For constructed inductively projections $p_i, q_i,r_i,$ we put $p := \sum _{i\in {\mathbb {N}}} p_i$, $q := \sum _{i\in {\mathbb {N}}} q_i,\, r := \sum _{i\in {\mathbb {N}}} r_i$. To show that p, q, r are well defined projections, we note that $p_i \perp p_j$ for $|i-j| \ge 2,$ since $p_i \le {{\,\mathrm{supp}\,}}(a_i + b_{i-1})$ for $i \in {\mathbb {N}}$, that is $p_i \le {{\,\mathrm{supp}\,}}(a_i + a_{i-1})$ for $i \ge 2$, $p_1 \le {{\,\mathrm{supp}\,}}a_1$. Moreover, $p_i \perp p_{i+1}$ since $p_i \perp b_i, a_{i+1}$, and $p_{i+1} \le {{\,\mathrm{supp}\,}}(a_{i+1} + b_i)$, for $i \in {\mathbb {N}}$. Hence projections $p_i$ are mutually orthogonal, and the same is true of $q_i$. Projections $r_i$ are also mutually orthogonal (since $r_i \le {{\,\mathrm{supp}\,}}a_i$ for $i \in {\mathbb {N}}$). We also have $\sum _{i\in {\mathbb {N}}} (a_i + (1 - \mu ) r_i) = \sum _{i\in {\mathbb {N}}} (-b_{i-1} + p_i + q_i + b_i) = \sum _{i\in {\mathbb {N}}} p_i + \sum _{i\in {\mathbb {N}}} q_i$, since $b_0 = 0$.

$\square$

A more precise description of the proof of Lemma D is as follows.

We will decompose the operator a into mutually orthogonal, commuting components $a', a''$ (so that $a=a'+a''$). We will obtain $a'=e-X+e'+X'$ for certain projections $e,e'$ satisfying $e\perp e',\,e\sim e'$, and correspondingly small perturbations $X\in {\mathcal {M}}_e^+,\,X'\in {\mathcal {M}}_{e'}^+$. We will develop a technique for constructing a couple of pairs of projections in generic position. The projections e, r will be in a generic position in ${\mathcal {M}}_{e+e'}$ and p, q in a generic position in ${\mathcal {M}}_{p\vee q}$. Then

$$\begin{aligned} \begin{aligned}&B:= e+\mu e'+(1-\mu )r\in ({\mathcal {M}}_{e+e'})_{(\mu ,1)\cup (1,2-\mu )}, \\&C:=p+q\in ({\mathcal {M}}_{p\vee q})_{(0,1)\cup (1,2)}. \end{aligned} \end{aligned}$$

(2)

Since $(\mu ^+,1)\cup (1,2-\mu ^+)=((\mu ,1)\cup (1,2-\mu ))\cap ((0,1)\cup (1,2))$, for an operator $b\in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2-\mu ^+)}$ with ${{\,\mathrm{supp}\,}}b\perp e+e'$ the operators B and C in (2) can be constructed in such a way that $B+b=C+c$ for some $c\perp p\vee q,\, c\in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2-\mu ^+)}$. The assumption

$$\begin{aligned} b\in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2-\mu ^+)} \end{aligned}$$

(3)

is here essential. We will indicate methods by which the construction of the operators p, q, r can be obtained also for non-zero perturbations $X,X'$. Assumption (3) is essential also in Corollary 6 (and has an impact on the form of the whole developed method).

3 An Inverse of an Equivalent Point

The main result of this section is contained in Theorem B. Here we deal with general Banach (or just normed) spaces. We will remind you of a notion of continuity that will be used intensively in the sequel.

Definition 7

Let ${\mathcal {Y}},{\mathcal {Z}}$ be normed (real or complex) spaces. A function $f:{\mathcal {D}}\rightarrow {\mathcal {Y}}$ with ${\mathcal {D}}\subset {\mathcal {Z}}$ is locally Lipschitzian if for all $b_0 \in {\mathcal {D}}$ there are $\delta> 0,\,K > 0$ such that for any $b,c \in {\mathcal {D}}$, if $\Vert b - b_0 \Vert <\delta$ and $\Vert c - b_0 \Vert < \delta ,$ then $\Vert f(b) - f(c)\Vert \le K \Vert b - c \Vert$.

We note a well known fact (proved here for completeness):

Lemma 8

If ${\mathcal {Y}}$ is a normed space and if $u,v:{\mathcal {Y}}\rightarrow {\mathcal {Y}}$ are (linear surjective) isometries, then $\Vert u-v\Vert =\Vert u^{-1}-v^{-1}\Vert$.

Proof

We have, for all $y\in {\mathcal {Y}},$

$$\begin{aligned} \Vert u(y) - v(y) \Vert&=\Vert u^{-1} (u(y)) - u^{-1} (v(y)) \Vert \\&=\Vert v^{-1} (v(y)) - u^{-1} (v(y)) \Vert \\&\le \Vert u^{-1} - v^{-1}\Vert \Vert v\Vert \Vert y \Vert , \end{aligned}$$

so that $\Vert u - v \Vert \le \Vert u^{-1} - v^{-1} \Vert ,$ and (replacing u and v by their inverses) $\Vert u - v \Vert \ge \Vert u^{-1} - v^{-1} \Vert$. $\square$

We shall need the following corollary of Banach’s contraction principle:

Lemma 9

Fix $\varepsilon >0$. Let $f : {\mathbf {B}}(b_0; \varepsilon ) \rightarrow {\mathcal {D}}$ be a function defined on the closed ball ${\mathbf {B}}(b_0; \varepsilon )$ contained in a Banach space ${\mathcal {D}}$. If

$$\begin{aligned} \Vert f(b) - f(c) - (b-c) \Vert \le \Vert b - c \Vert /3 \text { for } b, c \in {\mathbf {B}}(b_0; \varepsilon ) \end{aligned}$$

and

$$\begin{aligned} \Vert f(b_0) - b_0 \Vert \le \varepsilon /3, \end{aligned}$$

then for $b \in {\mathbf {B}}(b_0; \varepsilon /3)$ there exists an $a \in {\mathbf {B}}(b_0; \varepsilon )$ such that $f(a) = b$.

Proof

Let $\Vert b - b_0 \Vert \le \varepsilon /3$. The function $h_b$ defined by ${\mathbf {B}}(b_0;\varepsilon )\ni c\mapsto b-f(c)+ c\in {\mathcal {D}}$ satisfies $\Vert h_b(c) - h_b(d) \Vert \le \Vert c - d \Vert /3$ for $c, d \in {\mathbf {B}}(b_0; \varepsilon )$, so that it is a contraction. For $c \in {\mathbf {B}}(b_0; \varepsilon )$ we have $\Vert h_b(c) - b_0 \Vert \le \Vert h_b(c) - h_b(b_0) \Vert + \Vert h_b(b_0) - b \Vert + \Vert b - b_0 \Vert \le \Vert c - b_0 \Vert /3 + \Vert -f(b_0) + b_0 \Vert + \Vert b - b_0 \Vert \le \varepsilon /3 +\varepsilon /3 + \varepsilon /3$, so that $h_b(c) \in {\mathbf {B}}(b_0; \varepsilon )$. By Banach’s contraction principle, there exists an $a \in {\mathbf {B}}(b_0; \varepsilon )$ satisfying $h_b(a) = a$, so that $f(a) = b$. $\square$

Set up for Theorem B: We fix a (real or complex) Banach space ${\mathcal {Y}}$ and a group G of (linear surjective) isometries of ${\mathcal {Y}},$ with the unit of G denoted by $\mathrm{id}$. Further, ${\mathcal {W}}$, ${\mathfrak {X}}$ are subspaces of ${\mathcal {Y}}$ with ${\mathcal {W}}\cap {\mathfrak {X}}= \{0\}$, and e is a projection of ${\mathfrak {X}}+ {\mathcal {W}}$ onto ${\mathfrak {X}}$ of the form $e(b+d)=b$ for $b\in {\mathfrak {X}}$ and $d\in {\mathcal {W}}$. We assume that $\Vert e \Vert = 1$. Finally, we fix three sets: ${\mathfrak {X}}_0$ open in ${\mathfrak {X}}, \,{\mathcal {W}}_0$ open in ${\mathcal {W}},$ and then ${\mathcal {Y}}_0$ open in ${\mathcal {Y}}$ and such that ${\mathfrak {X}}_0 + {\mathcal {W}}_0 \subset {\mathcal {Y}}_0$.

Definition 10

For $b\in {\mathfrak {X}}_0,\, C \in {\mathcal {Y}}$, we say that C is represented by b and write $C\vartriangleright b$ if, for some $v \in G$, $v(C) \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$ and $b = ev(C)$.

Note that $C\vartriangleright b$ means that $v(C)= b+d$ for some $v\in G$ and $d\in {\mathcal {W}}_0$.

We shall need results of the following type: For a $b \in {\mathfrak {X}}_0,$ a function $g: {\mathfrak {X}}_0\rightarrow {\mathcal {Y}}$ and a perturbation X from ${\mathcal {Y}}$ there is an $a\in {\mathfrak {X}}_0$ (depending on X and b) satisfying

$$\begin{aligned} g(a)+X \vartriangleright b\qquad \qquad \qquad \qquad \qquad (*) \end{aligned}$$

(cf. Theorem B below).

The application of Definition 10 and Theorem B in constructions of projections will be described in Sect. 4. Some generalizations and connections between the existence of points a solving (*) and the existence of fixed points will be explained in Sect. 6.

Theorem B uses fairly special assumptions on functions g and the group G.

Definition 11

Assume that a family of isometries ${\mathcal {U}}\subset G$ is such that the system of sets $\{u^{-1} ({\mathfrak {X}}_0 + {\mathcal {W}}_0):u \in {\mathcal {U}}\}$ is a partition of ${\mathcal {Y}}_0$ and $\mathrm{id}\in {\mathcal {U}}$ (so that ${\mathfrak {X}}_0 + {\mathcal {W}}_0$ belongs to the elements of the partition). Then one has a well-defined map $\Phi :{\mathcal {Y}}_0 \ni y \mapsto \Phi _y \in {\mathcal {U}}$, $\Phi _y = u$ if $y \in u^{-1} ({\mathfrak {X}}_0 + {\mathcal {W}}_0)$. (In particular, $\Phi _y = \mathrm{id}$ for $y \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$). We say that ${\mathcal {U}}$ partitions ${\mathcal {Y}}_0$ if $\Phi$ is locally Lipschitzian.

Definition 12

A function $g : {\mathfrak {X}}_0 \rightarrow {\mathcal {Y}}_0$ agrees with a partitioning family ${\mathcal {U}}$ if, for each $b\in {\mathfrak {X}}_0,$ $b=e\Phi _{g(b)} (g(b)),$ and the function g is locally Lipschitzian. Then in particular $g(b) \vartriangleright b$.

The main result of the section is

TheoremB

If a function $g : {\mathfrak {X}}_0 \rightarrow {\mathcal {Y}}_0$ agrees with some partitioning family ${\mathcal {U}}$, then it satisfies the following property:

$$\begin{aligned}&\text {For any }b_0 \in {\mathfrak {X}}_0 \text { there is a }\delta > 0 \text { such that}\\&\quad \text { for all }b \in {\mathfrak {X}}_0,\, X \in {\mathcal {Y}}\text { with } \Vert b - b_0 \Vert , \Vert X \Vert < \delta \qquad \qquad \qquad (**)\\&\quad \text { there is an }a \in {\mathfrak {X}}_0 \text { such that }g(a) + X \vartriangleright b. \end{aligned}$$

To make following the text easier, we collect first basic properties of a partitioning family and a function agreeing with it.

Lemma 13

If ${\mathcal {U}}$ is a partitioning family, then

• $1^\circ$ for any $a \in {\mathcal {Y}}_0$ there are $\delta , K > 0$ such that

  $$\begin{aligned} b, c \in {\mathcal {Y}}\text { and } \Vert b - a \Vert ,\Vert c - a \Vert < \delta \end{aligned}$$

  implies

  $$\begin{aligned} b, c \in {\mathcal {Y}}_0 \text { and }\Vert \Phi _b - \Phi _c \Vert ,\Vert \Phi _b^{-1} - \Phi _c^{-1} \Vert \le K \Vert b - c \Vert ; \end{aligned}$$

• $2^\circ$ $\Phi _a (a) \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$ for $a \in {\mathcal {Y}}_0$;

• $3^\circ$ $\Phi _a = \mathrm{id}$ for $a \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$;

• $4^\circ$ for $a \in {\mathcal {Y}}_0$, $c \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$ we have $\Phi ^{-1}_a (c) \in {\mathcal {Y}}_0$ and $\Phi _{\Phi ^{-1}_a (c)} = \Phi _a$.

If additionally a function $g : {\mathfrak {X}}_0 \rightarrow {\mathcal {Y}}_0$ agrees with ${\mathcal {U}}$, then

• $1^{\circ \circ }$ for $b_0 \in {\mathfrak {X}}_0$ there exist $\delta , K_1 > 0$ such that for $b, c \in {\mathfrak {X}}$, $\Vert b - b_0 \Vert$, $\Vert c - b_0 \Vert < \delta$ implies $b, c \in {\mathfrak {X}}_0$ and $\Vert g(b) - g(c) \Vert \le K_1 \Vert b - c\Vert$;

• $2^{\circ \circ }$ for $b \in {\mathfrak {X}}_0$ there exists a $d \in {\mathcal {W}}_0$ such that $\Phi _{g(b)} (g(b)) = b+ d$.

The lemma follows directly from Definitions 11, 12 and Lemma 8.

Proof of Theorem B

We assume that the function g under consideration agrees with a partitioning family ${\mathcal {U}}$. To facilitate the proof of Theorem B, we fix the function g and define functions: $\Psi : {\mathfrak {X}}_0\rightarrow {\mathfrak {X}}_0+{\mathcal {W}}_0$ by $\Psi : a\mapsto \Phi _{g(a)}(g(a))$ and $\Omega : {\mathcal {Y}}\times {\mathfrak {X}}_0\rightarrow {\mathcal {Y}}$ by $\Omega :(X,a)\mapsto \Phi _{g(a)}(X)$. Note that by $2^\circ$ and our assumption on g, both $\Psi$ and $\Omega$ are well defined. For the arguments that follow, we introduce the following simplifying notations: For all $b_0, b, c \in {\mathfrak {X}}_0$ and all $X \in {\mathcal {Y}},$

$$\begin{aligned} \begin{aligned} B_0 := \Psi (b_0),\, B := \Psi (b),\, C := \Psi (c),\\ Y_0:=\Omega (X,b_0),\,Y := \Omega (X,b),\, Z := \Omega (X,c). \end{aligned} \end{aligned}$$

(4)

Note that $Y_0, Y$ and Z depend on X and $\Vert Y_0\Vert =\Vert Y\Vert =\Vert Z\Vert =\Vert X\Vert ,$ and by $2^\circ$ of the previous lemma, $B_0, B$ and C belong to ${\mathfrak {X}}_0+{\mathcal {W}}_0$. Moreover, since the function g agrees with the partitioning family ${\mathcal {U}},$ we have $e\Psi (a)=a$ for all $a\in {\mathfrak {X}}_0$.

Step 1. For $b_0 \in {\mathfrak {X}}_0$ there exist $\varepsilon ,K_2 > 0$ such that for $b, c \in {\mathfrak {X}}_0,\, X \in {\mathcal {Y}}$ with $\Vert b-b_0 \Vert , \Vert c-b_0 \Vert , \Vert X \Vert < \varepsilon$, we have

$$\begin{aligned} B + Y, C + Z \in {\mathcal {Y}}_0 \end{aligned}$$

(5)

and

$$\begin{aligned} \Vert \Phi _{B+Y} - \Phi _{C+Z}\Vert \le K_2 \Vert X \Vert \Vert b-c \Vert . \end{aligned}$$

(6)

Proof

Let us fix $b_0 \in {\mathfrak {X}}_0,$ and choose $\delta > 0$ so small that for some K, $K_1>0$ we have the implication in $1^{\circ \circ },$ and the implication in $1^\circ$ both for $a:=g(b_0)$ and for $a:=B_0$.

Note that

$$\begin{aligned} {\mathcal {Y}}_0 \ni a \mapsto \Phi _a (a) \in {\mathcal {Y}}\text { is continuous;} \end{aligned}$$

(7)

in fact, $\Vert \Phi _a(a)-\Phi _{a'}(a') \Vert \le \Vert (\Phi _a - \Phi _{a'})(a) \Vert + \Vert \Phi _{a'} \Vert \Vert a-a'\Vert$. By the continuity of composition of continuous mappings, both $\Psi$ and $\Omega$ are continuous, and there exists an $\varepsilon > 0$ such that for $X \in {\mathcal {Y}}$, $b,c \in {\mathfrak {X}}_0$ with $\Vert X\Vert ,\Vert b-b_0 \Vert ,\Vert c-b_0 \Vert < \varepsilon$, we have $b,c \in {\mathbf {B}}(b_0; \delta ) \cap {\mathfrak {X}}_0$, as well as

$$\begin{aligned}&g(b), g(c), g(b)+X, g(c)+X \in {\mathbf {B}}(g(b_0); \delta ) \cap {\mathcal {Y}}_0, \\&\quad B+Y, C+Z, B+Y+\Phi _{B+Y}^{-1}(C-B) \in {\mathbf {B}}(B_0; \delta ) \cap {\mathcal {Y}}_0, \end{aligned}$$

which contains (5), and, by $2^\circ$,

$$\begin{aligned} \Phi _{B+Y}(B+Y),\Phi _{B+Y}(B+Y)+C-B\in {\mathbf {B}}(B_0;\delta )\cap ({\mathfrak {X}}_0+{\mathcal {W}}_0). \end{aligned}$$

In particular, for $\Vert b-b_0 \Vert ,\Vert c-b_0 \Vert ,\Vert X \Vert < \varepsilon$ we have, by $4^\circ$,

$$\begin{aligned} \Phi _{ \Phi _{B+Y}^{-1}(\Phi _{B+Y}(B+Y)+C-B)}=\Phi _{B+Y} \end{aligned}$$

(8)

and

$$\begin{aligned} \Vert \Phi _{B+Y} - \Phi _{C+Z} \Vert&= \Vert \Phi _{B+Y+\Phi _{B+Y}^{-1}(C-B)} - \Phi _{C+Z}\Vert \text { by }(4)\\&\le K \Vert B+Y+\Phi _{B+Y}^{-1}(C-B)-C-Z \Vert \text { by } 1^\circ \\&\le K\Vert Y-Z\Vert +K \Vert B-C+\Phi _{B+Y}^{-1}(C-B)\Vert \\&\le K\Vert Y-Z\Vert +K \Vert B-C+\Phi _B^{-1}(C-B)\Vert \\&\quad +K\Vert (\Phi _{B+Y}^{-1}-\Phi _B^{-1})(C-B)\Vert \\&\le K\Vert Y-Z\Vert +0+K \Vert \Phi _{B+Y}^{-1}-\Phi _B^{-1}\Vert \Vert C-B\Vert \,\text { by } 3^\circ \\&\le K\Vert Y-Z\Vert +K^2 \Vert Y\Vert \Vert C-B\Vert \text { by } 1^\circ . \end{aligned}$$

Moreover,

$$\begin{aligned} \begin{aligned} \Vert Y-Z\Vert&\le \Vert \Phi _{g(b)}-\Phi _{g(c)}\Vert \Vert X\Vert \\&\le K\Vert g(b)-g(c)\Vert \Vert X\Vert \text { by } 1^\circ \\&\le KK_1 \Vert b-c\Vert \Vert X\Vert \text { by } 1^{\circ \circ } \end{aligned} \end{aligned}$$

(9)

and

$$\begin{aligned} \begin{aligned} \Vert B-C\Vert&\le \Vert (\Phi _{g(b)}-\Phi _{g(c)})(g(b))\Vert +\Vert \Phi _{g(c)}(g(b)-g(c))\Vert \\&\le \Vert \Phi _{g(b)}-\Phi _{g(c)}\Vert \Vert g(b)\Vert +\Vert g(b)-g(c)\Vert \\&\le (K\Vert g(b)\Vert +1) \Vert g(b)-g(c)\Vert \text { by } 1^\circ \\&\le [K(\Vert g(b_0)\Vert + K_1 \varepsilon )+1] \cdot K_1 \Vert b-c\Vert \text { by } 1^{\circ \circ }. \end{aligned} \end{aligned}$$

(10)

Now, both the existence of a constant $K_2>0$ depending only on $b_0$ and inequality (6) are proved. $\square$

Step 2. For $b_0 \in {\mathfrak {X}}_0$ there exists an $\varepsilon _1 > 0$ such that for $b,c \in {\mathfrak {X}}_0,\,X \in {\mathcal {Y}}$ with $\Vert b-b_0\Vert ,\Vert c-b_0\Vert ,\Vert X\Vert < \varepsilon _1$, (5) holds (with notation (4)) and

$$\begin{aligned} \Vert \Phi _{B+Y}(B+Y)-\Phi _{C+Z}(C+Z)-B+C\Vert \le \Vert b-c\Vert /3. \end{aligned}$$

Proof

By Step 1, $1^\circ$ and $1^{\circ \circ }$, we can choose $\varepsilon _1 > 0$ so small that there exist constants $K, K_1, K_2 >0$ (depending only on $b_0$) such that for $\Vert b-b_0\Vert ,\Vert c-b_0\Vert ,\Vert X\Vert <\varepsilon _1$

$$\begin{aligned} \Vert \Phi _{B+Y}-\Phi _{C+Z}\Vert&\le K_2 \Vert X\Vert \Vert b-c\Vert ,\\ \Vert g(b)-g(c)\Vert&\le K_1 \Vert b-c\Vert , \end{aligned}$$

and $\Vert g(b)-g(b_0)\Vert , \Vert g(c)-g(b_0)\Vert$ are small enough to guarantee that

$$\begin{aligned} \Vert \Phi _{g(b)}-\Phi _{g(c)}\Vert \le K \Vert g(b)-g(c)\Vert . \end{aligned}$$

Moreover, we can assure estimations (9) and (10) and, using (10), we can get $\Vert C-B_0\Vert$, $\Vert (C+Z)-B_0\Vert$ so small that, by $1^\circ$,

$$\begin{aligned} \Vert \Phi _C-\Phi _{C+Z}\Vert \le K \Vert Z\Vert . \end{aligned}$$

Then we have

$$\begin{aligned}&\Vert \Phi _{B+Y}(B+Y)-\Phi _{C+Z}(C+Z)-B+C\Vert \\&\quad \le \Vert (\Phi _{B+Y}-\Phi _{C+Z})(B+Y)\Vert +\Vert \Phi _{C+Z}(B+Y-C-Z)-B+C\Vert \\&\quad \le K_2 \Vert X\Vert \Vert b-c\Vert \Vert B+Y\Vert +\Vert Y-Z\Vert \\&\qquad +\Vert \Phi _{C+Z}(B-C)-\Phi _C(B-C)\Vert \end{aligned}$$

(by Step 1 and $3^\circ , 2^\circ$)

$$\begin{aligned}&\le K_2(\Vert B_0\Vert +\Vert B-B_0\Vert +\Vert Y\Vert ) \Vert X\Vert \Vert b-c\Vert +KK_1 \Vert X\Vert \Vert b-c\Vert \\&\quad + K \Vert Z\Vert \Vert B-C\Vert \text { by }(9)\text { and } 1^\circ . \end{aligned}$$

Inequality (10) may be applied to $\Vert B-C\Vert$, and (in particular) to $\Vert B-B_0\Vert$. Hence there exists a constant $K_3$ depending on $b_0$ such that for some $\varepsilon _1 >0$ (depending on $b_0$) we have

$$\begin{aligned}&\Vert \Phi _{B+Y}(B+Y)-\Phi _{C+Z}(C+Z)-B+C\Vert \le K_3 \Vert X\Vert \Vert b-c\Vert . \end{aligned}$$

It is enough to require additionally $\varepsilon _1 < 1/(3 K_3)$. $\square$

Step 3. For $b_0 \in {\mathfrak {X}}_0$ and $\varepsilon _1 > 0$ chosen in Step 2, one can find an $\varepsilon _2 \in (0, \varepsilon _1)$ such that $\Vert X\Vert < \varepsilon _2$ implies

$$\begin{aligned} \Vert \Phi _{B_0+Y_0}(B_0+Y_0)-B_0\Vert < (1/3) \varepsilon _1. \end{aligned}$$

Proof

By $3^\circ$, we have $\Phi _{B_0}(B_0)=B_0$. Hence it is enough to use (7).

Step 4. For $b_0 \in {\mathfrak {X}}_0$, for $\varepsilon _1 >0$ as in Step 2, $\varepsilon _2 >0$ as in Step 3 and for $X \in {\mathcal {Y}}$ satisfying $\Vert X\Vert < \varepsilon _2$, the following condition holds: for $b \in {\mathfrak {X}}_0$ with $\Vert b-b_0\Vert < \varepsilon _1/3$, there exists an $a \in {\mathfrak {X}}_0$ such that we have

$$\begin{aligned} e(\Phi _{\Phi _{g(a)}(g(a)+X)} \circ \Phi _{g(a)})(g(a)+X)=b. \end{aligned}$$

(11)

Proof

For a fixed $b_0 \in {\mathfrak {X}}_0$, for $\varepsilon _1 >0$ as in Step 2, for $\varepsilon _2$ as in Step 3 and for a fixed $X \in {\mathcal {Y}}$ satisfying $\Vert X\Vert < \varepsilon _2$, we define a function $f_X:{\mathbf {B}}(b_0; \varepsilon _1) \rightarrow {\mathfrak {X}}_0$ on ${\mathbf {B}}(b_0; \varepsilon _1) \subset {\mathfrak {X}}_0$ by $f_X: a\mapsto e(\Phi _A(A)),$ where $A:=\Psi (a)+\Omega (X,a)$. By $2^\circ$, the function $f_X$ is well defined for all $X\in {\mathcal {Y}},\,\Vert X\Vert <\varepsilon _2$ (since $A \in {\mathcal {Y}}_0$). Note that $f_X(b_0)=e\Phi _{B_0+Y_0}(B_0+Y_0),\,f_X(b)=e\Phi _{B+Y}(B+Y)$ and $f_X(c)=e\Phi _{C+Z}(C+Z)$. Moreover, for $b,c\in {\mathbf {B}}(b_0;\varepsilon _1)$ we have $e(B-C)=b-c$, by $2^{\circ \circ };$ hence Step 2 yields

$$\begin{aligned}&\Vert f_X(b)-f_X(c)-(b-c)\Vert \\&\quad =\Vert e(\Phi _{B+Y}(B+Y)-\Phi _{C+Z}(C+Z)-(B-C))\Vert \\&\quad \le \Vert b-c\Vert /3 \end{aligned}$$

(we assumed that $\Vert e\Vert =1$ in the space ${\mathfrak {X}}+ {\mathcal {W}}$). Similarly, by $2^{\circ \circ }$ Step 3 gives

$$\begin{aligned} \Vert f_X(b_0)-b_0\Vert&= \Vert e(\Phi _{B_0+Y_0}(B_0+Y_0)-B_0)\Vert \\&\le \Vert \Phi _{B_0+Y_0}(B_0+Y_0)-B_0\Vert < \varepsilon _1/3. \end{aligned}$$

By Lemma 9, for $b \in {\mathfrak {X}}_0$ with $\Vert b-b_0\Vert \le \varepsilon _1/3,$ there exists a $a \in {\mathfrak {X}}_0$ satisfying $f_X(a)=b$. This is equivalent to (11). $\square$

Step 4 finishes the proof of Theorem B. In fact, (11) means that $g(a)+X\vartriangleright b;$ it is enough to take $\delta :=\min (\varepsilon _2,\varepsilon _1/3)$. $\square$

4 Inverse to an Equivalent Element in Construction of Operators

In this section ${\mathcal {M}}$ is an arbitrary von Neumann algebra. Let us fix a projection e satisfying $e \simeq e^{\perp }$ (or, equivalently, $e\sim e^\perp$) and a number $\eta \in (0,1]$.

We shall make frequent use of $2\times 2$ matrix units. We start with gathering various properties of a pair of projections, most of them certainly known, and given here for completeness. Take first any $p,q\in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$. Then we easily check that

$$\begin{aligned} qpq + q^{\perp }p^{\perp }q^{\perp } = \mathbb {1}- (p-q)^2, \end{aligned}$$

(12)

so that, in particular, $(p-q)^2\le \mathbb {1}$ and $\Vert p-q\Vert \le 1$. We put

$$\begin{aligned} \begin{aligned} p_0&:=p-p\wedge q- p\wedge q^\perp ,\\ q_0&:=q-p\wedge q-p^\perp \wedge q, \end{aligned} \end{aligned}$$

(13)

so that $p_0,q_0$ are in generic position in ${\mathcal {M}}_{p_0\vee q_0}$. In particular, $p_0\sim q_0$ and $q_0\sim p_0\vee q_0-q_0$. The latter equivalence allows us to treat operators from ${\mathcal {M}}_{p_0\vee q_0}$ as $2\times 2$ matrices over the reduced algebra ${\mathcal {M}}_{q_0}$. Thus

$$\begin{aligned} \left[ \begin{array}{cc} a_{11} &{} a_{12} \\ a_{21} &{} a_{22} \\ \end{array} \right] = a_{11}+w_0a_{22}w_0^{*}+a_{12}w_0^{*}+w_0a_{21} \end{aligned}$$

(14)

for $a_{ij}\in {\mathcal {M}}_{q_0}$ (note that $\mathbb {1}_{{\mathcal {M}}_{q_0}} = q_0),$ where $w_0 \in {\mathcal {M}}$ is a partial isometry satisfying $w_0^* w_0=q_0\sim p_0\vee q_0-q_0 = w_0w_0^*$. Then

$$\begin{aligned} q_0=\left[ \begin{array}{cc} 1&{} 0\\ 0 &{} 0 \end{array} \right] ,\,p_0\vee q_0-q_0=\left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} 1\end{array} \right] \text { and } w_0=\left[ \begin{array}{cc} 0 &{} 0\\ 1&{} 0 \end{array} \right] . \end{aligned}$$

(15)

(where $1:=\mathbb {1}_{{\mathcal {M}}_{q_0}}$). Moreover, for a suitably chosen $w_0,$

$$\begin{aligned} p_0= \left[ \begin{array}{cc} c^{2} &{} cs \\ cs &{} s^{2} \end{array} \right] \end{aligned}$$

(16)

for some commuting $c, s\in {\mathcal {M}}_{q_0}$ with $0\le c,s\le \mathbb {1}_{{\mathcal {M}}_{q_0}}$ and $c^2+s^2=\mathbb {1}_{{\mathcal {M}}_{q_0}},$ with $e_c(\{0\})=e_s(\{0\})=0$.

Note that

$$\begin{aligned} \begin{aligned} \Vert p-q\Vert&=\Vert p_0-q_0 + p\wedge q^\perp - p^\perp \wedge q\Vert \\&=\max \{\Vert p_0-q_0\Vert ,\Vert p\wedge q^\perp \Vert ,\Vert p^\perp \wedge q\Vert \}, \end{aligned} \end{aligned}$$

(17)

hence

$$\begin{aligned} \Vert p-q\Vert <1 \text { implies }p\wedge q^\perp = p^\perp \wedge q=0. \end{aligned}$$

(18)

Lemma 14

For any $p, q \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$,

$$\begin{aligned} \Vert p-q \Vert ^2 = \max \{\Vert p - pqp \Vert , \Vert q - qpq\Vert \}. \end{aligned}$$

(19)

Proof

Note that

$$\begin{aligned} \max \{\Vert p - pqp \Vert , \Vert q - qpq\Vert \}=\max \{\Vert pq^\perp p \Vert , \Vert qp^\perp q\Vert \}\le 1. \end{aligned}$$

With (13),

$$\begin{aligned} \Vert p-pqp\Vert =\Vert p_0-p_0q_0p_0+p\wedge q^\perp \Vert \ge \Vert p\wedge q^\perp \Vert , \end{aligned}$$

so by (12) and (18), if $p\wedge q^\perp \ne 0$ or $p^\perp \wedge q\ne 0,$ then both sides of (19) are equal to 1. If $p\wedge q^\perp = p^\perp \wedge q=0$, then with (15) and (16), we have by (12) and (17),

$$\begin{aligned} \Vert q - q p q \Vert&= \Vert q_0 - q_0 p_0 q_0 \Vert = \Vert \left[ \begin{array}{cc} s^2 &{} 0\\ 0 &{} 0 \end{array} \right] \Vert = \Vert \left[ \begin{array}{cc} s^2 &{} 0\\ 0 &{} s^2 \end{array} \right] \Vert \\&= \Vert q_0 - q_0 p_0 q_0 + q_0^{\perp } - q_0^{\perp } p_0^{\perp } q_0^{\perp } \Vert = \Vert (p_0 - q_0)^2 \Vert =\Vert p-q\Vert ^2. \end{aligned}$$

The exchange of p and q finishes the proof. $\square$

If $p\in {{\,\mathrm{Proj}\,}}{\mathcal {M}},$ then $p_0,$ defined as in (13) with $q=e$ is in generic position with

$$\begin{aligned} e_0:=e-e\wedge p - e\wedge p^\perp \end{aligned}$$

(20)

in ${\mathcal {M}}_{e_0\vee p_0},$ and we have (16) for some $c,s\in {\mathcal {M}}_{e_0}$ with $0\le c,s\le \mathbb {1}_{{\mathcal {M}}_{e_0}}$ and $c^2+s^2=\mathbb {1}_{{\mathcal {M}}_{e_0}}$.

Notation 15

Let $p \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ with $epe+e^\perp p^\perp e^\perp$ invertible. We put

$$\begin{aligned} V_p := (pe+p^{\perp }e^{\perp })(epe+e^\perp p^\perp e^\perp )^{-1/2} \end{aligned}$$

(21)

and

$$\begin{aligned} v_p(\cdot ):= {{\,\mathrm{Ad}\,}}(V_p)(\cdot )=V_p (\cdot ) V^*_p\in {{\,\mathrm{Aut}\,}}({\mathcal {M}}). \end{aligned}$$

(22)

Lemma 16

If $p \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and $\Vert p-e\Vert < 1,$ then $V_p$ is a unitary from ${\mathcal {M}}$ and $v_p$ is an (inner) automorphism of ${\mathcal {M}}$ such that $v_p(e)=p$.

Proof

Note first that $\Vert p-e\Vert <1$ implies $p^\perp \wedge e=p\wedge e^\perp =0$, so that e is the right support of pe and p is the left support of pe. The same condition implies that the absolute value of pe, namely $|pe|=(epe)^{1/2}$, is invertible in ${\mathcal {M}}_e$, and we denote the inverse by $(epe)^{-1/2}$. Then for $u:=(pe)(epe)^{-1/2}$ we have $pe=u|pe|$ and $u^*u=e$. Hence u is a partial isometry from the polar decomposition of pe, with right support e and left support p. Similarly, $(p^\perp e^\perp )(e^\perp p e^\perp )^{-1/2}$ is a partial isometry with right support $e^\perp$ and left support $p^\perp$. The lemma is a simple consequence of these observations. $\square$

Notation 17

In the sequel we use the $2\times 2$ matrix representation of ${\mathcal {M}}$ over ${\mathcal {M}}_e,$ so from now on $w\in {\mathcal {M}}$ will denote a partial isometry such that $w^*w=e$ and $ww^*=e^\perp$. The equalities (15) will be replaced by

$$\begin{aligned} e=\left[ \begin{array}{cc} 1&{} 0\\ 0 &{} 0 \end{array} \right] ,\,e^\perp =\left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} 1\end{array} \right] \text { and } w=\left[ \begin{array}{cc} 0 &{} 0\\ 1&{} 0 \end{array} \right] \end{aligned}$$

(23)

(here $1:=\mathbb {1}_{{\mathcal {M}}_e}$), and w replaces $w_0$ in (14).

Set up for Theorem C: To apply Theorem B, we need to identify objects appearing in the set up of the previous section. For this section, we fix an arbitrary $\eta \in (0,1],$ put ${\mathcal {Y}}: = {\mathcal {M}}_h$ and denote by ${\mathfrak {X}}_0$ the interior of the set $\{b \in {\mathcal {M}}_h:0\le b \le \eta e\}$ in the subspace ${\mathfrak {X}}:=({\mathcal {M}}_e)_h=e{\mathcal {M}}_he$ of ${\mathcal {Y}}$, and by ${\mathcal {W}}_0$ the interior of the set $\{d \in {\mathcal {M}}_h; - \eta e^{\perp } \le d \le 0\}$ in the subspace ${\mathcal {W}}:= ({\mathcal {M}}_{e^\perp })_h$ of ${\mathcal {Y}}$. Operation ${\mathfrak {X}}+ {\mathcal {W}}\ni b+d \mapsto e(b+d)=b \in {\mathfrak {X}}$ is then a projection (an idempotent) of norm 1.

We identify b with $\left[ \begin{array}{cc} b &{} 0\\ 0 &{} 0 \end{array} \right]$ and d with $\left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} d \end{array} \right]$ for $b \in {\mathfrak {X}}_0$, $d \in {\mathcal {W}}_0$, and for $b\in {\mathfrak {X}}_0$ we introduce a projection

$$\begin{aligned} r(b)&:= \left[ \begin{array}{cc} 1-b^2 &{} \sqrt{1-b^2}b\\ \sqrt{1-b^2}b &{} b^2 \end{array} \right] \end{aligned}$$

(24)

and a unitary

$$\begin{aligned} V(b)&:= \frac{1}{\sqrt{2}} \left[ \begin{array}{cc} \sqrt{1+b} &{} \sqrt{1-b}\\ -\sqrt{1-b} &{} \sqrt{1+b} \end{array} \right] . \end{aligned}$$

(25)

TheoremC

For a fixed operator $b_0 \in {\mathfrak {X}}_0$ there is a $\delta > 0$ such that for all $b \in {\mathfrak {X}}_0,\,X \in {\mathcal {M}}_h$ with $\Vert X\Vert , \Vert b-b_0\Vert < \delta$, there exist $a\in {\mathfrak {X}}_0$ and $d \in {\mathcal {W}}_0$ satisfying

$$\begin{aligned} e+X-r(a) \simeq b+d. \end{aligned}$$

The theorem will be used only in the case $b=b_0$, but the more general formulation will not lengthen the proof.

Remark 18

In Sect. 5 we indeed use Theorem C for all $\eta \in (0,1]$. Replacing $\eta =1$ by $\eta \in (0,1]$ implies a stronger assumption about the operator b, but also stronger conclusions on the operators a and d.

Notation 19

In the group $G={{\,\mathrm{Inn}\,}}({\mathcal {M}})$ of inner automorphisms of ${\mathcal {M}}$, using (22), we introduce the family

$$\begin{aligned} {\mathcal {U}}:= \bigg \{ v_p^{-1}:p \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}, \Vert e-p \Vert < \frac{1}{\sqrt{2}} \bigg \}. \end{aligned}$$

(Note that $\mathrm{id}=v_e^{-1}\in {\mathcal {U}}$.) Then we define the set

$$\begin{aligned} {\mathcal {Y}}_0 := \bigcup _{u \in {\mathcal {U}}} u^{-1} ({\mathfrak {X}}_0 + {\mathcal {W}}_0). \end{aligned}$$

For $p, q \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ with $\Vert p-e \Vert ,\Vert q-e \Vert < 1$ and for $b, c \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$ we have ${{\,\mathrm{supp}\,}}b^{+} = {{\,\mathrm{supp}\,}}c^{+} = e$, so that ${{\,\mathrm{supp}\,}}v_p (b)^+ = p$, ${{\,\mathrm{supp}\,}}v_q (c)^+ = q$. Thus $p \ne q$ implies $v_p ({\mathfrak {X}}_0 + {\mathcal {W}}_0) \cap v_q ({\mathfrak {X}}_0 + {\mathcal {W}}_0) = \emptyset$. We will see shortly (in Lemma 24) that the family ${\mathcal {U}}$ partitions ${\mathcal {Y}}_0$ according to Definition 11, and $\Phi _a = v_{{{\,\mathrm{supp}\,}}a^+}^{-1}$ for $\Phi$ as in Definition 11, that is $\Phi _a = u$ if $a \in u^{-1} ({\mathfrak {X}}_0 + {\mathcal {W}}_0)$.

Lemma 20

For any $b\in {\mathfrak {X}}_0,\,V(b)$ is a unitary and r(b) is a projection; moreover, we have $e-r(b) \in {\mathcal {Y}}_0,\,b - wbw^* \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$ and

$$\begin{aligned}&\Phi _{e-r(b)}(\cdot ) ={{\,\mathrm{Ad}\,}}(V(b)^*)(\cdot )= V(b)^* (\cdot ) V(b)\in {{\,\mathrm{Inn}\,}}({\mathcal {M}}),\\&\Phi _{e-r(b)}(e-r(b))=b-wbw^*. \end{aligned}$$

Proof

Let

$$\begin{aligned} p(b) := \frac{1}{2} \left[ \begin{array}{cc} 1 +b &{}-\sqrt{1-b^2}\\ sqrt{1-b^2} &{} 1-b \end{array} \right] . \end{aligned}$$

Since b is in the interior of the set $\{ c \in ({\mathcal {M}}_{e})_h:0 \le c \le \eta \mathbb {1}_{{\mathcal {M}}_e} \},$ the operator $b^{-1}$ is well defined and bounded, and

$$\begin{aligned} \Vert p(b)-e \Vert&= \Vert (p(b)-e)^2 \Vert ^{1/2}=\big \Vert \frac{1}{2}\left[ \begin{array}{cc} 1-b &{} 0\\ 0 &{} 1-b \end{array} \right] \big \Vert ^{1/2}\\&\le \frac{1}{\sqrt{2}}\Vert 1- b \Vert ^{1/2} < \frac{1}{\sqrt{2}}. \end{aligned}$$

For $V_{p(b)}$ defined as in (21), it is easy to calculate that

$$\begin{aligned} V_{p(b)} = \frac{1}{\sqrt{2}} \left[ \begin{array}{cc} \sqrt{1+b} &{} \sqrt{1-b}\\ sqrt{1-b} &{} \sqrt{1+b} \end{array} \right] =V(b), \end{aligned}$$

and $v_{p(b)}^{-1} (e-r(b)) = \left[ \begin{array}{cc} b &{} 0\\ 0 &{} -b \end{array} \right]$, so that $e - r(b)\in {\mathcal {M}}_{(-\eta , 0)\cup (0, \eta )}$ and $e_{e - r(b)}((0, \infty ))= v_{p(b)}(e)=p(b)$. Thus we have $\Phi _{e-r(b)} = v_{p(b)}^{-1},$ and the lemma is proved. $\square$

To use Theorem B, we need to show that the set ${\mathcal {Y}}_0$ is open in ${\mathcal {Y}}$ and that the functions $g:{\mathfrak {X}}_0\ni a\mapsto a-r(a)$ and ${\mathcal {Y}}_0\ni a \mapsto \Phi _a$ are locally Lipschitzian. We will give full proofs of the facts.

Lemma 21

For operators $b, c\in {\mathcal {M}}_e$ satisfying $\Vert c\Vert + \Vert b-c\Vert <1,$ the following holds:

$$\begin{aligned} \Vert (1-b)^{\pm 1/2} - (1-c)^{\pm 1/2}\Vert \le \frac{\Vert b-c\Vert }{2(1-\Vert c\Vert )(1-\Vert c\Vert -\Vert b-c\Vert )}. \end{aligned}$$

Proof

The convergence of the series used below implies (using the convention $0!! = (-1)!! = 1$)

$$\begin{aligned} \Vert (1-b)^{-1/2} - (1-c)^{-1/2}\Vert&= \Vert \sum _{k=0}^{\infty } \frac{(2k-1)!!}{(2k)!!} (b^k - c^k)\Vert \\&\le \frac{1}{2} \sum _{k=0}^{\infty } \Vert b^k - c^k\Vert . \end{aligned}$$

For $X=b-c$,

$$\begin{aligned} \Vert b^k - c^k\Vert&= \Vert (c+X)^k - c^k \Vert \le \sum _{1 \le j \le k}\left( {\begin{array}{c}k\\ j\end{array}}\right) \Vert X\Vert ^j \Vert c\Vert ^{k-j}\\&= (\Vert X\Vert + \Vert c\Vert )^k -\Vert c\Vert ^k \end{aligned}$$

for $k \ge 0$. It is enough to note that

$$\begin{aligned} \sum _{k=0}^{\infty } (\Vert X\Vert + \Vert c\Vert )^k - \sum _{k=0}^{\infty } \Vert c\Vert ^k&= \frac{1}{1 - (\Vert X\Vert + \Vert c\Vert )} - \frac{1}{1 - \Vert c\Vert }\\&\le \frac{\Vert X\Vert }{(1 - \Vert c\Vert )(1 - \Vert c\Vert - \Vert X\Vert )}. \end{aligned}$$

For $\Vert ({1- b})^{1/2} - ({1- c})^{1/2} \Vert$ we follow a similar path. $\square$

Lemma 22

If $p,q \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ satisfy $\Vert e-p\Vert ^2 + \Vert p-q\Vert < 1/2$, then $\Vert V_p - V_q\Vert < 4 \Vert p-q\Vert$, so that $\Vert v_p-v_q\Vert <8 \Vert p-q\Vert$, for $V_p, V_q$ defined by ( 21) and $v_p, v_q$ by ( 22).

Proof

Observe that

$$\begin{aligned}&\Vert epe + e^{\perp }p^{\perp }e^{\perp }- eqe - e^{\perp }q^{\perp }e^{\perp }\Vert \\&\quad = \Vert e(p-q)e + e^{\perp }(p^{\perp }-q^{\perp })e^{\perp }\Vert \\&\quad \le \max (\Vert p-q\Vert , \Vert p^{\perp }-q^{\perp }\Vert ) = \Vert p-q\Vert , \end{aligned}$$

so that by (12) applied twice with proper replacements,

$$\begin{aligned} \Vert (p-e)^2 - (q-e)^2 \Vert \le \Vert p-q\Vert . \end{aligned}$$

The assumption implies $1-\Vert e-p\Vert ^2,1-\Vert e-p\Vert ^2 - \Vert p-q\Vert \ge 1/2$. Hence, by Lemma 21 and (12),

$$\begin{aligned}&\Vert (epe + e^{\perp }p^{\perp }e^{\perp })^{-1/2} - (eqe + e^{\perp }q^{\perp }e^{\perp })^{-1/2}\Vert \\&\quad \le \frac{\Vert p-q\Vert }{2(1-\Vert e-p\Vert ^2)(1-\Vert e-p\Vert ^2 - \Vert p-q\Vert )} \le 2\Vert p-q\Vert . \end{aligned}$$

For the isometries $V_p$, $V_q$ we have, by (21),

$$\begin{aligned} \begin{aligned} \Vert V_p-V_q\Vert&\le \Vert (pe + p^{\perp }e^{\perp } - qe - q^{\perp }e^{\perp })(epe + e^{\perp }p^{\perp }e^{\perp })^{-1/2}\Vert \\&\quad +\Vert (qe + q^{\perp }e^{\perp }) \big [(epe + e^{\perp }p^{\perp }e^{\perp })^{-1/2} - (eqe + e^{\perp }q^{\perp }e^{\perp })^{-1/2}]\Vert . \end{aligned} \end{aligned}$$

Again by assumption, $\mathbb {1}-(e-p)^2\ge \mathbb {1}-\Vert e-p\Vert ^2\mathbb {1}\ge (1/2)\mathbb {1}$. Hence, by (12), $\Vert (epe + e^{\perp }p^{\perp }e^{\perp })^{-1/2}\Vert = \Vert (\mathbb {1}-(p-e)^2)^{-1/2}\Vert < 2$. Since eqe and $e^{\perp }q^{\perp }e^{\perp }$ are orthogonal, $\Vert qe+q^{\perp }e^{\perp }\Vert = \Vert (qe+q^{\perp }e^{\perp })^* (qe+q^{\perp }e^{\perp })\Vert ^{1/2} = \Vert eqe+e^{\perp }q^{\perp }e^{\perp }\Vert ^{1/2} \le 1$. Using

$$\begin{aligned}&\Vert pe + p^{\perp }e^{\perp } - qe - q^{\perp }e^{\perp }\Vert \\&\quad = \Vert (p-q)e + (-p+q)e^{\perp }\Vert \\&\quad \le \Vert p-q\Vert \Vert e-e^{\perp }\Vert = \Vert p-q\Vert , \end{aligned}$$

we get $\Vert V_p-V_q\Vert < 4\Vert p-q\Vert ,$ and $\Vert v_p-v_q\Vert < 8\Vert p-q\Vert$ easily follows. $\square$

Lemma 23

If $B, C \in {\mathcal {M}}_{(- \infty , 0)\cup (\varepsilon , \infty )}$ for some $\varepsilon >0$, then

$$\begin{aligned} \Vert e_B((0, \infty )) - e_C((0, \infty )) \Vert \le \Vert B-C\Vert /\varepsilon . \end{aligned}$$

(26)

Proof

Put $p := e_B((0, \infty ))$, $q := e_C((0, \infty ))$. It is enough to prove the lemma for $p,q\ne 0$. We are going to show that

$$\begin{aligned} \Vert p-pqp\Vert ,\, \Vert q-qpq\Vert \le \big ( \Vert B-C\Vert /\varepsilon \big )^2 \end{aligned}$$

(27)

(cf. Lemma 14). It is enough to show that $pqp \ge \big (1- \big (\Vert B-C\Vert /\varepsilon \big )^2 \big )p$. Assume that for some $\alpha \in (0,1)$ the inequality $pqp \ge \alpha p$ does not hold. Hence, pqp must have part of its spectrum below $\alpha$. Thus, we can find vectors $\xi _n$ with $\Vert \xi _n\Vert =1$ such that $\xi _n \in pe_{pqp}\big (\big [\alpha _n, \alpha _n + \frac{1}{n} \big ]\big )(H)$ for some numbers $\alpha _n \le \alpha$, $n \in {\mathbb {N}}$. We put $\zeta _n:= q \xi _n$. We have

$$\begin{aligned} \Vert p \zeta _n - \alpha _n \xi _n \Vert = \Vert pqpe_{pqp} \big (\big [\alpha _n, \alpha _n + \frac{1}{n} \big ]\big ) \xi _n -\alpha _n\xi _n\Vert \le \frac{1}{n}. \end{aligned}$$

(28)

Since $\xi _n \in p(H)$, for $\xi _n' \in H$, $\beta _n \in {\mathbb {C}}$ defined by

$$\begin{aligned} B \xi _n = \beta _n \xi _n + \xi _n',\, \xi _n' \perp \xi _n \end{aligned}$$

(29)

we have $\beta _n \ge \varepsilon$, $\Vert \xi _n' \Vert \le \Vert B\Vert$, $\xi '_n\in p(H)$ (since p commutes with B). By (28),

$$\begin{aligned} \langle \xi _n', \zeta _n \rangle = \langle \xi _n', p\zeta _n \rangle = \langle \xi _n', p\zeta _n-\alpha _n\xi _n \rangle \rightarrow 0. \end{aligned}$$

(30)

Similarly, since $\zeta _n \in q(H),$ for $\zeta _n' \in H,\,\gamma _n \in {\mathbb {C}}$ defined by

$$\begin{aligned} C\zeta _n = \gamma _n\zeta _n+\zeta _n', \zeta _n' \perp \zeta _n, \end{aligned}$$

(31)

we have $\gamma _n\ge \varepsilon , \,\zeta _n' \in q(H)$ and

$$\begin{aligned} \zeta _n' \perp \xi _n - q\xi _n = \xi _n - \zeta _n. \end{aligned}$$

(32)

Again analogously, since $\xi _n - \zeta _n \in e_C((- \infty ,0])(H),$ for $\zeta _n'' \in H,\,\delta _n \in {\mathbb {C}}$ defined by

$$\begin{aligned} C(\xi _n - \zeta _n) = \delta _n (\xi _n - \zeta _n) + \zeta _n '',\, \zeta _n'' \perp \xi _n - \zeta _n \end{aligned}$$

(33)

we have $\delta _n \le 0$, $\zeta _n'' \in e_C((-\infty ,0))(H)$; in particular,

$$\begin{aligned} \zeta _n'' \perp \zeta _n. \end{aligned}$$

(34)

Consequently,

$$\begin{aligned} \sup _n \Vert B \xi _n - C \xi _n \Vert&= \sup _n \Vert \beta _n \xi _n + \xi _n' - (\gamma _n \zeta _n + \zeta _n' + \delta _n(\xi _n - \zeta _n)+ \zeta _n'')\Vert \end{aligned}$$

(by (29), (31) and (33))

$$\begin{aligned}&= \sup _n \Vert \beta _n \xi _n - \gamma _n \zeta _n - \delta _n (\xi _n - \zeta _n) + \xi _n' - \zeta _n' - \zeta _n'' \Vert \\&\quad \ge \limsup _{n \rightarrow \infty } \Vert \beta _n \xi _n - \gamma _n \zeta _n - \delta _n (\xi _n - \zeta _n) \Vert , \end{aligned}$$

since $\xi _n' - \zeta _n' - \zeta _n'' \perp \xi _n$ (by (29), (31), (32), (33) and (34)) and $\langle \xi _n' - \zeta _n' - \zeta _n'', \zeta _n \rangle \rightarrow 0$ (by (30), (31)and (34)). We also have

$$\begin{aligned} \Vert \beta _n \xi _n - \gamma _n \zeta _n - \delta _n (\xi _n - \zeta _n)\Vert&= \Vert (\beta _n - \delta _n)(\xi _n - \zeta _n) + (\beta _n - \gamma _n) \zeta _n \Vert \\&\ge \varepsilon \Vert \xi _n - \zeta _n \Vert \end{aligned}$$

(since $\beta _n - \delta _n \ge \beta _n \ge \varepsilon$ and $\zeta _n=q\xi _n\perp \xi _n-\zeta _n$) and $\limsup \limits _{n \rightarrow \infty } \Vert \xi _n - \zeta _n \Vert \ge \sqrt{1 - \alpha }$ (since the convergence in (28) means that $\widehat{\xi _n} \zeta _n - \alpha _n \xi _n \rightarrow 0$ and $\Vert \widehat{\xi _n} \zeta _n \Vert = 1 - \Vert \xi _n - \zeta _n \Vert ^2$ for the projection $\widehat{\xi _n} =\langle \cdot , \xi _n \rangle \xi _n$ on the direction of the vector $\xi _n$.)

We have shown that $\Vert B-C \Vert /\varepsilon \ge \sqrt{1 - \alpha }$, so that $\alpha \ge 1-(\Vert B-C\Vert /\varepsilon )^2$. Since $pqp\ge \alpha p$ for all $0<\alpha < 1-(\Vert B-C\Vert /\varepsilon )^2,$ we also have $pqp\ge (1-(\Vert B-C\Vert /\varepsilon )^2)p$. By Lemma 14, the proof of (26) is finished. $\square$

Lemma 24

The set ${\mathcal {Y}}_0$ is an open subset of ${\mathcal {Y}}$ and it contains ${\mathfrak {X}}_0+{\mathcal {W}}_0$. Moreover, G is a group of isometries of ${\mathcal {Y}}$ and ${\mathcal {U}}\subset G$ is a partitioning family for ${\mathcal {Y}}_0$. The values of the function $g: {\mathfrak {X}}_0\ni b\mapsto e-r(b)$ (with r(b) given by ( 24)) lie in the set ${\mathcal {Y}}_0$ and the function agrees with the family ${\mathcal {U}}$.

Proof

According to the definition, ${\mathfrak {X}}_0,{\mathcal {W}}_0$ are the interiors of the sets $\{ b \in {\mathcal {M}}_h:0 \le b \le \eta e \}$, $\{ d \in {\mathcal {M}}_h :- \eta e^{\perp } \le d \le 0 \}$ in spaces $({\mathcal {M}}_{e})_h$, $({\mathcal {M}}_{e^{\perp })h},$ respectively. This means that $B \in {\mathcal {M}}_{(- \eta + \varepsilon , - \varepsilon ) \cup (\varepsilon , \eta - \varepsilon )},\,{{\,\mathrm{supp}\,}}B=\mathbb {1}$ for $B \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$, for $\varepsilon > 0$ depending on B. If, additionally, $\Vert p - e \Vert < 1/\sqrt{2}$ for some $p \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, then also $v_p (B) \in {\mathcal {M}}_{(- \eta + \varepsilon , - \varepsilon ) \cup (\varepsilon , \eta - \varepsilon )},\,{{\,\mathrm{supp}\,}}v_p(B)=\mathbb {1}$. Choose $\delta \in (0,\varepsilon )$ so that $\delta /(\varepsilon -\delta )<\frac{1}{\sqrt{2}} - \Vert p - e \Vert$. If $C \in {\mathcal {M}}_h$ is such that $\Vert C - v_p (B) \Vert < \delta$, then

$$\begin{aligned} C \in {\mathcal {M}}_{(- \eta + \varepsilon - \delta , - \varepsilon + \delta ) \cup (\varepsilon - \delta , \eta - \varepsilon + \delta )},\, {{\,\mathrm{supp}\,}}C=\mathbb {1}\end{aligned}$$

and

$$\begin{aligned} \Vert e_C((0, \infty )) - e_{v_p (B)}((0, \infty )) \Vert < \frac{1}{\sqrt{2}} - \Vert p - e \Vert , \end{aligned}$$

according to Lemma 23. Then $C = v_q (D)$ for $q = e_C((0, \infty ))$, satisfying $\Vert q - e \Vert < \frac{1}{\sqrt{2}}$, for some $D \in {\mathfrak {X}}_0 + {\mathcal {W}}_0$. The openness of the set ${\mathcal {Y}}_0$ follows. Using Notation 19, we see that for $B \in {\mathcal {Y}}_0$ we have $B \in {\mathcal {M}}_{(- \eta + \varepsilon , - \varepsilon ) \cup (\varepsilon , \eta - \varepsilon )},\,{{\,\mathrm{supp}\,}}B=\mathbb {1}$, for $\varepsilon > 0$ depending on B. By Lemmas 23 and 22 we get that the mapping ${\mathcal {Y}}_0 \ni B \mapsto \Phi _B$ is locally Lipschitzian (cf. Lemma 20). This establishes that ${\mathcal {U}}$ is indeed a partitioning family for ${\mathcal {Y}}_0$.

The fact that the function $g(b) = e - r(b)$ is locally Lipschitzian follows directly from Lemma 21 and the form of the operator r(b),  given by (24). Moreover, $e\Phi _{g(b)}(g(b))=b$ by Lemma 20.

We showed that the function g agrees with the partitioning family ${\mathcal {U}}$. Hence, the lemma is proved. $\square$

Proof of Theorem C

By Lemma 24, we can apply Theorem B from Sect. 3; now, condition (**) implies Theorem C. $\square$

5 Proof of Lemma D on the Construction of  3 Projections

In this section ${\mathcal {M}}$ is a factor of type III.

Definition 25

For $e \in {{\,\mathrm{Proj}\,}}{\mathcal {M}},$ an operator $a \in {\mathcal {M}}$ of type $\mu \in [0,1)$ (see Definition 1) is said to be covered by e if $e \ge {{\,\mathrm{supp}\,}}a$ and, whenever $\mu >0,$ we have $e\ne {{\,\mathrm{supp}\,}}a$.

We can rephrase the condition by saying that e covers $a\in {\mathcal {M}}_{[0,1]}$ if and only if $e \ge {{\,\mathrm{supp}\,}}a$ and $ee_a([0, \delta )),\,ee_a((1 - \delta , 1]) \ne 0$ for each $\delta > 0$.

We will use the following properties of operators covered by e.

Lemma 26

A. If an operator a is covered by e, then there exist a partition $e = e_1 + e_2$, $ae_i = e_ia$, $e_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ for $i = 1, 2$ such that $ae_1$ is covered by $e_1$ and $ae_2$ is covered by $e_2$.

B. If an operator a is of type $\mu$, $\mu \in [-1,1)$ (see Definition  1), then $\frac{1}{1- \mu } (a - \mu {{\,\mathrm{supp}\,}}a)$ is covered by any $e\ge {{\,\mathrm{supp}\,}}a$.

C. If a is covered by e for $e \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and $\delta _i > 0$ for $i \in {\mathbb {N}}$, then $e = \sum _{i \ge 0} e_i$ for (mutually orthogonal) projections $e_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}},\, i \ge 0$, satisfying $ae_0 \in {\mathcal {M}}_{[0, 1]}$, $ae_i\in {\mathcal {M}}_{[0, \delta _i)\cup (1-\delta _i, 1]}$, with $e_ie_{a}([0, \delta _i)),e_ie_{a}((1 - \delta _i, 1]) \ne 0$ for $i \in {\mathbb {N}}$ (in particular, $ae_i = e_ia$).

Proof

B. For $\mu \ne 0$, the statement is obvious. For $\mu = 0,$ we have in particular $e_a ((0, \delta )) \ne 0$ for $\delta > 0$ and B is again easy to see.

C. If $ee_a(\{0\}) \ne 0$, we take a decomposition of $ee_a(\{0\})$ into a sum $\sum _{i \in {\mathbb {N}}} f_i$ of non-zero mutually orthogonal projections $f_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$. Whenever $ee_a(\{0\}) = 0$, we will inductively find $\varepsilon _1> \varepsilon _2> \ldots > 0$ satisfying $\varepsilon _i < \delta _i$, $f_i := e_a([\varepsilon _{i +1}, \varepsilon _i)) \ne 0$ for $i \in {\mathbb {N}}$ and $\varepsilon _1 < 1 / 2$. Thus we have: $f_i$ are mutually orthogonal, $af_i \in {\mathcal {M}}_{[0, \delta _i)}$ and $0 \ne f_i\le e_a([0, 1 / 2))$. Analogously, there exist mutually orthogonal $g_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, satisfying $ag_i \in {\mathcal {M}}_{(1 - \delta _i, 1]}$, $0 \ne g_i \le e_a((1 / 2, 1])$. It is enough to put $e_i := f_i + g_i$ (a sum of orthogonal terms) for $i \in {\mathbb {N}}$, $e_0 = e - \sum _{i \in {\mathbb {N}}}e_i$.

The statement A follows from C. $\square$

The essence of the next corollary is a generalization of Theorem C (Section 4) to a wider class of operators b.

Corollary 27

If a is covered by e, and $b \in {\mathcal {M}}_{(0, \eta )}$ for some $\eta \in (0,1],$ then there exist projections $e'$, $r \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $r \le e' \le e$ satisfying

$$\begin{aligned} ae' = e'a, \,ae' - r = c + d \end{aligned}$$

for some operators $c \sim b$, $d \in {\mathcal {M}}_{(-\eta , 0)}$ with ${{\,\mathrm{supp}\,}}c + {{\,\mathrm{supp}\,}}d = e'$.

Proof

We can assume that ${{\,\mathrm{supp}\,}}b$, $({{\,\mathrm{supp}\,}}b)^{\perp } \ne 0$. Indeed, if $b = 0$, $r = e' = 0$ ends the proof. If, on the other hand, ${{\,\mathrm{supp}\,}}b=1,$ we can replace b with an equivalent operator $ubu^*$ (still in ${\mathcal {M}}_{(0,\eta )}$), where $u\in {\mathcal {M}}$ is a non-surjective isometry, that is $u^*u=\mathbb {1},\, uu^*\ne \mathbb {1}$. Thus $({{\,\mathrm{supp}\,}}b)^{\perp } = \sum _{i\in {\mathbb {N}}} f_i$ for mutually orthogonal non-zero projections $f_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$; furthermore, ${{\,\mathrm{supp}\,}}b = \sum _{i\in {\mathbb {N}}} g_i$ for mutually orthogonal non-zero projections $g_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ satisfying additionally $g_ib = bg_i$ with $bg_i$ belonging to the interior of the set $\{ c \in ({\mathcal {M}}_{g_i})_h; 0 \le c \le \eta g_i \}$ in the space $({\mathcal {M}}_{g_i})_h$, for the reduced algebra ${\mathcal {M}}_{g_i}$, $i \in {\mathbb {N}}$. To obtain it, we can proceed as in the proof of part C of Lemma 26.

Let us apply Theorem C in the reduced algebras ${\mathcal {M}}_{f_i+g_i}$, with ${\mathfrak {X}}_{0i}$, ${\mathcal {W}}_{0i}$ being the interiors of the sets $\{ c \in ({\mathcal {M}}_{g_i})_h:0 \le c \le \eta g_i \}$ and $\{ d \in ({\mathcal {M}}_{f_i})_h:{- \eta f_i }\le d \le 0 \}$ in the spaces $({\mathcal {M}}_{g_i})_h, ({\mathcal {M}}_{f_i})_h$, respectively. We will find $\delta _i > 0$ matched to $b_i \in {\mathfrak {X}}_{0i}$, according to Theorem C (with $b_i$ in place of $b_0$). Next, by Lemma 26C, we will find a partition $e = \sum _{i \ge 0} e_i$, and then, for $i\in {\mathbb {N}},$ partial isometries $u_i, w_i \in {\mathcal {M}}$ satisfying

$$\begin{aligned} u_iu_i^* = g_i,\, u_i^*u_i = e_ie_{a }((1 - \delta _i, 1]),\,w_i w_i^* = f_i,\, w_i^* w_i = e_ie_{a}([0, \delta _i)), \end{aligned}$$

so that we have $v_iv_i^* = g_i + f_i$, $v_i^*v_i = e_i$ for $v_i = u_i + w_i$.

Note that the operator $v_iae_iv_i^*$ is of the form $g_i + X_i$ for some $X_i \in ({\mathcal {M}}_{g_i + f_i})_h$, $\Vert X_i \Vert \le \delta _i$. By Theorem C, $g_i + X_i - r(a_i) \sim b_i + d_i$ for a projection $r(a_i) \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}_{g_i + f_i}$ and an operator $d_i \in {\mathcal {W}}_{0i}$. We can define now projections $e' = \sum _{i \in {\mathbb {N}}} e_i$, $r = \sum _{i \in {\mathbb {N}}} v_i^* r(a_i) v_i$, and operators $c = \sum _{i \in {\mathbb {N}}}v_i^*b_iv_i$, $d= \sum _{i \in {\mathbb {N}}} v_i^*d_iv_i$, satisfying the statement from Corollary 27. $\square$

Corollary 27 leads to simple corollaries obtained by linear transformations of operators $b \in {\mathcal {M}}_h$. Finally, they give a proof of Lemma D, and consequently end the proof of Theorem A and Main Theorem from Sect. 2.

Proof of Lemma D

It is enough to conduct the proof for $b \ne 0$. Assume first (Steps 1, 2, 3) that a is covered by $e \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$.

In Steps 1, 2, 3, $\eta \in (0,1]$.

Step 1. Suppose that $b \in {\mathcal {M}}_{(1, 1+\eta )}$. Then there exist projections $r, e' \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and operators $c, d \in {\mathcal {M}}_h$ satisfying

$$\begin{aligned} \begin{aligned} r \le e' \le e, \quad ae' = e'a, ae' + r = c + d,\\ \quad c \perp d, \quad c \sim b, \quad {{\,\mathrm{supp}\,}}c + {{\,\mathrm{supp}\,}}d = e' \end{aligned} \end{aligned}$$

(35)

and $d \in {\mathcal {M}}_{(1-\eta , 1)}$.

Proof

We apply Corollary 27 to $b' = b - {{\,\mathrm{supp}\,}}b \in {\mathcal {M}}_{(0, \eta )}$. There exist $r', e' \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $r' \le e' \le e$, $ae' = e'a$ satisfying

$$\begin{aligned} ae' - r' = c' + d',\, c' \perp d',\, c' \sim b',\, d' \in {\mathcal {M}}_{(-\eta , 0)} \end{aligned}$$

(36)

and ${{\,\mathrm{supp}\,}}c + {{\,\mathrm{supp}\,}}d = e'$. Equivalently, for $r = e' - r'$, $c = c' + {{\,\mathrm{supp}\,}}c'$, $d = d' + {{\,\mathrm{supp}\,}}d'$ we have (35) and $d \in {\mathcal {M}}_{(1-\eta , 1)}$. $\square$

Step 2. Suppose that $b \in {\mathcal {M}}_{(1-\eta , 1)}$. There exist $r, e' \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $c, d \in {\mathcal {M}}_h$ satisfying (35) and $d \in {\mathcal {M}}_{(1, 1+\eta )}$.

Proof

We apply Corollary 27 to $a' = e - a$ (being an operator covered by e), $b'={ {{\,\mathrm{supp}\,}}b - b} \in {\mathcal {M}}_{(0, \eta )}$. We find $e', r' \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $c', d' \in {\mathcal {M}}_h$ satisfying (36) for $a'$ instead of a and ${{\,\mathrm{supp}\,}}c' + {{\,\mathrm{supp}\,}}d' = e'$. Hence (multiplying $a'e' - r' = c' + d'$ by $-1$ and adding $e'$) we get (35) for $c = -c' + {{\,\mathrm{supp}\,}}c',\,d = -d' + {{\,\mathrm{supp}\,}}d'$, and at the same time $d \in {\mathcal {M}}_{(1, 1+\eta )}$. $\square$

Step 3. Suppose that $b \in {\mathcal {M}}_{(1 - \eta , 1)\cup (1, 1 + \eta )}$. There exist $e', r \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$, $c, d \in {\mathcal {M}}_h$ satisfying (35) and $d \in {\mathcal {M}}_{(1 - \eta , 1)\cup (1, 1 + \eta )}$.

Proof

By Steps 1 and 2, we can assume that $e_b((1,1+\eta ))$ and $e_b((1-\eta ,1))$ are non-zero. Using Lemma 26A, we choose projections $e_1, e_2 \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ satisfying $e_1 + e_2 = e$, $ae_i = e_ia$ such that $ae_i$ is covered by $e_i$ for $i = 1, 2$, and put $b_1 := be_b((1, 1 + \eta )),\, b_2 := b e_b((1 - \eta , 1))$. Using Steps 1 and 2, we find $r_i, e'_i \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and $c_i, d_i \in {\mathcal {M}}_h$ satisfying

$$\begin{aligned} \begin{aligned} r_i \le e'_i \le e_i,\, ae'_i = e'_ia,\, ae_i' + r_i = c_i + d_i,\,c_i \perp d_i,\\ c_i \sim b_i,\,{{\,\mathrm{supp}\,}}c_i+{{\,\mathrm{supp}\,}}d_i=e_i' \text { for }i = 1, 2,\\ d_1 \in {\mathcal {M}}_{(1-\eta , 1)},\, d_2 \in {\mathcal {M}}_{(1, 1+\eta )}. \end{aligned} \end{aligned}$$

It is enough to define $r := r_1 + r_2$, $c := c_1 + c_2$, $d := d_1 + d_2$. $\square$

From this point on a will be an operator of type $\mu ,\,\mu \in [-1,1)$. To make the proof uniform and treat cases $\mu \ge 0$ and $\mu <0$ at the same time, we put $\eta := \frac{1}{1+ \mu ^{-}}=\frac{1-\mu ^+}{1-\mu }$, so that $\eta \in \big [ \frac{1}{2}, 1 \big ]$.

Step 4. If a is of type $\mu ,\,\mu \in [-1, 1),$ and $b \in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2-\mu ^+)}$, then there exist projections $r, e' \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and operators $c, d \in {\mathcal {M}}_h$ satisfying

$$\begin{aligned} \begin{aligned} r \le e' \le {{\,\mathrm{supp}\,}}a,\, ae' = e'a, \,ae' + (1 - \mu )r = c + d,\\ c \perp d,\, c \sim b,\, {{\,\mathrm{supp}\,}}c + {{\,\mathrm{supp}\,}}d = e' \end{aligned} \end{aligned}$$

(37)

and $d \in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2-\mu ^+)}$.

Proof

For $e = {{\,\mathrm{supp}\,}}a$, $\frac{1}{1 - \mu } (a - \mu e)$ is covered by e, by Lemma 26B. For $b' := \frac{1}{1 - \mu } (b - \mu {{\,\mathrm{supp}\,}}b) \in {\mathcal {M}}_{\frac{1}{1 - \mu }[((\mu ^+,1)\cup (1,2-\mu ^+))-\mu ]}= {\mathcal {M}}_{( 1- \eta , 1)\cup (1,1+ \eta )}$, by Step 3 there exist projections $r\le e'\le e$ and operators $c',d'$, satisfying

$$\begin{aligned} \begin{aligned} \frac{1}{1 - \mu } (a - \mu e) e' + r = c' + d',\\ c' \perp d',\, c' \sim b',\,{{\,\mathrm{supp}\,}}c' + {{\,\mathrm{supp}\,}}d' = e',\\ ae'=e'a\text { and }d' \in {\mathcal {M}}_{(1 - \eta , 1)\cup (1, 1+ \eta )}. \end{aligned} \end{aligned}$$

Hence for $c = (1 - \mu )c' + \mu {{\,\mathrm{supp}\,}}c'$, $d = (1 - \mu ) d' + \mu {{\,\mathrm{supp}\,}}d'$ we have (37) (in particular, it is easy to see that ${{\,\mathrm{supp}\,}}c = {{\,\mathrm{supp}\,}}c'$, ${{\,\mathrm{supp}\,}}d = {{\,\mathrm{supp}\,}}d'$). Further, $d \in {\mathcal {M}}_{(1 - \mu ) \cdot [(1 - \eta , 1)\cup (1, 1 + \eta )] + \mu } = {\mathcal {M}}_{(\mu ^+,1)\cup (1,2-\mu ^+)}$. $\square$

Step 5. If a is of type $\mu$, $b \in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2 - \mu ^+)}$ and additionally $b \perp a$, then there exist $e', r \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ satisfying

$$\begin{aligned} \begin{aligned} r \le e' \le {{\,\mathrm{supp}\,}}a, \, ae' = e'a,\, ae' + (1 - \mu )r + b = p + q +d,\\ d\perp p\vee q,\, p\vee q\le e'+{{\,\mathrm{supp}\,}}b,\, {{\,\mathrm{supp}\,}}d\le e', \end{aligned} \end{aligned}$$

for some $p, q \in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and an operator $d \in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2 - \mu ^+)}$. Additionally, $p\wedge q=p\wedge q^\perp =p^\perp \wedge q=0$.

Proof

We apply Step 4 to $b' = -b + 2{{\,\mathrm{supp}\,}}b \in {\mathcal {M}}_{(\mu ^+,1)\cup (1,2 - \mu ^+)}$. In particular, we have $ae' + (1 - \mu )r = c + d$, $c \sim b'$. For the algebra ${\mathcal {M}}_{{{\,\mathrm{supp}\,}}b + {{\,\mathrm{supp}\,}}c}$ and a fixed partial isometry u, we have $u^*u = {{\,\mathrm{supp}\,}}b,\, uu^* = {{\,\mathrm{supp}\,}}c,\, c = u(-b + 2{{\,\mathrm{supp}\,}}b)u^*$. Let us adopt a convention:

$$\begin{aligned} {{\,\mathrm{supp}\,}}b=\left[ \begin{array}{cc} 1&{} 0\\ 0 &{} 0 \end{array} \right] ,\, {{\,\mathrm{supp}\,}}c = \left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} 1\end{array} \right] , \, u = \left[ \begin{array}{cc} 0 &{} 0\\ 1&{} 0 \end{array} \right] \end{aligned}$$

and identify

$$\begin{aligned} b \text { with }\left[ \begin{array}{cc} b &{} 0\\ 0 &{} 0 \end{array} \right] \text { and }c \text { with }\left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} c \end{array} \right] =\left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} 2\cdot 1-b \end{array} \right] . \end{aligned}$$

Then $b + c = p + q$ for projections

$$\begin{aligned} p&= \frac{1}{2} \left[ \begin{array}{cc} b &{} \sqrt{b(2\cdot 1- b)}\\ \sqrt{b(2\cdot 1- b)} &{} 2\cdot 1- b \end{array} \right] ,\\ q&=\frac{1}{2} \left[ \begin{array}{cc} b &{} -\sqrt{b(2\cdot 1- b)}\\ -\sqrt{b(2\cdot 1- b)} &{} 2\cdot 1- b \end{array} \right] . \end{aligned}$$

Thus $ae'+(1-\mu )r+b=c+d+b=p+q+d$. The rest is easy to see. $\square$

To obtain Lemma D, put $a' := ae'$, $a'' := a - a'$ (observing that $a''\in {\mathcal {M}}_{[\mu ,1]}$) and change d to c. $\square$

6 Generalizations, Connections with Fixed Point Theory and Further Questions

In this section we will discuss some generalizations of Theorem B from Sect. 3. We will point out the usefulness of this generalization in operator theory (Section 6.2). We will also pose some open questions (Section 6.3).

6.1 Connection with Reversal of Mappings and Fixed Point Theory

In Sect. 3 we considered in particular Banach spaces ${\mathfrak {X}},{\mathcal {Y}}$ and an open subset ${\mathfrak {X}}_0\subset {\mathfrak {X}}$. The claim in Theorem B reduces to the following: For every $b_0\in {\mathfrak {X}}_0$, there exists an $\varepsilon >0$ and we can construct a special “local” function $(X,a)\mapsto f_X(a)\in {\mathfrak {X}}$ which is defined for $(X,a)\in {\mathcal {Y}}\times {\mathfrak {X}}$ for which $\Vert X\Vert ,\Vert a-b_0\Vert <\varepsilon$ and satisfying the condition

$$\begin{aligned}&\text {there exists a }\delta >0\text { such that}\\&\qquad \qquad \text { for } X\in {\mathcal {Y}}, b\in {\mathfrak {X}}\text { with }\Vert X\Vert ,\Vert b-b_0\Vert<\delta \text { there is} \quad \quad \quad \qquad \, (***)\\&\qquad \quad \qquad \text {an } a \in {\mathfrak {X}},\, \Vert a-b_0\Vert <\varepsilon \text { satisfying }f_X(a)=b. \end{aligned}$$

The function has been already used in Step 4 of the proof of Theorem B (Section 3). It has the following form:

$$\begin{aligned} f_{X}(a)=e(\Phi _{\Phi _{g(a)}(g(a)+X)} \circ \Phi _{g(a)})(g(a)+X). \end{aligned}$$

(38)

The proof of Theorem B has been elementary, but fairly complex. On the other hand, it is easy to see that the following remark follows from Lemma 13.

Remark 28

For $b\in {\mathfrak {X}}_0$ there is an $\varepsilon >0$ such that for $X\in {\mathcal {Y}},\, a\in {\mathfrak {X}}$ with $\Vert X\Vert ,\Vert a-b_0\Vert <\varepsilon$ and for $f_{X}(a)$ defined by (38) we have

• 1. the mapping $(X,a)\mapsto f_X(a)$ is well-defined and Lipschitzian;

• 2. $f_0(a)=a$.

This observation points to the importance of the following general problem:

Problem 29

Let ${\mathfrak {X}}$ be a Banach space, and ${\mathcal {N}}$ a normed space. Assume that for some $b_0\in {\mathfrak {X}}$ there is an $\varepsilon >0$ such that a function $(X,a)\mapsto f_X(a)$ defined for $X\in {\mathcal {N}},\, a\in {\mathfrak {X}}$ with $\Vert X\Vert , \Vert a-b\Vert <\varepsilon ,$ with values in ${\mathfrak {X}}$, satisfies conditions (1) and (2) of Remark 28. What additional conditions are needed to get condition (***), for ${\mathcal {N}}$ in place of ${\mathcal {Y}}$?

In this subsection we shall answer some elementary questions connected with Problem 29. Positive results will be obtained under the strong assumption

$$\begin{aligned} \Vert f_X(b)-f_X(c)-f_0(b)+f_0(c)\Vert \le K\Vert X\Vert \Vert b-c\Vert \end{aligned}$$

(39)

for $X\in {\mathcal {N}},\, b,c\in {\mathfrak {X}}$ with $\Vert X\Vert , \Vert b-b_0\Vert ,\Vert c-b_0\Vert <\varepsilon$. In Sect. 6.2 we will carry out a non-trivial consideration of condition (39). We will describe a special case of Problem 29, useful in operator theory.

We begin with elementary considerations related to Problem 29. Let $B(b_0, \varepsilon )$ denote a closed ball in ${\mathbb {R}}^n$. Assume that for $b \in B(b_0, \varepsilon )$, the function $X \mapsto f_X (b) \in {\mathbb {R}}^n$ is defined on some neighbourhood of 0 in a normed space ${\mathcal {N}}$ (representing parameters).

Theorem 30

Suppose that

• 1. for $b_0 \in {\mathbb {R}}^n$ there exist $\varepsilon > 0$ such that the function $(X, b) \mapsto f_X (b)$ is defined and continuous for $X \in {\mathcal {N}}$, $b \in {\mathbb {R}}^n$ with $\Vert X\Vert$, $\Vert b - b_0\Vert < \varepsilon$;

• 2. $f_0 (b) = b$ for $b \in B(b_0, \varepsilon )$.

Then for $b _0\in {\mathbb {R}}^n$, there exists $\delta > 0$ such that for $b \in {\mathbb {R}}^n,\,X \in {\mathcal {N}}$ with $\Vert b - b_0\Vert , \Vert X\Vert < \delta$ there exists $c \in B(b_0, \varepsilon )$ satisfying $f_X (c) = b$.

Proof

Fix $b \in {\mathbb {R}}^n$, $\Vert b - b_0\Vert < \varepsilon /2$. For $X \in {\mathcal {N}}$, $c \in {\mathbb {R}}^n$ satisfying $\Vert X\Vert , \Vert c - b_0\Vert <\varepsilon ,$ define a continuous mapping $(X, c) \mapsto g_X (c) = -f_X (c) + c + b$. By compactness of the closed ball $B(b_0, \varepsilon )$, for some $b_i \in B(b_0, \varepsilon )$, $\delta _i > 0$ one has $g_X (B(b_i, \delta _i)) \subset B(b, \frac{\varepsilon }{2}) \subset B(b_0, \varepsilon )$ for $\Vert X\Vert < \delta _i$, $i = 1, \ldots , k$ and $\bigcup _{1 \le i \le k} B(b_i, \delta _i) \supset B(b_0, \varepsilon )$. In particular, $g_X (B(b_0, \varepsilon )) \subset B(b_0, \varepsilon )$ for $\Vert X\Vert < \delta := \min (\frac{\varepsilon }{2}, \delta _1, \ldots , \delta _k)$. By Brouwer’s fixed point theorem, there exists $c \in B(b_0, \varepsilon )$ satisfying $g_X (c) = c$, so that $f_X (c) = b$. Since $\delta < \frac{\varepsilon }{2}$, the theorem is proved. $\square$

Theorem 30 is no longer true if we replace the continuity of $(X, b) \mapsto g_X (b)$ (with respect to both variables) by separate continuity (or even Lipschitz condition) for each variable.

Example 31

For $X \in {\mathbb {R}}$, $|X| < 1$, $b \in {\mathbb {R}}^2$ there exists a mapping $(X, b) \mapsto f_X (b)$ satisfying $f_0 (b) = b$ with $f_X (\cdot )$ Lipschitzian for a fixed X and $f_{\cdot } (b)$ Lipschitzian for a fixed b, but for $X \ne 0$, $|X| < 1$ and arbitrary $c \in {\mathbb {R}}^2$ we have $f_X (c) \ne 0$.

It is enough to put, for $c = (x,y) \in {\mathbb {R}}^2$, $X \ne 0$,

$$\begin{aligned} h_X(c): = \left\{ \begin{array}{rll} c &{} \text { for } &{} y \le \frac{1}{X^2} |x| - |X|\\ (x, \frac{1}{X^2}|x| - |X|) &{} \text {for} &{} y > \frac{1}{X^2} |x| - |X| \end{array}\right. \end{aligned}$$

and then $f_X (c) := e^{-iX} h_X (e^{iX}c)$. We identify here (x, y) with $x+yi \in {\mathbb {C}}$. For $X = 0$, we put $f_0(c) = c$ for all $c \in {\mathbb {R}}^2$. The separate Lipschitz condition, for $|X| > \delta$ with a fixed $\delta > 0$, follows from the Lipschitz condition for the function $y_X(x) = \frac{1}{X^2} |x| - X$ (separately for x and X). Moreover, for $X \ne 0$ we have $f_X(c) \ne 0$ (since $h_X (c) \ne 0$ for $c \in {\mathbb {R}}^2$). It is enough to note that for any $c \in {\mathbb {R}}^2 \smallsetminus \{0\}$ there exists $\delta > 0$ satisfying $f_X (c) = c$ for $|X| < \delta$. In fact, for $c = x + iy \ne 0$ and for $x_1 + iy_1 = e^{iX} c$, we have $\frac{1}{X^2} |x_1| - |X| = \frac{1}{X^2} | x \cos X - y \sin X | - |X| \rightarrow \infty$ for $X \rightarrow 0$, so that $y_1 \le \frac{1}{X^2} |x_1| - |X|$ and $f_X (c) = c$ when $|X| < \delta$, for a suitable $\delta > 0$. For $c = 0$, we have a Lipschitzian $X \mapsto h_X (0) = (0,-|X|)$.

Theorem  30 is also not true if we take an infinite dimensional Hilbert space in place of ${\mathbb {R}}^n$. It is enough to take ${\mathcal {N}}={\mathbb {R}}$, ${\mathfrak {X}}=l_2$.

Example 32

For the norm $\Vert (X,b)\Vert =|X|+\Vert b\Vert$ in ${\mathbb {R}}\times l_2$, there is a Lipschitzian mapping ${\mathbb {R}}\times l_2\ni (X, b) \mapsto f_X (b) \in l_2$ satisfying $f_0 (b) = b$ for $b \in l_2$, but $f_X (b) \ne 0$ for $|X| > 0$ (and an arbitrary $b \in l_2$), as seen from the following.

For $b = (b_1, b_2, \ldots ) \in l_2$, put $Tb:=(0,b_1,b_2,\ldots )$ and $e := (1, 0, 0, \ldots )$, $s \vee t := \max (s, t)$, $s \wedge t := \min (s, t)$ for $s, t \in {\mathbb {R}}$. Instead of parameter $X \in {\mathbb {R}}$ it is enough to use $x = |X| \in [0, \infty )$. Define functions on $[0, \infty )$ by

$$\begin{aligned} \alpha _x (s) := \bigg ( \frac{2s - x}{x} \bigg )^+ \wedge 1 \ \text {for} \ x > 0, \ \alpha _0 (s) := 1, \end{aligned}$$

and then the required mapping

$$\begin{aligned} f_x (b) := \alpha _x (\Vert b\Vert )b + (\alpha _{x/2} (\Vert b\Vert ) - \alpha _x (\Vert b\Vert ) )Tb + (1 - \alpha _{x/2} (\Vert b\Vert )) xe. \end{aligned}$$

For $x = 0$ we immediately have $f_0(b) = b$ for $b \in l_2$. It is easy to see that $f_x(b) \ne 0$ for $x > 0$, $b \in l_2$. In fact, $\Vert b\Vert \ge x$ implies $f_x(b) = b \ne 0$; $\frac{x}{2}< \Vert b\Vert < x$ implies $f_x (b) = \alpha b + \beta Tb$ for some $\alpha , \beta > 0$, the assumption $\alpha b + \beta T b = 0$ (for $b = (b_1, b_2, \ldots )$) implies consecutively $b_1 = 0, b_2 = 0, b_3 = 0, \ldots$, hence contradiction; $\Vert b\Vert = \frac{x}{2}$ implies $f_x(b) = 0 \cdot b + Tb + 0 \cdot e \ne 0$; finally, $0\le \Vert b\Vert <\frac{x}{2}$ implies, for $c=(c_1,c_2,\ldots )=f_x(b)$, $c=\alpha Tb+\beta e$, for some $\alpha \in [0,1]$, $\beta \in (0,1]$, so that $c_1=\beta \ne 0$ and $f_x(b)\ne 0$.

We shall show in a few steps that the mapping $(x, b) \mapsto f_x (b)$ is Lipschitzian. Fix $x, y \ge 0$, $b, c \in l_2$. For $x = y = 0$, we have $\Vert f_0(b) - f_0(c)\Vert = \Vert b - c\Vert$. For $x > 0$, $y = 0$ consider the following cases: $\Vert b\Vert \ge x$ implies $\Vert f_x(b) - f_0(c)\Vert = \Vert b - c\Vert$; $\Vert b\Vert < x$, $\Vert c\Vert \ge 2x$ implies the existence of $\alpha , \beta , \gamma \in [0,1]$ satisfying $\Vert f_x(b) - f_0(c)\Vert = \Vert \alpha b + \beta Tb + \gamma xe - c\Vert \le \Vert c\Vert + 2\Vert b\Vert + x \le \Vert c\Vert - \Vert b\Vert + 4x \le 5(\Vert c\Vert - \Vert b\Vert ) \le 5\Vert b-c\Vert$; similarly, $\Vert b\Vert < x$, $\Vert c\Vert < 2x$ implies

$$\begin{aligned} \Vert f_x(b) - f_0(c)\Vert \le \Vert b\Vert + \Vert Tb\Vert + x + \Vert c\Vert \le 5(x - 0). \end{aligned}$$

To investigate the case $x, y > 0$, we will use the following lemma:.

Lemma 33

For $x,y>0$ we have

• (A) $|x(1 - \alpha _x (s)) - y(1 - \alpha _y (t))| \le 2|s - t| + 2|x - y|$, for $s,t\ge 0$.

• (B) $\Vert \alpha _x (\Vert b\Vert )b - \alpha _y (\Vert c\Vert )c\Vert \le 2|x-y| + 3\Vert b-c\Vert$, for $b,c\in l_2$.

Proof

(A) We have

$$\begin{aligned}&|\frac{d}{du}(x(1- \alpha _x(u)))| \in \{0, 2\} \text { for } u\in U:=\big (0,\frac{x}{2}\big )\cup \big (\frac{x}{2},x\big )\cup (x,\infty ); \\&\quad \big |\frac{d}{dz}(z(1-\alpha _z(t)))\big |\in \{0, 1, 2 \} \text { for } z\in Z:=(0,t)\cup (t,2t)\cup (2t,\infty ). \end{aligned}$$

In particular,

$$\begin{aligned}&|x (1-\alpha _x(s)) - y(1- \alpha _y(t))| \le |s-t| \max \limits _{u\in U} \bigg |\frac{d}{du}x(1-\alpha _x(u))\bigg | \\&\quad + |x-y| \max \limits _{z\in Z} \bigg |\frac{d}{dz}z(1-\alpha _z(t))\bigg | \le 2|s-t| + 2|x-y|. \end{aligned}$$

(B) We can assume that $\Vert b\Vert \le \Vert c\Vert$ and $\Vert b\Vert >0$ (the case of $b=0$ is obvious). With $s := \Vert b\Vert$, $t := \Vert c\Vert$, we have

$$\begin{aligned} \begin{aligned} \Vert \alpha _x (\Vert b\Vert ) b - \alpha _y (\Vert c\Vert ) c \Vert&\le \Vert (\alpha _x (\Vert b\Vert ) - \alpha _y (\Vert c\Vert ))b \Vert \\&\quad + \Vert \alpha _y (\Vert c\Vert )(b-c) \Vert \\&\le | \alpha _x (s) - \alpha _y (t)| s + \Vert b - c\Vert \\ {}&\le ( \alpha _{x\wedge y} (t) - \alpha _{x \vee y} (s)) s + \Vert b - c\Vert \end{aligned} \end{aligned}$$

(40)

since $\alpha _z (s)$ decreases as z increases. One can suppose that $x \le y$ and use the estimation:

$$\begin{aligned} (\alpha _x (t) - \alpha _y (s))s \le (\alpha _x (t) - \alpha _x (s))s + (\alpha _x (s) - \alpha _y (s))s; \end{aligned}$$

$$\begin{aligned} (\alpha _x (t) - \alpha _x (s))s&\le (t-s) \max _{u>0} \frac{d}{du} (\alpha _x(u)s)\\&=(t-s)\max _{x/2<u<x}\frac{d}{du}\bigg (\frac{2u-x}{x}s\bigg )\\&\le 2(t-s) \ \text {for} \ s\le x;\\ (\alpha _x(t)-\alpha _x(s))s&=0 \ \text {for} \ s\ge x;\\ \max _{s>0}(\alpha _x(s)-\alpha _y(s))s&\le \max _{s\in (0,y]}(\alpha _x(s)-\alpha _y(s))y\\&=(\alpha _x(x)-\alpha _y(x))y\\&=\bigg (1-\bigg (\frac{2x-y}{y} \bigg )^+\bigg )y\\&\le 2(y-x); \end{aligned}$$

$$\begin{aligned} t-s=\Vert c\Vert -\Vert b\Vert \le \Vert b-c\Vert . \end{aligned}$$

Finally, by (40), we get (B). $\square$

We return to Example 32. The mapping $(x,b)\mapsto f_x(b)$, for $x>0$ and $b\in l_2$ is Lipschitzian. In fact, T is an isometry, hence the mapping $(x,b)\mapsto \alpha _x(\Vert b\Vert )b+(\alpha _{x/2}(\Vert b\Vert )-\alpha _x(\Vert b\Vert ))Tb$ is Lipschitzian by part B of Lemma 33. The mapping $(x,b)\mapsto (1-\alpha _{x/2}(\Vert b\Vert ))xe$ is Lipschitzian by part A.

Example 32 shows that obtaining a result analogous to Theorem 30 for an arbitrary Banach space ${\mathfrak {X}}$ in place of ${\mathbb {R}}^n$ requires stronger assumptions on the mapping $(X,b)\mapsto f_X(b)\in {\mathfrak {X}}$, defined for $\Vert X\Vert$, $\Vert b-b_0\Vert <\varepsilon$, $X\in {\mathcal {N}}$, $b\in {\mathfrak {X}}$, for a fixed $\varepsilon >0$. The following observation proves useful:

Theorem 34

Let ${\mathcal {N}}$ be a normed space and ${\mathfrak {X}}$ a Banach space. If $b_0\in {\mathfrak {X}},\,\varepsilon >0$ and the mapping $X\mapsto f_X(b_0)\in {\mathfrak {X}}$ is continuous at $X=0,$ $f_0(b)=b$ for $\Vert b-b_0\Vert \le \varepsilon$ and, for some $K>0$, condition ( 39) is satisfied for $X\in {\mathcal {N}}$, $b,c\in {\mathfrak {X}}$, $\Vert X\Vert ,\Vert b-b_0\Vert ,\Vert c-b_0\Vert <\varepsilon ,$ then there exists $\delta >0$ such that for $b\in {\mathfrak {X}}$, $X\in {\mathcal {N}}$, $\Vert X\Vert , \Vert b-b_0\Vert <\delta$, there exists $c\in {\mathfrak {X}}$, $\Vert c-b_0\Vert <\varepsilon$ satisfying $f_X(c)=b$.

Proof

We choose $\delta \in \big (0,\varepsilon /3)$ so small that $\Vert X\Vert <\delta$ implies $\Vert f_X(b_0)-b_0\Vert =\Vert f_X(b_0)-f_0(b_0)\Vert <\varepsilon /3$, $K\Vert X\Vert \le 1/3$. Then $\Vert b-b_0\Vert <\varepsilon /3$, $\Vert X\Vert <\delta$ implies $f_X(c)=b$ for some $c\in {\mathfrak {X}}$, $\Vert c-b_0\Vert \le \varepsilon$, according to Lemma 9 (Section 3).

$\square$

6.2 Principle of Using Unitary Transformations

Theorem  34 follows easily from the Banach contraction principle. However, it uses a strong assumption (39) about the family $f_X : {\mathfrak {X}}\rightarrow {\mathfrak {X}}$ for $X \in {\mathcal {N}}$, $\Vert X\Vert <\varepsilon$ (for a Banach space ${\mathfrak {X}}$ and a normed space ${\mathcal {N}}$, real or complex). The usefulness of the theorem is a consequence of the following fact. Let a family $g_X : B(b_0, \varepsilon ) \rightarrow {\mathcal {Y}}$ be defined on a closed ball $B(b_0,\varepsilon )\subset {\mathfrak {X}}$, for normed spaces ${\mathfrak {X}},{\mathcal {Y}}$, and let as before $X\in {\mathcal {N}}$ with $\Vert X\Vert <\varepsilon$, for a normed space ${\mathcal {N}}$. Assume that the following condition (which is, in fact, condition (39) with $f_X$ replaced with $g_X$, but with a different constant) is fulfilled: For some $K_1>0$ we have

$$\begin{aligned} \Vert g_X(b)-g_X(c)-g_0(b)+g_0(c)\Vert \le K_1\Vert X\Vert \Vert b-c\Vert \end{aligned}$$

(41)

for $X\in {\mathcal {N}},\,b,c\in {\mathfrak {X}}$ with $\Vert X\Vert ,\Vert b-b_0\Vert ,\Vert c-b_0\Vert \le \varepsilon$. Then we can get condition (39) for the family

$$\begin{aligned} f_X(b)=(v_{v_{g_0(b)}(g_X(b))}\circ v_{g_0(b)})(g_X(b)), \end{aligned}$$

(42)

defined for some family of unitary transformations $\{v_B : {\mathcal {Y}}\rightarrow {\mathcal {Y}}:B\in {\mathcal {Y}}_0\}$, for a set of parameters ${\mathcal {Y}}_0\subset {\mathcal {Y}}$. The properties of (slightly generalized) formula (42) is given by the following theorem.

Theorem 35

Let ${\mathcal {N}}$, ${\mathfrak {X}}$ and ${\mathcal {Y}}$ be normed spaces, $b_0\in {\mathfrak {X}}$ and $\varepsilon >0$. Let the mapping $(X,b)\mapsto g_X(b)\in {\mathcal {Y}}$ be Lipschitzian and satisfy condition ( 41) (with some $K_1>0$) for $X\in {\mathcal {N}},\, b,c\in {\mathfrak {X}}$ with $\Vert X\Vert ,\Vert b-b_0\Vert , \Vert c-b_0\Vert \le \varepsilon$; suppose further both $b\mapsto v_b$ and $B\mapsto u_B$ to be Lipschitzian families of (linear surjective) isometries of ${\mathcal {Y}}$ for $b\in {\mathfrak {X}}$ and $B\in {\mathcal {Y}}$ with $\Vert b-b_0\Vert ,\,\Vert B-v_{b_0}(g_0(b_0))\Vert <\varepsilon$. We assume additionally that for B from a neighbourhood of $v_{b_0}(g_0(b_0)),$ and b, c from a neighbourhood of $b_0$ the two following formulas

• (i) $u_{v_b(g_0(b))}=\mathrm{id}$,

• (ii) $u_B=u_{B+u_B^{-1}(v_b(g_0(b))-v_c(g_0(c)))}$,

make sense and are satisfied. Then the family

$$\begin{aligned} f_X(b)=(u_{v_b(g_X(b))}\circ v_b)(g_X(b)) \end{aligned}$$

is well-defined and satisfies ( 39) for a constant K and $X\in {\mathcal {N}},\, b,c\in {\mathfrak {X}}$ with $\Vert X\Vert ,\Vert b-b_0\Vert , \Vert c-c_0\Vert <\delta ,$ for some $\delta >0$.

Proof

We will essentially repeat the idea of the proof of Theorem B from Sect. 3. We can choose $K_i>0$ so large (and also $\varepsilon _i>0$ so small), that for $Y,Z\in {\mathcal {N}},\, b,c\in {\mathfrak {X}}$ with $\Vert Y\Vert ,\Vert Z\Vert , \Vert b-b_0\Vert ,\Vert c-b_0\Vert <\varepsilon$ we will have

$$\begin{aligned} \Vert g_Y(b)-g_Z(c)\Vert&<K_2(\Vert Y-Z\Vert +\Vert b-c\Vert ), \end{aligned}$$

(43)

$$\begin{aligned} \Vert g_Y(b)-g_Y(c)-g_0(b)+g_0(c)\Vert&\le K_1\Vert Y\Vert \Vert b-c\Vert \qquad \text {(cf. }(41)\mathrm{)}. \end{aligned}$$

(44)

Moreover, for $B_0 :=v_{b_0}(g_0(b_0))$ and arbitrary $Y\in {\mathcal {N}},\, b,c\in {\mathfrak {X}},\,B,C\in {\mathcal {Y}}$ with $\Vert Y\Vert ,\Vert B-B_0\Vert ,\Vert C-B_0\Vert ,\Vert b-b_0\Vert , \Vert c-b_0\Vert <\varepsilon$ we will have

$$\begin{aligned} \Vert g_0(b)\Vert , \Vert B+Y\Vert&\le K_3, \end{aligned}$$

(45)

$$\begin{aligned} \Vert u_B-u_C\Vert&\le K_4 \Vert B-C\Vert , \end{aligned}$$

(46)

$$\begin{aligned} \Vert v_b-v_c\Vert&\le K_5 \Vert b-c\Vert . \end{aligned}$$

(47)

We can assume that $\varepsilon _1 < \varepsilon$ and that for $\Vert Y\Vert , \Vert B-B_0\Vert ,\Vert C-B_0\Vert ,\Vert b-b_0\Vert , \Vert c-b_0\Vert <\varepsilon _1$ the following formulas

$$\begin{aligned}&u_{v_b(g_0(b))}=\mathrm{id}\text { (cf. (i)),} \end{aligned}$$

(48)

$$\begin{aligned}&u_{B+Y+u_{B+Y}^{-1}(v_b(g_0(b))-v_c(g_0(c)))} = u_{B+Y} \text { (cf. (ii)),} \end{aligned}$$

(49)

$$\begin{aligned}&\Vert B+Y-B_0\Vert , \Vert B+Y+u_{B+Y}^{-1}(C-B)-B_0\Vert <\varepsilon \end{aligned}$$

(50)

make sense and are satisfied; additionally, we can suppose that $\varepsilon _2<\varepsilon _1$ and that, for $\Vert b-b_0\Vert <\varepsilon _2,$ we have

$$\begin{aligned} \Vert v_b(g_0(b))-v_{b_0}(g_0(b_0))\Vert <\varepsilon _1 \quad \text {(cf. } (47)\mathrm{)}. \end{aligned}$$

(51)

We can now take $0<\delta <\varepsilon _2$ so small that for $X\in {\mathcal {N}},\, b_1, c_1\in {\mathfrak {X}}$ with $\Vert X\Vert , \Vert b_1-b_0\Vert , \Vert c_1-b_0\Vert <\delta$, we have (cf. (43)), using $D_0:=g_0(b_0)$ (so that $B_0=v_{b_0}(D_0)$) and $D_1:=g_0(b_1),\,E_1:=g_0(c_1),\, B_1:=v_{b_1}(D_1),\,C_1:=v_{c_1}(E_1),\, Y_1:=v_{b_1}(g_X(b_1)-D_1),\, Z_1:=v_{c_1}(g_X(c_1)-E_1)$,

$$\begin{aligned} \Vert b_1-b_0\Vert , \Vert c_1-b_0\Vert , \Vert Y_1\Vert , \Vert Z_1\Vert <\varepsilon _1. \end{aligned}$$

By (51) we have as well $\Vert B_1-B_0\Vert ,\Vert C_1-B_0\Vert <\varepsilon _1$.

Conditions (43)–(50) may be now used as follows:

$$\begin{aligned} \Vert Y_1\Vert , \Vert Z_1\Vert&<K_2\Vert X\Vert \qquad \text {(cf. }(43)\mathrm{)};\\ \Vert D_1-E_1\Vert&\le K_2\Vert b_1-c_1\Vert \qquad \text {(cf. }(43)\mathrm{)};\\ \Vert Y_1-Z_1\Vert&\le \Vert v_{b_1}-v_{c_1}\Vert \cdot \Vert g_X(b_1)-D_1\Vert \\&\quad + \Vert v_{c_1}\Vert \cdot \Vert g_X(b_1)-D_1-g_X(c_1)+E_1\Vert \\&\le K_5 \Vert b_1-c_1\Vert \cdot K_2\Vert X\Vert \qquad \text {(cf. }(47), (43)\mathrm{)}\\&\quad + K_1\Vert X\Vert \Vert b_1-c_1\Vert \qquad \text {(cf. }(44)\mathrm{)}\\&= K_6\Vert X\Vert \Vert b_1-c_1\Vert ; \end{aligned}$$

$$\begin{aligned} \Vert B_1-C_1\Vert&\le \Vert v_{b_1}-v_{c_1}\Vert \Vert D_1\Vert +\Vert v_{c_1}\Vert \Vert D_1-E_1\Vert \\&\le K_5 K_3\Vert b_1-c_1\Vert +\Vert D_1-E_1\Vert \text { (cf. }(45), (47)\mathrm{)}\\&\le (K_5 K_3+K_2) \Vert b_1-c_1\Vert \\&= K_7 \Vert b_1-c_1\Vert ; \end{aligned}$$

$$\begin{aligned}&\Vert u_{B_1+Y_1}^{-1}(C_1-B_1)-C_1+B_1\Vert \\&\quad = \Vert u_{B_1+Y_1}^{-1}(C_1-B_1)-u_{B_1}^{-1}(C_1-B_1)\Vert \text { (cf. } (48)\mathrm{)}\\&\quad = \Vert u_{B_1+Y_1}-u_{B_1}\Vert \cdot \Vert C_1-B_1\Vert \\&\quad \le K_4\Vert Y_1\Vert \cdot K_7\Vert b_1-c_1\Vert \quad \text {(cf. }(46), (50)\mathrm{)}\\&\quad \le K_4K_2K_7\Vert X\Vert \Vert b_1-c_1\Vert \\&\quad =K_8\Vert X\Vert \Vert b_1-c_1\Vert ; \end{aligned}$$

$$\begin{aligned} \Vert u_{B_1+Y_1}-u_{C_1+Z_1}\Vert&= \Vert u_{B_1+Y_1+u_{B_1+Y_1}^{-1}(C_1-B_1)}-u_{C_1+Z_1}\Vert \text { (cf. } (49)\mathrm{)}\\&\le K_4\Vert B_1+Y_1+u_{B_1+Y_1}^{-1}(C_1-B_1)\\&\quad -(C_1+Z_1)\Vert \text { (cf. }(46), (50)\mathrm{)}\\&\le K_4\Vert Y_1-Z_1\Vert +K_4\Vert u_{B_1+Y_1}^{-1}(C_1-B_1)-(C_1-B_1)\Vert \\&\le (K_4K_6+K_4K_8)\Vert X\Vert \Vert b_1-c_1\Vert \\&=K_9\Vert X\Vert \Vert b_1-c_1\Vert . \end{aligned}$$

Note that with our notation we have

$$\begin{aligned} f_X(b_1)=u_{v_{b_1}(g_X(b_1))}(v_{b_1}(g_X(b_1)))=u_{B_1+Y_1}(B_1+Y_1) \end{aligned}$$

and similarly $f_X(c_1)=u_{C_1+Z_1}(C_1+Z_1)$ as well as $f_0(b_1)=u_{B_1}(B_1)=B_1$, $f_0(c_1)=C_1$ (cf. (48)), hence

$$\begin{aligned}&\Vert f_X(b_1)-f_X(c_1)-f_0(b_1)+f_0(c_1)\Vert \\&\quad = \Vert u_{B_1+Y_1}(B_1+Y_1)-u_{C_1+Z_1}(C_1+Z_1)-B_1+C_1\Vert \\&\quad \le \Vert u_{B_1+Y_1}-u_{C_1+Z_1}\Vert \Vert C_1+Z_1\Vert \\&\qquad + \Vert u_{B_1+Y_1}(B_1+Y_1-C_1-Z_1)-B_1+C_1\Vert \\&\quad \le K_9\Vert X\Vert \Vert b_1-c_1\Vert \Vert C_1+Z_1\Vert \\&\qquad + \Vert Y_1-Z_1\Vert + \Vert u_{B_1+Y_1}(B_1-C_1)-(B_1-C_1)\Vert \\&\quad \le K_9K_3\Vert X\Vert \Vert b_1-c_1\Vert \qquad \text {(cf. }(45)\mathrm{)} \\&\qquad + K_6\Vert X\Vert \Vert b_1-c_1\Vert + K_8\Vert X\Vert \cdot \Vert b_1-c_1\Vert \\&\quad = K\Vert X\Vert \Vert b_1-c_1\Vert \quad (\text {for some } K>0). \end{aligned}$$

$\square$

Corollary 36

Let ${\mathcal {N}}, {\mathfrak {X}}, {\mathcal {Y}}, b_0, \varepsilon ,$ function $(X,b)\mapsto g_X(b)$ and families $B \mapsto u_B,\, b \mapsto v_b$ of isometries of ${\mathcal {Y}}$ be as in Theorem  35. Assume additionally that ${\mathfrak {X}}$ is a Banach space and that $e:{\mathcal {Y}}\rightarrow {\mathfrak {X}}$ is a linear contraction satisfying $ev_{b}(g_0(b))=b$ for $b\in {\mathfrak {X}}$ with $\Vert b-b_0\Vert <\varepsilon$. Then there exists $\delta >0$ such that for $X\in {\mathcal {N}},\,b\in {\mathfrak {X}}$ with $\Vert X\Vert , \Vert b-b_0\Vert <\delta$ there exists $c\in {\mathfrak {X}}$ with $\Vert c-b_0\Vert <\varepsilon$ such that

$$\begin{aligned} f_X(c):=eu_{v_{c}(g_X(c))}(v_{c}(g_X(c))) \end{aligned}$$

is well-defined and $f_X(c)=b$.

Proof

For $\Vert b-b_0\Vert <\varepsilon _1$ with sufficiently small $\varepsilon _1>0,$ the difference $g_0(b)-g_0(b_0)$ is so small that the point $B=v_{b}(g_0(b))$ satisfies $u_B=\mathrm{id}$ (by condition (i) in Theorem 35). This implies $f_0(b)=ev_{b}(g_0(b))=b$ for $\Vert b-b_0\Vert <\min (\varepsilon _1,\varepsilon )$. By Theorem 35, for some $\delta _1>0,$ inequalities $\Vert X\Vert ,\Vert b-b_0\Vert ,\Vert c-b_0\Vert <\delta _1$ imply (39) for function

$$\begin{aligned} \tilde{f_X}(b)=u_{v_b(g_X(b))}(v_{b}(g_X(b))) \end{aligned}$$

in place of $f_X(b)$. Hence (39) is satisfied for $f_X(b)=e{\tilde{f}}_X(b),$ so that we can apply Theorem 34 with $\min (\varepsilon ,\varepsilon _1,\delta _1)$ in place of $\varepsilon$ to end the proof. $\square$

Corollary 36 is useful if a locally Lipschitzian family of isometries $u_B$ satisfying conditions (i) and (ii) of Theorem 35 surface in a natural way. One example is that of automorphisms of a von Neumann algebra ${\mathcal {M}}$ naturally connected with projections of the algebra. Let us remind the reader that for a fixed projection $e\in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ and an arbitrary $p\in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ with $\Vert e-p\Vert <1,\,V_p=(pe+p^\perp e^\perp )(epe+e^\perp p^\perp e^\perp )^{-1/2}$ is a unitary in ${\mathcal {M}}$ (see Lemma 16). This time we define an inner automorphism $w_p(\cdot ):={{\,\mathrm{Ad}\,}}(V_p^*)(\cdot )=V_p^*(\cdot ) V_p,$ satisfying $w_p(p)=e$ and also $w_e=\mathrm{id},$ and such that $p\mapsto w_p$ is locally Lipschitzian (cf. Lemma 22 with $v_p=w_p^{-1})$.

We will describe now a method of investigation of spectral windows of hermitian operators B. Note that an automorphism u of ${\mathcal {M}}$ preserves spectral windows: We have $u(B)_{|Z}=u(B_{|Z})$ for $B\in {\mathcal {M}}_h$.

Let e be a fixed projection in ${\mathcal {M}},\,b_0$ an operator from $({\mathcal {M}}_e)_h,\,{\mathcal {N}}$ a normed space, $Z\subset {\mathbb {R}}$ a Borel set and $\varepsilon >0$. Suppose that for $X\in {\mathcal {N}},\,b\in ({\mathcal {M}}_e)_h,\, p\in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$ satisfying $\Vert X\Vert ,\Vert b-b_0\Vert ,\Vert p-e\Vert <\varepsilon ,$ the operators $B=g_X(b)\in {\mathcal {M}}_h$ and the automorphisms $v_b, w_p$ of the algebra ${\mathcal {M}}$ satisfy, as always,

$$\begin{aligned}&w_p(p)=e, \end{aligned}$$

(52)

$$\begin{aligned}&w_e=\mathrm{id}, \end{aligned}$$

(53)

$$\begin{aligned}&p\mapsto w_p \text { is Lipschitzian,} \end{aligned}$$

(54)

and additionally

$$\begin{aligned} (X,b)\mapsto g_X(b)\text { is Lipschitzian} \end{aligned}$$

(55)

and (41) holds. Suppose also that

$$\begin{aligned}&b\mapsto v_b\text { is Lipschitzian,} \end{aligned}$$

(56)

$$\begin{aligned}&v_b(g_0(b))_{|Z}=b, \end{aligned}$$

(57)

$$\begin{aligned}&e_{v_b(g_0(b))}(Z)=e. \end{aligned}$$

(58)

Thus we assume a natural strengthening of the condition $g_0(b)_{|Z}\simeq b$.

Theorem 37

If $B_0:=v_{b_0}(g_0(b_0))$ is such that for $B\in {\mathcal {M}}_h,\Vert B-B_0\Vert <\varepsilon$ we have

$$\begin{aligned}&B\mapsto e_B(Z)\text { is Lipschitzian,} \end{aligned}$$

(59)

$$\begin{aligned}&e_B(Z)=e \text { whenever }eB=Be, \end{aligned}$$

(60)

then there exists $\delta >0$ such that for $X\in {\mathcal {N}},\,b\in ({\mathcal {M}}_e)_h$ with $\Vert X\Vert ,\Vert b-b_0\Vert <\delta$ there exists $c\in ({\mathcal {M}}_e)_h$ with $\Vert c-b_0\Vert <\varepsilon$ satisfying $g_X(c)_{|Z}\simeq b$.

Proof

By (60), we have in particular $e_{B_0}(Z)=e$. By (54), for B from some neighbourhood of $B_0$ in ${\mathcal {M}}_h,$

$$\begin{aligned} \text {the mapping }B\mapsto u_B:=w_{e_B(Z)}\text { is Lipschitzian.} \end{aligned}$$

(61)

We will denote by ${\hat{e}}$ contractions from ${\mathcal {M}}_h$ into $({\mathcal {M}}_e)_h$ defined by ${\hat{e}}(B):=eBe$. To use Corollary 36 for $g_X(b),v_b,u_B$ given above and the contraction ${\hat{e}}$ in place of e,  one has, by (56), (55), (41) and (61), to show additionally that

$$\begin{aligned}&{\hat{e}}v_b(g_0(b))=b, \end{aligned}$$

(62)

$$\begin{aligned}&u_{v_b(g_0(b))}=\mathrm{id}, \end{aligned}$$

(63)

$$\begin{aligned}&u_B=u_{B+u_B^{-1}(v_b(g_0(b))-v_c(g_0(c)))} \end{aligned}$$

(64)

for B from some neighbourhood of $B_0$ in ${\mathcal {M}}_h$ and b, c from some neighbourhood of $b_0$ in $({\mathcal {M}}_e)_h$. By (58), we have

$$\begin{aligned} v_b(g_0(b))_{|Z}=ev_b(g_0(b))e={\hat{e}}(v_b(g_0(b))), \end{aligned}$$

so that, by (57), we get (62). Condition (63) we get, in turn, by (58), (61) and (53). Finally, by (61), condition (64) is equivalent to

$$\begin{aligned} e_B(Z)=e_{B+u_B^{-1}(v_b(g_0(b))-v_c(g_0(c)))}(Z), \end{aligned}$$

in other words

$$\begin{aligned} e_{u_B(B)}(Z)=e_{u_B(B)+v_b(g_0(b))-v_c(g_0(c))}(Z). \end{aligned}$$

(65)

Moreover, by (61) and (52), e commutes with $u_B(B),$ while by (58), e commutes with $v_b(g_0(b))$ and $v_c(g_0(c))$. By (60), both sides of (65) are equal to e whenever B is in a sufficiently small neighbourhood of $B_0$ in ${\mathcal {M}}_h$ and b, c are in a sufficiently small neighbourhood of $b_0$ in $({\mathcal {M}}_e)_h$. This ends the justification of (64).

Now, we can use Corollary 36 to get $\delta >0$ such that for $X\in {\mathcal {N}},\,b\in ({\mathcal {M}}_e)_h,\, \Vert X\Vert , \Vert b-b_0\Vert <\delta$ there is a $c\in ({\mathcal {M}}_e)_h$ satisfying $\Vert c-b_0\Vert <\varepsilon$ and

$$\begin{aligned} b={\hat{e}}(u_{v_c(g_X(c))}(v_c(g_X(c)))). \end{aligned}$$

(66)

By (61) and (52),

$$\begin{aligned} u_{v_c(g_X(c))}(e_{v_c(g_X(c))}(Z))=e, \end{aligned}$$

so that for $p:=u^{-1}_{v_c(g_X(c))}(e)$ we have

$$\begin{aligned}&e_{v_c(g_X(c))}(Z)=p,\\&v_c(g_X(c))_{|Z}=pv_c(g_X(c))p \end{aligned}$$

and finally

$$\begin{aligned} u_{v_c(g_X(c))}(v_c(g_X(c))_{|Z})&=eu_{v_c(g_X(c))}(v_c(g_X(c)))e\\&={\hat{e}}(u_{v_c(g_X(c))}(v_c(g_X(c))))\\&=b\qquad \text {(by }(66)\mathrm{)}, \end{aligned}$$

which ends the proof of the equivalence $g_X(c)_{|Z}\simeq b$. $\square$

Remark 38

Conditions (59), (60) are equivalent to the following condition: For $D\in {\mathcal {M}}_h,\, \Vert D-g_0(b_0)\Vert <\varepsilon$ we have:

$$\begin{aligned}&D\mapsto e_D(Z)\text { is Lipschitzian,} \\&e_D(Z)=e_{g_0(b_0)}(Z)\text { whenever }e_{g_0(b_0)}(Z)D=De_{g_0(b_0)}(Z). \end{aligned}$$

As seen from the above, Theorem 37 says that certain properties of the spectral window $g_X(a)_{|Z}$ of operators $g_X(a)$, which are perturbations of the operators $g_0(a)$ for a from a neighbourhood of $b_0,$ can be obtained by examining just the operator $g_0(b_0)$.

Example 39

Conditions (59), (60) are satisfied if $Z=(0,\eta )$ and the operator $B_0:=v_{b_0}(g_0(b_0))$ is of the form $b_0-wb_0w^*$ for $b_0$ from the interior of the set $({\mathcal {M}}_e)_{(0,\eta )}$ in the space $({\mathcal {M}}_e)_h$ and for a partial isometry w such that $w^*w=e,\,ww^*=e^\perp$ (see Lemma 23), if only B is sufficiently close to $B_0$.

6.3 Generalizations and Open Problems

The proof of Theorem A suggests the following generalization. Let ${\mathcal {M}}$ be an arbitrary infinite von Neumann factor (we do not assume separability of the Hilbert space).

Conjecture 40

For $a\in {\mathcal {M}}_h$, suppose that $\max {{\,\mathrm{sp}\,}}a=1$, $\min ({{\,\mathrm{sp}\,}}a\smallsetminus \{0\})=\mu$, $\mu \in [-1,1)$ and $e_a([\mu ,\mu +\delta )) \simeq e_a((1-\delta ,1])\sim 1_{{\mathcal {M}}}$ for all $\delta >0$. Then $a+(1-\mu )r=p+q$ for some $p,q,r\in {{\,\mathrm{Proj}\,}}{\mathcal {M}}$.

Let us stop here to see when, for some positive integer m,  for each operator $a\in {\mathcal {M}}_h,$ the following representation is possible:

$$\begin{aligned} a=\sum _{1\le i\le m}\alpha _i p_i, \ p_i\in {{\,\mathrm{Proj}\,}}{\mathcal {M}}, \ \alpha \in {\mathbb {R}}. \end{aligned}$$

(67)

Of course, we are mainly interested in the smallest m, such that there is an operator $a\in {\mathcal {M}}_h$ which is not a real linear combination of $m-1$ projections. Just now we know that the choice $m=4$ is good enough for (67) to hold for any factor besides factors of type $II_1$. In fact, for type $I_n$ that was shown by Nakamura [16], and in all infinite factors (in particular, in factors of type III) in our previous paper [14, Theorem 3.5]. Up to now, the best m for factors of type $II_1$ is 12. The following claim is much more difficult to establish, and the proof uses methods from Sect. 6.2 in an essential way:

Claim 41

(see [17]) For a factor ${\mathcal {M}}$ of type $II_1$ and $a\in {\mathcal {M}}_h$, the representation ( 67) is possible with $m=4$.

For factors ${\mathcal {M}}$ of finite type I, the situation is unclear. It is known from [16] that $m=3$ for a factor of type $I_n$ with $n\le 7$ and from [13] that $m=4$ whenever ${\mathcal {M}}$ is of type $I_n$ with $n\ge 76$ or an infinite type I factor. In our previous paper [14, Corollary 4.21 and Theorem 5.3] it was shown that $m\ge 4$ for factors of type II. To complete the picture (disregarding factors of type $I_n$ with $7<n<76$) we need the following fact:

Conjecture 42

If ${\mathcal {M}}$ is a non-$\sigma$-finite factor of type III,  then for some $a\in {\mathcal {M}}_h$, the representation (67) is not possible with $m=3$.

It is obvious that if ${\mathcal {M}}$ is a type $I_n$ factor, one can write any $a\in {\mathcal {M}}$ as a complex linear combination of 8 projections. The following conjecture strengthens the statement that 7 projections do not suffice for the purpose.

Conjecture 43

For any k there is n such that for some $a_1,\dots ,a_k\in {\mathcal {M}}_h,$ where ${\mathcal {M}}$ is a factor of type $I_n,$ and for arbitrary projections $p_1, \ldots , p_m$ with $m\le 4k-1$, there is an operator $a_l$, $1\le l\le k$ that is not their linear combination (clearly impossible if $m \ge 4k$).

An analogous question can obviously be discussed for factors of other types.