Total Curvature and the Isoperimetric Inequality in Cartan–Hadamard Manifolds

Abstract

We obtain an explicit formula for comparing total curvature of level sets of functions on Riemannian manifolds and develop some applications of this result to the isoperimetric problem in spaces of nonpositive curvature.

Appendices

Appendix A: Smoothing the Distance Function

In this section we discuss how to smooth the (signed) distance function ${\widehat{d}}_\Gamma$ of a hypersurface $\Gamma$ in a Riemannian manifold M via inf-convolution. We also derive some basic estimates for the derivatives of the smoothing via the associated proximal maps. For $t>0$, the inf-convolution (or more precisely Moreau envelope or Moreau-Yosida regularization) of a function $u:M\rightarrow {\mathbf {R}}$ is given by

$$\begin{aligned} {{\widetilde{u}}}^{\,t}(x):=\inf _y\left\{ u(y)+\frac{d^2(x,y)}{2t}\right\} . \end{aligned}$$

(49)

It is well-known that $\widetilde{u}^{\,t}$ is the unique viscosity solution of the Hamilton–Jacobi equation $f_t+(1/2)|\nabla f|^2=0$ for functions $f:{\mathbf {R}}\times M\rightarrow {\mathbf {R}}$ satisfying the initial condition $f(0,x)=u(x)$. Furthermore, when $M={\mathbf {R}}^n$, ${{\widetilde{u}}}^{\,t}$ is characterized by the fact that its epigraph is the Minkowski sum of the epigraphs of u and $|\cdot |^2/(2t)$ [132, Thm. 1.6.17]. The following properties are well-known,

$$\begin{aligned} \widetilde{(\widetilde{u}^{\,t})}^s={{\widetilde{u}}}^{\,t+s},\quad \quad \;\;\text {and}\quad \quad \;\; \widetilde{\lambda u}^{\,t}=\lambda {{\widetilde{u}}}^{\lambda t}, \end{aligned}$$

(50)

e.g., see [18, Prop. 12.22]. A simple but highly illustrative example of inf-convolution occurs when it is applied to $\rho (x):=d(x_0,x)$, the distance from a single point $x_0\in M$. Then

$$\begin{aligned} {{\widetilde{\rho }}}^{\,t}(x)= \left\{ \begin{array}{ll} \rho ^2(x)/(2t), &{}\,\, \hbox { if}\ \rho (x) \le t,\\ \rho (x)-t/2, &{} \,\,\hbox { if}\ \rho (x)>t , \end{array} \right. \end{aligned}$$

(51)

which is known as the Huber function; See Fig. 3 which shows the graph of ${{\widetilde{\rho }}}^{\,t}$ when $M={\mathbf {R}}$ and $x_0=0$. Note that ${{\widetilde{\rho }}}^{\,t}$ is ${\mathcal {C}}^{1,1}$ and convex, $\inf ({{\widetilde{\rho }}}^{\,t})=\inf (\rho )$, $|\nabla {{\widetilde{\rho }}}^{\,t}|\le 1$ everywhere, $|\nabla {{\widetilde{\rho }}}^{\,t}|=1$ when $\rho >t$, and $|\nabla ^2{{\widetilde{\rho }}}^{\,t}|\le C/t$. Remarkably enough, all these properties are shared by the inf-convolution of ${\widehat{d}}_\Gamma$ when $\Gamma$ is d-convex, as we demonstrate below.

Fig. 3

Huber function

Some of the following observations are well-known or easy to establish in ${\mathbf {R}}^n$ or even Hilbert spaces [18, 37]. In the absence of a linear structure, however, finer methods are required to examine the inf-convolution on Riemannian manifolds, especially with regard to its differential properties [9, 10, 19, 23, 65]. First let us record that, by [9, Cor. 4.5]:

Lemma A.1

([9]) Let u be a convex function on a Cartan–Hadamard manifold. Then for all $t>0$ the following properties hold:

• (i) ${{\widetilde{u}}}^{\,t}$ is ${\mathcal {C}}^1$ and convex.

• (ii) $t\mapsto {{\widetilde{u}}}^{\,t}(x)$ is non-increasing, and $\lim _{t\rightarrow 0} {{\widetilde{u}}}^{\,t}(x)=u(x).$

• (iii) $inf ({{\widetilde{u}}}^{\,t})=inf (u)$, and minimum points of ${\widetilde{u}}^{\,t}$ coincide with those of u.

See also [19, Ex. 2.8] for part (i) above. Next let us rewrite (49) as

$$\begin{aligned} {{\widetilde{u}}}^{\,t}(x)=\inf _y F(y), \quad \quad \quad F(y)=F(x,y):=u(y) +\frac{d^2(x,y)}{2t}. \end{aligned}$$

Since $d^2(x,y)$ is strongly convex and u is convex, F(y) is strongly convex and thus its infimum is achieved at a unique point

$$\begin{aligned} x^*:=prox ^u_{t}(x), \end{aligned}$$

which is called the proximal point [18] or resolvent [19] of ${{\widetilde{u}}}^{\,t}$ at x. In other words,

$$\begin{aligned} {{\widetilde{u}}}^{\,t}(x)=F\big (x^*). \end{aligned}$$

The next estimate had been observed earlier [10, Prop. 2.1] for 2tL.

Lemma A.2

Let u be an L-Lipschitz function on a Riemannian manifold. Then

$$\begin{aligned} d(x, x^*)\le tL. \end{aligned}$$

Proof

Suppose, toward a contradiction, that $d(x^*, x)>tL$. Then there exists an $\varepsilon >0$ such that

$$\begin{aligned} d(x^*, x)\ge (1+\varepsilon )tL. \end{aligned}$$

Choose a point $x'$ on the geodesic segment between x and $x^*$ with

$$\begin{aligned} d(x, x')=d(x,x^*)-\varepsilon tL. \end{aligned}$$

Since $\varepsilon$ may be chosen arbitrarily small, we may assume that $x'$ is arbitrarily close to $x^*$. Thus by the local L-Lipschitz assumption, $u(x')-u(x^*)\le Ld(x^*,x')$. Consequently,

$$\begin{aligned} F(x')-F(x^*)= & {} u(x')-u(x^*)+\frac{d^2(x, x')-d^2(x, x^*)}{2t}\\\le & {} Ld(x^*,x')+ \frac{ (d(x,x')-d(x, x^*))(d(x,x')+d(x, x^*))}{2t}\\\le & {} Ld(x^*,x')-d(x^*,x') \,\, \frac{d(x,x')+d(x,x^*)}{2t}\\= & {} d(x^*,x') \left( L- \frac{2d(x,x^*)-d(x^*,x')}{2t}\right) \\\le & {} d(x^*,x')\left( L- \frac{2(1+\varepsilon )tL-d(x^*,x')}{2t}\right) \\= & {} d(x^*, x')\left( \frac{d(x^*,x')-2\varepsilon tL}{2t}\right) \;\;=\;\;-\frac{1}{2} \varepsilon ^2 tL. \end{aligned}$$

So $F(x')<F(x^*)$ which contradicts the minimality of $x^*$, and completes the proof. $\square$

Part (i) below, which shows that the proximal map is nonexpansive, is well-known [19], and part (ii) follows from [9, Prop. 3.7]. Recall that $d_x(\,\cdot \,):=d(x,\,\cdot \,).$

Lemma A.3

([9, 19]) Let u be a convex function on a Cartan–Hadamard manifold. Then

• (i) $d(x^*_1, x^*_2) \le d(x_1,x_2),$

• (ii) If $x^*$ is a regular point of u, then

  $$\begin{aligned} \nabla u(x^*)=-\frac{d(x,x^*)}{t} \nabla d_x(x^*),\quad \quad \text {and} \quad \quad \nabla {{\widetilde{u}}}^{\,t}(x)=\frac{d(x,x^*)}{t} \nabla d_{x^*}(x). \end{aligned}$$

• (iii) $\nabla u (x^*)$ and $\nabla {{\widetilde{u}}}^{\,t} (x)$ are tangent to the geodesic connecting $x^*$ to x, and

  $$\begin{aligned} |\nabla u (x^*)|=|\nabla {{\widetilde{u}}}^{\,t} (x)|. \end{aligned}$$

• (iv) If u is L-Lipschitz, then so is ${{\widetilde{u}}}^{\,t}$.

Proof

For part (i) see [19, Thm. 2.2.22]. For part (ii) note that by definition $F(y)\ge F(x^*)$. Furthermore, $x^*$ is a regular point of F, since by assumption $x^*$ is a regular point of u. Consequently,

$$\begin{aligned} 0= \nabla F(x^*)= \nabla _y F(x,y)\big |_{y=x^*}=\nabla u(x^*)+\frac{d(x,x^*)}{t} \nabla d_x(x^*), \end{aligned}$$

which yields the first equality in (ii). Next we prove the second inequality in (ii) following [9, Prop. 3.7]. To this end note that

$$\begin{aligned} {{\widetilde{u}}}^{\,t}(z)= & {} \inf _y F(z,y)\le F(z,x^*)= u(x^*)+\frac{d^2(z,x^*)}{2t},\\ {{\widetilde{u}}}^{\,t}(x)= & {} \inf _y F(x,y)=F(x,x^*)=u(x^*)+ \frac{d^2(x,x^*)}{2t}. \end{aligned}$$

So it follows that

$$\begin{aligned} {{\widetilde{u}}}^{\,t}(z)-\frac{d(z,x^*)^2}{2t}\le u(x^*)=\widetilde{u}^{\,t}(x)-\frac{d^2(x,x^*)}{2t}. \end{aligned}$$

Hence $g(\cdot ):={{\widetilde{u}}}^{\,t}(\cdot )-d(\cdot ,x^*)^2/(2t)$ achieves its maximum at x. Further note that g is ${\mathcal {C}}^1$ since $\widetilde{u}^{\,t}$ is ${\mathcal {C}}^1$ by Lemma A.1. Thus

$$\begin{aligned} 0=\nabla g(x)= \nabla {{\widetilde{u}}}^{\,t}(x)-\frac{d(x,x^*)}{t} \nabla d_{x^*}(x), \end{aligned}$$

which yields the second equality in (ii). Next, to establish (iii), let $\alpha :[0, s_0]\rightarrow M$ be the geodesic with $\alpha (0)=x^*$ and $\alpha (s_0)=x$. Then, by Lemma 2.2,

$$\begin{aligned} \nabla d_x(x^*)=-\alpha '(0),\quad \quad \text {and}\quad \quad \nabla d_{x^*}(x)=\alpha '(s_0). \end{aligned}$$

So $\nabla u(x^*)$ and $\nabla {{\widetilde{u}}}^t(x)$ are tangent to $\alpha$ and

$$\begin{aligned} |\nabla u(x^*)|=\left| \frac{d(x,x^*)}{t}\right| =|\nabla \widetilde{u}^t(x)| \end{aligned}$$

as desired. Finally, to establish (iv), note that if u is L-Lipschitz, then $|\nabla u|\le L$ almost everywhere. Thus by part (iii), $|\nabla {{\widetilde{u}}}^t(x)|=|\nabla u(x^*)|\le L$ for almost every $x\in M$. So ${{\widetilde{u}}}^t$ is L-Lipschitz. $\square$

Recall that we say a function $u:M\rightarrow {\mathbf {R}}$ is locally ${\mathcal {C}}^{1,1}$ provided that it is ${\mathcal {C}}^{1,1}$ in some choice of local coordinates around each point. There are other notions of ${\mathcal {C}}^{1,1}$ regularity [10, 65] devised in order to control the Lipschitz constant; however, all these definitions yield the same class of locally ${\mathcal {C}}^{1,1}$ functions; see [10]. The ${\mathcal {C}}^{1,1}$ regularity of functions is closely related to the more robust notion of semiconcavity which is defined as follows. We say that u is C-semiconcave (or is uniformly semiconcave with a constant C) on a set $\Omega \subset M$ provided that there exists a constant $C>0$ such that for every $x_0\in \Omega$, the function

$$\begin{aligned} x\;\mapsto \; u(x)-Cd^2(x,x_0) \end{aligned}$$

(52)

is concave on $\Omega$. Furthermore, we say u is C-semiconvex, if $-u$ is semiconcave.

Lemma A.4

([10, 37]) If a function u on a Riemannian manifold is both C/2-semiconvex and C/2-semiconcave on some bounded domain $\Omega$, then it is locally ${\mathcal {C}}^{1,1}$ on $\Omega$. Furthermore $|\nabla ^2 u|\le C$ almost everywhere on $\Omega$.

The above fact is well-known in ${\mathbf {R}}^n$, see [37, Cor. 3.3.8] (note that the constant C in the book of Cannarsa and Sinestrari [37] corresponds to 2C in this work due to a factor of 1/2 in their definition of semiconcavity.) The Riemannian analog follows from the Euclidean case via local coordinates to obtain the ${\mathcal {C}}^{1,1}$ regularity (since semiconcavity is preserved under ${\mathcal {C}}^2$ diffeomorphisms), and then differentiating along geodesics to estimate the Hessian, see the proof of [37, Cor. 3.3.8], and using Rademacher’s theorem. The above lemma has also been established in [10, Thm 1.5]. The next observation, with a nonexplicit estimate for C, has been known [10, Prop. 7.1(2)]. Here we provide another argument via Lemma A.2.

Proposition A.5

Suppose that u is a convex function on a bounded domain $\Omega$ in a Riemannian manifold. Then for all $0<t\le t_0$, $\widetilde{u}^{\,t}$ is C/(2t)-semiconcave on $\Omega$ for

$$\begin{aligned} C\ge \sqrt{-K_0}\,3t_0L\, \coth \left( \sqrt{-K_0}\, 3t_0L\right) , \end{aligned}$$

(53)

where $K_0$ is the lower bound for the curvature of $B_{t_0L}(\Omega )$, and L is the Lipschitz constant of u on $\Omega$. In particular, ${{\widetilde{u}}}^{\,t}$ is locally ${\mathcal {C}}^{1,1}$, and

$$\begin{aligned} |\nabla ^2 {{\widetilde{u}}}^{\,t}|\le \frac{C}{t} \end{aligned}$$

(54)

almost everywhere on $\Omega$.

Proof

Since by Lemma A.1, ${{\widetilde{u}}}^{\,t}$ is convex, it is C/(2t)-semiconvex. Thus as soon as we show that $\widetilde{u}^{\,t}$ is C/(2t)-semiconcave, ${{\widetilde{u}}}^{\,t}$ will be ${\mathcal {C}}^{1,1}$ and (54) will hold by Lemma A.4, which will finish the proof. To establish the semiconcavity of ${{\widetilde{u}}}^{\,t}$ note that, by Lemma A.2,

$$\begin{aligned} {{\widetilde{u}}}^{\,t}(x)=\inf _{y\in B_{tL}(x)}\left( u(y)+\frac{1}{2t}d^2(x,y)\right) . \end{aligned}$$

Let C be as in (53) and, according to (52), set

$$\begin{aligned} f(x):={{\widetilde{u}}}^{\,t}(x)-\frac{C}{2t}d^2(x,x_0)=\inf _{y\in B_{tL}(x)}\left( u(y)- \frac{1}{2t}\Big (Cd^2(x,x_0)-d^2(x,y)\Big )\right) . \end{aligned}$$

We have to show that f is concave on $\Omega$. To this end it suffices to show that f is locally concave on $\Omega$, since a locally concave function is concave. Indeed suppose that f is locally concave on $\Omega$ and let $\alpha :[a,b]\rightarrow \Omega$ be a geodesic. Then $-f\circ \alpha$ is locally convex. Thus, since $-f\circ \alpha$ is ${\mathcal {C}}^1$, $-(f\circ \alpha )'$ is nondecreasing, which yields that $-f\circ \alpha$ is convex [132, Thm. 1.5.10]. Now, to establish that f is locally concave on $\Omega$, set

$$\begin{aligned} r:=t_0L. \end{aligned}$$

We claim that f is concave on $B_r(p)$, for all $p\in \Omega$. To see this first note that if $x\in B_{r}(p)$ then $B_{tL}(x)\subset B_{r}(x)\subset B_{2r}(p)$. So, for $x\in B_{r}(p)$,

$$\begin{aligned} f(x)=\inf _{y\in B_{2r}(p)}\left( u(y)- \frac{1}{2t}\Big (Cd^2(x,x_0)-d^2(x,y)\Big )\right) . \end{aligned}$$

(55)

Since the infimum of a family of concave functions is concave, it suffices to check that the functions on the right hand side of (55) are concave on $B_{2r}(p)$ for each y. So we need to show that

$$\begin{aligned} g(x):=\frac{1}{2}\big (Cd^2(x,x_0)- d^2(x,y)\big ) \end{aligned}$$

is convex on $B_{2r}(p)$ for each y. To this end note that the eigenvalues of $\nabla ^2d^2_{x_0}(x)/2$ are bounded below by 1 [97, Thm. 6.6.1]. Furthermore, since $x\in B_{r}(p)$, and $y\in B_{2r}(p)$, we have $x\in B_{3r}(y)$. Thus the eigenvalues of $\nabla ^2 d^2_y(x)/2$ are bounded above by

$$\begin{aligned} \lambda :=\sqrt{-K_0}\,3r \coth \left( \sqrt{-K_0}\, 3r\right) , \end{aligned}$$

by [97, Thm. 6.6.1]. So the eigenvalues of $\nabla ^2g$ on $B_{2r}(p)$ are bounded below by $C-\lambda$. Hence g is convex on $B_{2r}(p)$ if $C\ge \lambda$, which is indeed the case by (53). So f is concave on $B_{2r}(p)$ which completes the proof. $\square$

Proposition A.6

Let $\Gamma$ be a closed hypersurface in a Cartan–Hadamard manifold M and set $u:={\widehat{d}}_\Gamma$. Then

• (i) ${{\widetilde{u}}}^{\,t}=u-t/2$ on $M \setminus U_t(cut (\Gamma ))$.

• (ii) $|\nabla {{\widetilde{u}}}^{\,t}|\equiv 1$ on $M\setminus U_t(cut (\Gamma ))$.

• (iii) $|\nabla {{\widetilde{u}}}^{\,t}|\le 1$ on M if $\Gamma$ is d-convex.

Proof

Let $x\in M\setminus {{\,\mathrm{cl}\,}}(U_t(cut (\Gamma )))$. Then $x^*$ is a regular point of u by Lemma A.2. So

$$\begin{aligned} 0=\nabla F(x^*)=\nabla u(x^*) +\frac{d(x,x^*)}{t} \nabla d_x(x^*). \end{aligned}$$

Since $|\nabla u(x^*)|=|\nabla d_x(x^*)|=1$, it follows that

$$\begin{aligned} d(x,x^*)=t. \end{aligned}$$

(56)

Furthermore, we obtain $\nabla u(x^*) =-\nabla d_x(x^*)$. But $\nabla d_x(x^*)$ is tangent to the geodesic which passes through $x^*$ and x, while $\nabla u(x^*)$ is tangent to the flow line of $\nabla u$ through $x^*$, which is also a geodesics. So $x^*$ lies on the geodesic $\alpha$, given by $\alpha (0)=x$ and $\alpha '(0)=\nabla u(x)$. Note that $u(\alpha (t))=u(x)+t.$ Furthermore, by (56), either $x^*=\alpha (t)$ or $x^*=\alpha (-t)$. If $x^*=\alpha (-t)$, then

$$\begin{aligned} u(x^*)=u(\alpha (-t))=u(x)-t. \end{aligned}$$

Hence

$$\begin{aligned} {{\widetilde{u}}}^t(x)=F(x^*)=u(x^*)+\frac{t^2}{2t}=u(x)-\frac{t}{2}, \end{aligned}$$

as desired. If on the other hand $x^*=\alpha (t)$, then a similar computation yields that ${{\widetilde{u}}}^t(x)=u(x)+t/2>u(x)$, which is not possible. So we have established part (i) of the proposition. To see part (ii) note that $|\nabla u|\equiv 1$ on $M\setminus U_t(cut (\Gamma ))$. Thus by (i) $|\nabla \widetilde{u}^t|\equiv |\nabla u|\equiv 1$ on $M\setminus U_t(cut (\Gamma ))$. Finally, part (iii) follows immediately from part (iv) of Lemma A.3. $\square$

Appendix B: Cut Locus of Convex Hypersurfaces

Recall that a hypersurface is d-convex if its distance function is convex, as we discussed in Sect. 3. Here we will study the cut locus of d-convex hypersurfaces and establish the following result:

Theorem B.1

Let $\Gamma$ be a d-convex hypersurface in a Cartan–Hadamard manifold M, and let $\Omega$ be the convex domain bounded by $\Gamma$. Then for any point $x\in \Omega$ and any of its footprints $x^\circ$ in $cut (\Gamma )$,

$$\begin{aligned} d_\Gamma (x^\circ ) \ge d_\Gamma (x). \end{aligned}$$

Throughout this section we will assume that M is a Cartan–Hadamard manifold. In particular the exponential map $\exp _p:T_p M\rightarrow {\mathbf {R}}^n$ will be a global diffeomorphism. The proof of the above theorem is based on the notion of tangent cones, which we defined in Sect. 6. Another approach to proving this result is discussed in a recent work of Kapovitch and Lytchak [98]. We start by recording that for a given a set $X\subset {\mathbf {R}}^n$ and $p\in X$, $T_p X$ is the limit of dilations of X based at p [75, Sect. 2]. More precisely, if we identify p with the origin o of ${\mathbf {R}}^n$, and for $\lambda \ge 1$ set $\lambda X:=\{\lambda x\mid x\in X\},$ then $T_o X$ is the outer limit [129] of the sets $\lambda X$:

$$\begin{aligned} T_o X=\limsup _{\lambda \rightarrow \infty } \lambda X. \end{aligned}$$

(57)

This means that for every $x\in T_o X\setminus \{o\}$ there exists a sequence of numbers $\lambda _i\rightarrow \infty$ such that $\lambda _iX$ eventually intersects any open neighborhood of x. Equivalently, we may record that:

Lemma B.2

([75]) Let $X\subset {\mathbf {R}}^n$ and $o\in X$. Then $x\in T_o X\setminus \{o\}$ if there exists a sequence of points $x_i\in X\setminus \{o\}$ such that $x_i\rightarrow o$ and $x_i/| x_i|\rightarrow x/|x|$.

The last lemma yields:

Lemma B.3

Let $\Gamma \subset {\mathbf {R}}^n$ be a closed hypersurface, and $o\in cut (\Gamma )\cap \Gamma$. Suppose that $T_o\Gamma$ bounds a convex cone containing $\Gamma$. Then

$$\begin{aligned} cut (T_{o}\Gamma )\subset T_{o}cut (\Gamma ). \end{aligned}$$

Proof

By (3) $cut (T_{o}\Gamma )={{\,\mathrm{cl}\,}}(\text {medial}(T_o\Gamma ))$. So it suffices to show that $\text {medial}(T_o\Gamma )\subset T_{o}cut (\Gamma )$, since $cut (T_{o}\Gamma )$ is closed by definition. Let $x\in \text {medial}(T_o\Gamma )$. Then there exists a sphere S centered at x which is contained in (the cone bounded by) $T_{o}\Gamma$, and touches $T_{o}\Gamma$ at multiple points. Suppose that S has radius r. Then, by (57), for each natural number i we may choose a number $\lambda _i$ so large that the sphere $S_i$ of radius $r-(1/i)$ centered at x is contained in ${\lambda _i}\Gamma$. Let $S_i'$ be the largest sphere contained in ${\lambda _i}\Gamma$ centered at x which contains $S_i$. Then $S_i'$ must intersect ${\lambda _i}\Gamma$ at some point y. Let $S_i''$ be the largest sphere contained in $\lambda _i\Gamma$ which passes through y. Then the center $c_i$ of $S_i''$ lies in $skeleton (\lambda _i\Omega )$, and therefore belongs to $cut ({\lambda _i}\Gamma )$, by Lemma 2.4. Now note that $cut ({\lambda _i}\Gamma )=\lambda _icut (\Gamma )$. So

$$\begin{aligned} x_i:=\frac{c_i}{\lambda _i}\in \frac{cut ({\lambda _i}\Gamma )}{\lambda _i}\in cut (\Gamma ). \end{aligned}$$

Furthermore, note that $c_i\rightarrow x$, since $S_i''$ and $S_i$ have a point in common, $S_i''$ is a maximal sphere in $\lambda _i\Gamma$, $S_i$ is a maximal sphere in $T_o\Gamma$, and $\lambda _i\Gamma \rightarrow T_o\Gamma$ according to (57). Thus $x_i\rightarrow o$, and $x_i/|x_i|\rightarrow x/|x|$. So $x\in T_{o}cut (\Gamma )$ by Lemma B.2, which completes the proof. $\square$

For any set $X\subset {\mathbf {R}}^n$ we define ${{\,\mathrm{cone}\,}}(X)$ as the set of all rays which emanate from the origin o of ${\mathbf {R}}^n$ and pass through a point of X. Furthermore we set

Lemma B.4

Let X be the boundary of a proper convex cone C with interior points in ${\mathbf {R}}^n$ and apex at o. Suppose that X is not a hyperplane. Then

where denotes the portion of the cut locus of as a hypersurface in ${\mathbf {S}}^{n-1}$, which is contained in C.

Proof

Let . Then, since X is not a hyperplane, there exists a sphere S centered at x which is contained inside the cone bounded by X and touches X at multiple points, or else x is a limit of the centers of such spheres, by (3). Consequently, forms a sphere in ${\mathbf {S}}^{n-1}$, centered at x, which is contained inside and touches at multiple points, or is the limit of such spheres, respectively. Thus x belongs to , which yields that The reverse inequality may be established similarly. $\square$

Using the last lemma, we next show:

Lemma B.5

Let X be as in Lemma B.4. Suppose that X is not a hyperplane. Then for every point $x\in X$, there exists a point $s\in cut (X)$ such that

$$\begin{aligned} \langle s, x\rangle >0. \end{aligned}$$

Proof

We may replace x by x/|x|. Then, by Lemma B.4, it is enough to show that $\langle s,x\rangle > 0$ for some , or equivalently that $\delta _{{\mathbf {S}}^{n-1}}(s,x)< \pi /2$, where $\delta _{{\mathbf {S}}^{n-1}}$ denotes the distance in ${\mathbf {S}}^{n-1}$. To this end let s be a footprint of x on . Suppose toward a contradiction that $\delta _{{\mathbf {S}}^{n-1}}(s,x)\ge \pi /2$. Consider the great sphere G in ${\mathbf {S}}^{n-1}$ which passes through s and is orthogonal to the geodesic segment xs; See Fig. 4. Let $G^+$ be the hemisphere bounded by G which contains x. Then the interior of $G^+$ is disjoint from , since . Next note that the intersection of the convex cone bounded by X with ${\mathbf {S}}^{n-1}$ is a convex set in ${\mathbf {S}}^{n-1}$. Thus G divides this convex set into two subregions. Consider the region, say R, which contains x, or lies in $G^+$, and let S be a sphere of largest radius in R. Then S must touch the boundary of R at least twice. Since S cannot touch G more than once, it follows that S must touch , because the boundary of R consists of a part of G and a part of . First suppose that S touches multiple times. Then the center of S belongs to . But this is impossible, since $S\subset R\subset G^+$. We may suppose then that S touches only once, say at a point y.

Fig. 4

Diagram for proof of Lemma B.5

Now we claim that the diameter of S is $\ge \pi /2$. Indeed let $G'$ be the great sphere which passes through y and is tangent to S. Then $G'$ supports , and R is contained entirely between G and $G'$. The maximum length of a geodesic segment orthogonal to both G and $G'$ is then equal to the diameter of S, since the points where S touches G and $G'$ must be antipodal points of S. In particular the length of the diameter of S must be greater than $\delta _{{\mathbf {S}}^{n-1}}(x,s)$ as desired.

Finally let $S'$ be the largest sphere contained in which passes through y. Then the center, say z, of $S'$ belongs to by Lemma 2.4. But the diameter of $S'$ is $< \pi$, since X is not a hyperplane by assumption. So, since the diameter of S is $\ge \pi /2$, it follows that z is contained in the interior of S and therefore in the interior R. Hence we reach the desired contradiction since, as we had noted earlier, R does not contain points of in its interior. $\square$

For $x\in \Omega$, set

$$\begin{aligned} \Omega _x:=\big \{y\in \Omega \mid d_\Gamma (y)> d_\Gamma (x)\big \},\quad \quad \text {and}\quad \quad \Gamma _x:=\partial \Omega _x. \end{aligned}$$

Lemma B.6

$cut (\Gamma _{x})\subset cut (\Gamma ).$

Proof

As we discussed in the proof of Lemma B.3, it suffices to show that $\text {medial}(\Gamma _x)\subset cut (\Gamma )$ by (3) . Let $y\in \text {medial}(\Gamma _x)$. Then there exists a sphere $S\subset {{\,\mathrm{cl}\,}}(\Omega _x)$ centered at y which intersects $\Gamma _x$ in multiple points. Let $S'$ be the sphere centered at y with radius equal to the radius of S plus $d(x,\Gamma )$. Then $S'\subset {{\,\mathrm{cl}\,}}(\Omega )$ and it intersects $\Gamma$ in multiple points. So, again by (3), $y\in cut (\Gamma )$ as desired. $\square$

We need to record one more observation, before proving Theorem B.1. An example of the phenomenon stated in the following lemma occurs when $\Gamma$ is the inner parallel curve of a (noncircular) ellipse in ${\mathbf {R}}^2$ which passes through the foci of the ellipse, and p is one of the foci.

Lemma B.7

Let $\Gamma$ be a d-convex hypersurface in a Cartan–Hadamard manifold M, and $p\in \Gamma \cap cut (\Gamma )$. Suppose that $T_p\Gamma$ is a hyperplane. Then $T_p cut (\Gamma )$ contains a ray which is orthogonal to $T_p\Gamma$.

Proof

Let $\alpha (t)$, $t\ge 0$, be the geodesic ray, with $\alpha (0)=p$, such that $\alpha '(0)$ is orthogonal to $T_p \Gamma$ and points toward $\Omega$. We have to show that $\alpha '(0)\in T_pcut (\Gamma )$. To this end we divide the argument into two cases as follows.

First suppose that there exists a sphere in ${{\,\mathrm{cl}\,}}(\Omega )$ which touches $\Gamma$ only at p. Then the center of that sphere coincides with $\alpha (t_0)$ for some $t_0>0$. We claim that then $\alpha (t)\in cut (\Gamma )$ for all $t\le t_0$. To see this note that $\alpha (t)$ has a unique footprint on $\Gamma$, namely p, for all $t\le t_0$. For $0<t\le t_0$, let $\Gamma ^{t}:=({\widehat{d}}_\Gamma )^{-1}(-t)$ be the inner parallel hypersurface of $\Gamma$ at distance t. Suppose, toward a contradiction, that $\alpha (t)\not \in cut (\Gamma )$. Then, by Lemma 2.2, ${\widehat{d}}_\Gamma$ is ${\mathcal {C}}^1$ near $\alpha (t)$, which in turn yields that $\Gamma ^t$ is ${\mathcal {C}}^1$ in a neighborhood $U^t$ of $\alpha (t)$. Furthermore, $\Gamma ^t$ is convex by the d-convexity assumption on $\Gamma$. So, by Lemma 6.4, the outward geodesic rays which are perpendicular to $U^t$ never intersect, and thus yield a homeomorphism between $U^t$ and a neighborhood U of p in $\Gamma$. Furthermore, since ${\widehat{d}}_\Gamma$ is ${\mathcal {C}}^1$ near $U^t$, each point of $U^t$ has a unique footprint on $\Gamma$ by Lemma 2.2. Thus there exists a sphere centered at each point of $U^t$ which lies in ${{\,\mathrm{cl}\,}}(\Omega )$ and passes through a point of U. Furthermore each point of U is covered by such a sphere. So it follows that a ball rolls freely on the convex side of U, and therefore U is ${\mathcal {C}}^{1,1}$, by the same argument we gave in the proof of Lemma 2.6. But, again by Lemma 2.6, if U is ${\mathcal {C}}^{1,1}$, then ${\widehat{d}}_\Gamma$ is ${\mathcal {C}}^1$ near U, which is not possible since $p\in U$ and $p\in cut (\Gamma )$. Thus we arrive at the desired contradiction. So we conclude that $\alpha (t)\in cut (\Gamma )$ as claimed, for $0<t\le t_0$, which in turn yields that $\alpha '(0)\in T_pcut (\Gamma )$ as desired.

So we may assume that there exists no sphere in ${{\,\mathrm{cl}\,}}(\Omega )$ which touches $\Gamma$ only at p. Now for small $\varepsilon >0$ let $S_\varepsilon$ be a sphere of radius $\varepsilon$ in ${{\,\mathrm{cl}\,}}(\Omega )$ whose center $c_\varepsilon$ is as close to p as possible, among all spheres of radius $\varepsilon$ in ${{\,\mathrm{cl}\,}}(\Omega )$. Then $S_\varepsilon$ must intersect $\Gamma$ in multiple points, since $\Gamma$ is convex and $S_\varepsilon$ cannot intersect $\Gamma$ only at p. Thus $c_\varepsilon \in cut (\Gamma )$. Let v be the initial velocity of the geodesic $c_\varepsilon p$, and $\theta (\varepsilon )$ be the supremum of the angles between v and the initial velocities of the geodesics connecting $c_\varepsilon$ to each of its footprints on $\Gamma$. We claim that $\theta (\varepsilon )\rightarrow 0$, as $\varepsilon \rightarrow 0$. To see this let $(T_{c_\varepsilon }M)^1$ denote the unit sphere in $T_{c_\varepsilon }M$, centered at $c_\varepsilon$. Furthermore, let $X\subset (T_{c_\varepsilon }M)^1$ denote the convex hull spanned by the initial velocities of the geodesics connecting $c(\varepsilon )$ to its footprints. Then v must lie in X, for otherwise $S_\varepsilon$ may be pulled closer to p. Indeed if $v\not \in X$, then v is disjoint from a closed hemisphere of $(T_{c_\varepsilon }M)^1$ containing X. Let w be the center of the opposite hemisphere. Then $\langle v, w\rangle >0$. Thus perturbing $c(\varepsilon )$ in the direction of w will bring $S_\varepsilon$ closer to p without leaving ${{\,\mathrm{cl}\,}}(\Omega )$, which is not possible. So $v\in X$ as claimed. Now note that the footprints of $c_\varepsilon$ converge to p, since $c_\varepsilon$ converges to p. Furthermore, since $T_p\Gamma$ is a hyperplane, it follows that the angle between every pair of geodesics which connect $c_\varepsilon$ to its footprints vanishes. Thus X collapses to a single point, which can only be v. Hence $\theta (\varepsilon )\rightarrow 0$ as claimed. Consequently $c_\varepsilon p$ becomes arbitrarily close to meeting $\Gamma$ orthogonally, or more precisely, the angle between $\alpha '(0)$ and the initial velocity vector of $pc_\varepsilon$ vanishes as $\varepsilon \rightarrow 0$. Hence, since $c_\varepsilon \in cut (\Gamma )$, it follows once again that $\alpha '(0)\in T_pcut (\Gamma )$ which completes the proof. $\square$

Fig. 5

Diagram for proof of Theorem B.1

Finally we are ready to prove the main result of this section:

Proof of Theorem B.1

Suppose, toward a contradiction, that $d(x,\Gamma )>d(x^\circ ,\Gamma )$ for some point $x\in \Omega$. Then

$$\begin{aligned} x\in \Omega _{x^\circ }, \end{aligned}$$

(58)

See Fig. 5. Since $x^\circ$ is a footprint of x on $\Gamma$, $cut (\Gamma )$ lies outside a sphere of radius $d(x^\circ ,x)$ centered at x. So if we let v be the initial velocity of the geodesic $x^\circ x$, then $\langle y, v\rangle \le 0,$ for all $y\in T_{x^\circ }cut (\Gamma )$, where we identify $T_{x^\circ }cut (\Gamma )$ with ${\mathbf {R}}^n$ and $x^\circ$ with the origin of ${\mathbf {R}}^n$. By Lemma B.6, $T_{x^\circ }cut (\Gamma _{x^\circ })\subset T_{x^\circ }cut (\Gamma ).$ Thus $\langle y, v\rangle \le 0,$ for all $y\in T_{x^\circ }cut (\Gamma _{x^\circ })$. Furthermore, by Lemma B.3, $cut (T_{x^\circ }\Gamma _{x^\circ })\subset T_{x^\circ }cut (\Gamma _{x^\circ }).$ So

$$\begin{aligned} \langle s, v\rangle \le 0, \quad \text {for all}\quad s\in cut (T_{x^\circ }\Gamma _{x^\circ }). \end{aligned}$$

(59)

Furthermore, since $\Gamma$ is d-convex, $T_{x^\circ }\Gamma _{x^\circ }$ bounds a convex cone by Lemma 6.1. Thus, since $T_{x^\circ }\Gamma _{x^\circ }$ contains v, it must be a hyperplane, by Lemma B.5. Consequently, by Lemma B.7, $T_{x^\circ }cut (\Gamma _{x^\circ })$ contains a ray which is orthogonal to $T_{x^\circ }\Gamma _{x^\circ }$. By (59), v must be orthogonal to that ray. So $v\in T_{x^\circ }\Gamma _{x^\circ }$, which in turn yields that $x\in \Gamma _{x^\circ }$. The latter is impossible by (58). Hence we arrive at the desired contradiction. $\square$

Having established Theorem B.1, we record the following consequence of it. Set

$$\begin{aligned} {{\widehat{r}}}(\,\cdot \,):=d\big (\,\cdot \,, cut (\Gamma )\big ). \end{aligned}$$

Recall that, by Lemma 2.1, ${{\widehat{r}}}$ is Lipschitz and thus is differentiable almost everywhere.

Corollary B.8

Let $\Gamma$ be a d-convex hypersurface in a Cartan–Hadamard manifold M, and set $u:={\widehat{d}}_\Gamma$. Suppose that ${\widehat{r}}$ is differentiable at a point $x\in M\setminus cut (\Gamma )$. Then

$$\begin{aligned} \big \langle \nabla u(x),\nabla {\widehat{r}}(x)\big \rangle \ge 0. \end{aligned}$$

In particular (since ${\widehat{r}}$ is Lipschitz), the above inequality holds for almost every $x\in M\setminus cut (\Gamma )$.

Proof

Since ${{\widehat{r}}}$ is differentiable at x, x has a unique footprint $x^\circ$ on $cut (\Gamma )$, by Lemma 2.2(i). Let $\alpha$ be a geodesic connecting x to $x^\circ$. Then, by Lemma 2.2(ii), $\alpha '(0)=-\nabla {\widehat{r}}(x)$. Furthermore, by Theorem B.1, $u\circ \alpha =-{\widehat{d}}_\Gamma \circ \alpha$ is non-increasing. Finally, recall that by Proposition 2.7, u is ${\mathcal {C}}^1$ on $M\setminus cut (\Gamma )$, and therefore, $u\circ \alpha$ is ${\mathcal {C}}^1$ as well. Thus,

$$\begin{aligned} 0\ge (u\circ \alpha )'(0)=\big \langle \nabla u(\alpha (0)), \alpha '(0)\big \rangle =\big \langle \nabla u(x), -\nabla {\widehat{r}}(x)\big \rangle , \end{aligned}$$

as desired. $\square$