# Lipschitz constant of cental projection of unit ball to surface of convex body

Let $X$ be a real normed space equipped with $\lVert\cdot\rVert$ and $M$ be a convex set in $X$, such that $B(0,r) \subset M \subset B(0,R)$ for $r>0$. Here $B(a, t)$ stands for closed ball with center $a$ and radius $t$.

Further, let $\pi\colon S \rightarrow \partial M$ denote the central projection of the unit sphere onto boundary of $M$. We can show that the best Lipschitz constant $L(\pi)$ of the projection is at most $\frac{R(R+r)}{r}$. If $\mu$ is Minkowski functional of $M$, then $\pi(x) = \frac{x}{\mu(x)}$. Hence, for all $x,y \in S$ \begin{align*} \frac{\lVert \pi(x)-\pi(y) \rVert}{\lVert x-y \rVert} &= \frac{\lVert \mu(y)\,x-\mu(x)\,y \rVert}{\lVert x-y \rVert \,\mu(x)\mu(y)} \leqslant \frac{\lvert \mu(y)-\mu(x) \rvert \,\lVert x \rVert + \mu(x)\,\lVert x-y \rVert}{\lVert x-y \rVert \,\mu(x)\mu(y)} = \\ &= \frac{1}{\mu(x)\mu(y)} \frac{\lvert \mu(y)-\mu(x) \rvert}{\lVert x-y \rVert} + \frac{1}{\mu(y)} \leqslant \bigg( \frac{R}{r} \bigg)^2 \frac{1}{r} + R = \frac{R(R+r)}{r}, \end{align*} as for every $z \in X$ $$\frac{\lVert z \rVert}{R} \leqslant \mu(z) \leqslant \frac{\lVert z \rVert}{r},$$ in particular, $\mu$ is $1/r$-Lipschitz.

So, we're getting closer to the point. Is this an exact bound of $L(\pi)$? No, at least in the case of X being inner product space. The exact bound then is $\frac{R^2}{r}$, it's proved in the paper "A note on starshaped sets" by Siniša Vrećica.

In fact, when dealing with $\frac{\lVert \pi(x)-\pi(y) \rVert}{\lVert x-y \rVert}$ we work inside $Y=\mathrm{span} \{x,y\}$ as $\pi(x),\pi(y) \in Y$. That means we can consider only two-dimensional normed spaces $X$. The proof of euclidean (planar) case of the estimate pretty uses trigonometry. (Siniša's proof looked too technical to me, so I proved it myself using just sines' theorem and elementary facts, like monotonicity of sine.)

It's also sufficient to consider convex bodies of the form $M = \mathrm{conv} (B(0,r) \cup \{z\})$, as that sets are the "worst" for $L(\pi)$. Also, WLOG, we can take $r=1$.

For example, due to my humble computations, for $X = \ell_\infty^2$ and $$M_1 = \mathrm{conv} (B(0,1) \cup \{(0,R)\}), \quad M_2 = \mathrm{conv} (B(0,1) \cup \{(R,R)\}), \quad R\geqslant 2$$ we have $L(\pi_1) = R(R-1)$ and $L(\pi_2) = R(R+1)/2$.

Now, I have no idea how to deal with non-euclidean case in general. It seems to me that the Lipschitz constant must be less than for euclidean case. Anyway, how can one obtain some estimations better than $\frac{R(R+r)}{r}$ (say, norm given by radial function of unit ball)? Or there is example when we can't do better?

• You get actually estimate $R(R+r)/r$. (And it has to be homogeneous a priori.) – Fedor Petrov Feb 28 '16 at 9:05
• @FedorPetrov, oh, right, my mistake. – quartermind Feb 28 '16 at 10:30
• In general this is sharp estimate for all $r,R$. I have to draw a picture, it requires some time for me. – Fedor Petrov Feb 28 '16 at 17:50
• "Monotonicity of sine"? All this time I assumed it was periodic... – Nik Weaver Feb 28 '16 at 18:23
• On the Lipschitz constant of $\mu$, see jstor.org/stable/2316022 – Yoav Kallus Feb 28 '16 at 19:03

$r=1$, $R=OC/OB$. Unit ball is $DEBD'$ and symmetric part, $M$ is $DCD'$ and symmetric part, $x=B$, $y=G$, $\pi(x)=C$, $\pi(y)=H$, ray $OGH$ is very close to $OBC$.
We have $$\|C-H\|:\|B-G\|=\frac{CH}{OI}:\frac{BG}{OD}=\frac{CH\cdot BE}{BG\cdot CE}\cdot \frac{CE}{OI}\cdot \frac{OD}{BE}=\frac{OH}{OG}\cdot \frac{CE}{OI}\cdot\frac{OC}{CB}$$ (Menelaus theorem for triangle $EHG$ and line $CBO$ is used.) $OH/OG$ tends to $OC/OB$ when the ray $OGH$ tends to $OCB$. Next, $CE/OI=(CE/ED)\cdot (ED/OI)=(CB/BO)\cdot (1+OB/OC)$. Totally we get in the limit $(OC/OB)^2+(OC/OB)=R^2+R$.
We have used relation $ED/OI=1+OB/OC$, which follows, for example, from parallelogram $EBD'M$: $ED/OI=DM/OD'=(2DO-EB)/DO=2-EB/DO=2-CB/OC=1+OB/OC$.
• This works for me, thanks! Your example of norm is very specific, I mean, I was likely to consider situations when $R \gg r$. We may ask, what if we fix a norm first (and $r=1$), and then release $R$ to infinity? But that's may be not so interesting, because our initial estimate can be at least asymptotically exact in non-euclidean case. – quartermind Feb 28 '16 at 19:27
• It works for all $R>1$, I think. – Fedor Petrov Feb 28 '16 at 19:28