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Suppose we have a bounded, strictly convex domain $D\subset \mathbb{R}^2$ with smooth boundary with strictly positive curvature. Suppose further that the projection of $D$ onto the horizontal coordinate axis is given by the interval $[0,2]$. Now, for any $x\in [0,2]$ we can consider the vertical diameter $d_D(x)$ which is the length of the intersection of $D$ with a vertical ray through $x$.

For instance for $D$ the unit disc with midpoint $(1,0)^\top$ we have $d_D(x)= 2\sqrt{2x-x^2}$ for $x\in [0,2]$. Let us make the definition $d_0(x):=2\sqrt{2x-x^2}$.

For a convex set $D$ as above consider the quotient $$\delta_D(x):= \frac{d_D(x)}{d_0(x)}.$$

It can be checked that $\delta_D$ is bounded from above and below for any convex $D$. My question is whether it can be shown that $ \delta_D $ is actually $C^\infty$ on $[0,2]$, that is with all derivatives bounded as we approach the endpoints?

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Strict convexity plus smooth boundary is not enough for $\delta_D$ to be bounded. You also need that the curvature be strictly positive at those points where the tangent is vertical. – Denis Serre Mar 9 '12 at 10:06
I have modified the question. – pil Mar 9 '12 at 10:28
up vote 8 down vote accepted

Yes. Incidentally, just recently I had to write down a proof of a similar fact in one of my papers. It is quite technical.

Let us work at the endpoint $x=0$. We have to prove that the function $x\mapsto d_D(x)/\sqrt x$ is $C^\infty$. We need the following well-known facts about $C^\infty$ functions $f$ defined in a neighborhood of 0:

  1. If $f(0)=0$, then $f(x)=xg(x)$ for some $g\in C^\infty$.

  2. If $f(0)=f'(0)=0$, then $f(x)=x^2g(x)$ for some $g\in C^\infty$.

  3. If $f(x)=f(-x)$ for all $x$, then $f(x)=g(x^2)$ for some $g\in C^\infty$.

(See this and this MO questions.)

We may assume that the boundary of $D$ contains the origin $(0,0)$. Then the boundary of $D$ near the origin is a graph $x=f(y)$ of a function $f\in C^\infty$ satisfying $f(0)=f'(0)=0$ and $f''(0)>0$. By the 2nd item above, we can write $f(y)=y^2g(y)$ where $g\in C^\infty$ and $g(0)=\frac12 f''(0)>0$. Let $h(y)=y \sqrt{g(y)}$, then $f(y)=h(y)^2$. Observe that $h(0)=0$ and $h'(0)>0$, so $h$ is invertible near 0. Denote $\varphi=h^{-1}$.

We can write $d_D(x)=d^+(x)-d^-(x)$ where $d^+(x)$ and $d^-(x)$ are the $y$-coordinates of the highest and lowest intersection point of $D$ and the vertical line through $(x,0)$. The values $d^\pm(x)$ are the solutions of the equation $f(y)=x$ (in the variable $y$), or, equivalently, $h(y)=\pm\sqrt x$, so $$ d^\pm(x) = \varphi(\pm\sqrt x) . $$ It remains to prove that the function $$ x \mapsto \frac{\varphi(\sqrt x)-\varphi(-\sqrt x)}{\sqrt x} $$ is $C^\infty$ on $[0,\varepsilon)$. Define $\psi(x)=\phi(x)-\phi(-x)$. The function $\psi$ is smooth and odd (i.e. $\psi(-x)=-\psi(x)$), therefore, by 1 and 3 above, it can be written in the form $\psi(x)=x \lambda(x^2)$ where $\lambda\in C^\infty$. Now we have $$ \lambda(x) = \frac{\psi(\sqrt x)}{\sqrt x} = \frac{\varphi(\sqrt x)-\varphi(-\sqrt x)}{\sqrt x} $$ for all $x\ge 0$, and $\lambda\in C^\infty$, q.e.d.

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beautiful! thanks a lot!! one minor typo: it should be $h(y) = y \sqrt{g(y)}$. – pil Mar 11 '12 at 18:21
Thanks, I corrected it. – Sergei Ivanov Mar 11 '12 at 21:20
"quite technical" but very pleasant – Pietro Majer Mar 13 '12 at 7:12

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