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?



Vertical Diameter of Convex Domains
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](https://mathoverflow.net/questions/72497) and [this](https://mathoverflow.net/questions/77120) 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.


