Suppose we have a bounded, strictly convex domain $D\subset \mathbb{R}^2$ with smooth boundary. 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?

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?