2 corrected a typo

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 $h(y)=yg(y)$, 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. 1 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$h(y)=yg(y)$, 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.