Vertical Diameter of Convex Domains - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:31:26Z http://mathoverflow.net/feeds/question/90678 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90678/vertical-diameter-of-convex-domains Vertical Diameter of Convex Domains pil 2012-03-09T10:01:04Z 2012-03-11T21:20:10Z <p>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 $x$. </p> <p>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 $d_0(x):=2\sqrt{2x-x^2}$. </p> <p>For a convex set $D$ as above consider the quotient $$\delta_D(x):= \frac{d_D(x)}{d_0(x)}.$$</p> <p>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?</p> http://mathoverflow.net/questions/90678/vertical-diameter-of-convex-domains/90748#90748 Answer by Sergei Ivanov for Vertical Diameter of Convex Domains Sergei Ivanov 2012-03-09T18:47:06Z 2012-03-11T21:20:10Z <p>Yes. Incidentally, just recently I had to write down a proof of a similar fact in one of my papers. It is quite technical.</p> <p>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:</p> <ol> <li><p>If $f(0)=0$, then $f(x)=xg(x)$ for some $g\in C^\infty$.</p></li> <li><p>If $f(0)=f'(0)=0$, then $f(x)=x^2g(x)$ for some $g\in C^\infty$.</p></li> <li><p>If $f(x)=f(-x)$ for all $x$, then $f(x)=g(x^2)$ for some $g\in C^\infty$.</p></li> </ol> <p>(See <a href="http://mathoverflow.net/questions/72497" rel="nofollow">this</a> and <a href="http://mathoverflow.net/questions/77120" rel="nofollow">this</a> MO questions.)</p> <p>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 $\varphi=h^{-1}$.</p> <p>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 <code>$$\lambda(x) = \frac{\psi(\sqrt x)}{\sqrt x} = \frac{\varphi(\sqrt x)-\varphi(-\sqrt x)}{\sqrt x}$$</code> for all $x\ge 0$, and $\lambda\in C^\infty$, q.e.d.</p>