Question

Suppose $K$ is a centrally symmetric, strictly convex body in $\mathbb{R}^2$. Let denote the curvature and the support function of $\partial K$, boundary of $K$, respectively with $\kappa$ and $s$. If $m\le\frac{\kappa}{s^3}(K)\le M$ for some positive numbers $m$ and $M$, does it mean there are ellipsoids $E_1$ and $E_2$ such that $E_1\subseteq K\subseteq E_2$ and $$\frac{\kappa}{s^3}({E_1})=M,~~~ \frac{\kappa}{s^3}(E_2)=m ~~~~? $$

Answer 1

Yes this is true. Let me handle the inner ellipse, the outer one is similar.

For brevity, denote $\kappa/s^3$ by $a$. It is easy to see that $$ a = \frac{\dot\gamma\wedge\ddot\gamma}{(\gamma\wedge\dot\gamma)^3} $$ where $t\mapsto \gamma(t)$ is any counter-clockwise parametrization of the boundary curve. For an ellipse, this is a constant inverse proportional to the square of the area. So we have to prove that the maximum-area ellipse contained in $K$ has $a\le M$.

So let $E$ be the maximum-area ellipse contained in $K$. Since the problem is centro-affine invariant, we may assume that $E$ is the unit circle. Then we have to prove that $M\ge 1$.

Consider the points where $\partial K$ touches $\partial E$. One easily sees that the intervals between these points on the circle are no greater than $\pi/2$, otherwise $E$ could be made larger. Choose coordinates so that one of the touch points is $(1,0)$, then there is another touch point of the form $(\cos\theta,\sin\theta)$ where $0<\theta\le\pi/2$. Observe that the arc of $\partial K$ between these touch points is contained in the square $[0,1]^2$.

Parametrize $\partial K$ by spanned area, i.e. by a curve $t\mapsto\gamma(t)$ such that $\gamma\wedge\dot\gamma=1$. Then $\gamma\wedge\ddot\gamma=0$, hence $\ddot\gamma(t)=-a(t)\gamma(t)$ for all $t$ where $a(t)$ is the centro-affine curvature. Let $\gamma(t)=(x(t),y(t))$, then $\ddot x(t)=-a(t)x(t)$ and $\ddot y(t)=-a(t)y(t)$. Suppose that $M=\sup a(t)<1$. Since $x(0)=1$, $\dot x(0)=0$, $y(0)=0$ and $\dot y(0)=1$, a standard comparison theorem for equations of the form $\ddot x=-ax$ implies that $x(t)>\cos t$ and $y(t)>\sin t$ for all $t\in(0,\pi/2]$. Therefore $x(t)^2+y(t)^2>1$ for all $t\in(0,\pi/2]$ and $y(\pi/2)>1$. This means that $\gamma$ leaves the square $[0,1]^2$ before is has a chance to touch the circle again, a contradiction.