# How do I prove this Hessian is positive definite?

The setting is the same as in this question. Precisely, we have a strictly convex hypersurface $S\subseteq(\mathbb{R}^{2n},\omega_0)$, where $\omega_0$ is the standard symplectic form, and we define $F$ as follows. We assumed $0\in S$, so any ray issuing from the origin intersects $S$ in precisely one point $\xi$. Thus, for $x\neq0$, there is a unique $\lambda$ such that $\xi=\lambda^{-1}x\in S$. We define $F(x)=\lambda$ for $x\neq0$ and $F(0)=0$. We thus obtain a $\mathcal{C}^2$ function, which makes $H_F$ (the Hessian matrix) symmetric. As seen in the linked question, by Euler's theorem for homogeneous functions we deduce a formula which in turn gives $H_F(x)x=0$ for all $x$. So for no $x$ is this hessian nonsingular. However, according to Hofer-Zehnder (p. 25, preliminaries to the existence theorem for periodic orbits on convex regular compact energy surfaces), we can, by convexity of $S$, conclude that the restriction to the tangent space of $S$ of the bilinear form induced by the Hessian, i.e. $H_F|TS$, is positive. I assume this means $H_F|TS$ is symmetric and positive definite. We have seen it is symmetric by regularity and the Schwarz lemma. But how do I deduce it is positive definite? And above all, how does convexity enter the proof? I really don't know where to start here…

• Is the symplectic form relevant here? – Amitai Yuval Sep 9 '15 at 10:57
• @AmitaiYuval I mentioned it for the sake of completeness of context. I guess it might be irrelevant. – MickG Sep 9 '15 at 11:00

Sketch: We want to prove that for any $p\in S$ and $v\in T_pS$, the second derivative of $F$ at $p$ in the direction $v$ is positive. Note that for any given such $p,v,$ everything that matters to us happens in the $2$-dimensional space $\mathrm{span}(p,v).$ Hence, it is enough to solve the problem for $n=1$.

So, we have a convex curve $S$ in $\mathbb{R}^2$, and we have some $p\in S$. Let $$l=\{p+t\cdot v|t\in\mathbb{R}\}$$ denote the line tangent to $S$ at $p$. (In other words, $0\neq v\in T_pS$). By definition of $F$, we have $F(p)=1$. By convexity, we have $F(q)>1$ for every $p\neq q\in l$. This means that the function $F|_l$ admits a minimum at $p$, and hence, the second derivative in the direction $v$ is positive. (Well, non-negative, anyway. All you have to do now is use strict convexity in order to show that the second derivative does not vanish at $p$).

Edit: We show that strong convexity of a hypersurface is independent of the defining function. Let $S\subset\mathbb{R}^n$ be defined locally both by $\{f=0\}$ and $\{g=0\}$, where $f,g:U\to\mathbb{R}$ are smooth and both derivatives $df, dg$, don't vanish on $S$. Let $p\in S\cap U$. It follows that there is a smooth $h:U\to\mathbb{R}$, such that $g=hf$. Since $dg_p\neq0,$ it follows from the Leibniz rule that necessarily $h(p)\neq0$.

We assume now that $S$ is strongly convex with respect to $f$, that is, $\nabla^2f_p$ is positive definite on the tangent space $T_pS$. Let $v\in T_pS$. Then$$dg(v)=dh(v)f+hdf(v),$$and at $p$ we have$$\nabla^2g_p(v)=\nabla^2h_p(v)f(p)+2dh_p(v)df_p(v)+h(p)\nabla^2f_p(v).$$By the assumptions, the first and second terms vanish. The last term has the sign of $h(p)$. Thus, if $g$ is such that $h$ is positive, then $\nabla^2g_p$ is positive definite on $T_pS$, and if $h$ is negative, $\nabla^2g_p$ is negative definite on the tangent space.

Edit 2: The above argument works whenever $f$ and $g$ are at least $C^3$. We now treat the case where they are only $C^2$ (then $h$ is $C^2$ far away from $S$, but only $C^1$ on $S$). Since $g$ is $C^2$, we have$$\nabla^2g_p(v)=\lim_{x\to p}\nabla^2g_x(v)=\lim_{x\to p}\nabla^2h_x(v)f(x)+2dh_x(v)df_x(v)+h(x)\nabla^2f_x(v),$$ where we let $x$ approach $p$ from outside of $S$, and so the limit is well defined. It suffices then to prove$$\lim_{x\to p}\nabla^2h_x(v)f(x)=0.$$For that, we change the coordinates and assume $S=\{x_n=0\}.$ Note that when we do so, the expression for the Hessian tensor includes a correction involving the Christoffel symbols. However, this correction is bounded, as it involves only the first derivatives of $h$, and hence vanishes when multiplied by $f$, as $x$ gets closer to $p$. Consequently, we may consider the standard Hessian in our new coordinates.

Outside of $S$, we have $h=g/f$. Hence, for $1\leq i\leq n-1$,$$h_i=\frac{g_if-gf_i}{f^2}.$$Differentiating again, \begin{align} h_{ii}&=\frac{(g_{ii}f+g_if_i-g_if_i-gf_{ii})f^2-(g_if-gf_i)2ff_i}{f^4}\\ &=\frac{g_{ii}}{f}-\frac{gf_{ii}}{f^2}-\frac{2g_if_i}{f^2}+\frac{2gf_i^2}{f^3}, \end{align} and hence, \begin{align} h_{ii}f&=g_{ii}-\frac{g}{f}f_{ii}-2f_i\frac{g_if-gf_i}{f^2}\\ &=g_{ii}-hf_{ii}-2f_ih_i. \end{align} Since on $S$ both $f$ and $g$ are constant, all the terms vanish on $S$. It follows that$$\lim_{x\to p}\nabla^2h_x(v)f(x)=0,$$and the proof is complete.

• By convexity if $q\in l$ and $q\neq p$ $q$ is out of the curve, outside the region the curve bounds, and hence $F(q)>1$, OK. – MickG Sep 9 '15 at 12:05
• So we have a minimum, thus non-negative derivative. But I can't really see how the fact that any segment with endpoints on $S$ must be contained in the interior of the region $S$ bounds except for the endpoints, i.e. the strict convexity of $S$, should help in proving the second derivative doesn't vanish at $p$. I mean, I can tell that the level sets of this function, being it homogeneous of rank 1 for positive multipliers, are simply dilations of $S$, so all strictly convex. But even this doesn't seem useful for what I want to prove. Strict convexity implies the tangent line to $p$ has to… – MickG Sep 9 '15 at 12:10
• …intersect $S$ only at $p$, and never the interior. Which gives us $F(q)>1$ for all $q\neq p$ in $l$, which we already know. What more can I extract out of strict convexity that convexity doesn't give me? – MickG Sep 9 '15 at 12:12
• @MickG This is not needed. We say the vector $v$ is mapped under the coordinate change to $\partial/\partial x_i$. Then the Hessian of $v$ is equal to the corresponding Hessian of $\partial/\partial x_i$. No need to consider any mixed derivatives. – Amitai Yuval Oct 18 '15 at 16:41
• @MickG Yes, right. – Amitai Yuval Oct 18 '15 at 16:45