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convex. Indeed, differentiating Euler's foruma $\langle F_x,x\rangle=F$ gives $F_{xx}x=0$. By the convexity assumption $F_{xx}\,\Big|\,TS>0$. Therefore, we define the function

From Hofer-Zehnder, p. 25, preliminaries to proof of thm 5 in paragraph 1.5. $F$ is a function in $\mathbb{R}^{2n}$, defined in the following way. $S$ is a strictly convex hypersurface. We can assume the interior of the region $S$ is the boundary of contains the origin. Then given $x\neq0$ the ray from the origin to $x$ intersects $S$ in a single point $\xi$, which we define to be $\lambda^{-1}x$. Then we set $F(x)=\lambda$, so that $S=\{x:F(x)=1\}$. That's the context. What is he referring to as «Euler's formula»? What is $\langle F_x,x\rangle$? How do I prove the formula?

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MickG
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I am not sure if this is what is meant here but there is Euler's theorem on homogenous functions which says that given function homogenous of rank rr (I think there is restriction on r but don't remember exactly) that is $f(\lambda x) = \lambda ^r f(x)$, we have $r f(x)=x \cdot \nabla f (x)$. This looks like your formula with $r=1$ and $F_x$ denoting partial derivative w.r.t. x (or gradient if x is a vector) and bracket denoting scalar product. Does this interpretation make sense in this context?

Blazej
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  • Well in this case $F(\lambda x)=\lambda F(x)$, I think. Because If $\xi$ must be $F(x)^{-1}x$ for all $x$ on that ray, then it is also $F(\lambda x)^{-1}\lambda x$, hence $F(x)^{-1}x=F(\lambda x)^{-1}\lambda x$, meaning $F(x)^{-1}=F(\lambda x)^{-1}\lambda$, so $F(\lambda x)\lambda^{-1}=F(x)$, giving linearity of $F$. So $F$ has rank 1 and Euler's theorem should be precisely what is given right after «Euler's formula». How do I prove it? – MickG Sep 08 '15 at 14:32
  • Differentiating both sides of the homogeneity equation w.r.t. $\lambda$. OK. – MickG Sep 08 '15 at 14:37
  • The homogeneity only works for positive multipliers, as $F(-x)\neq-F(x)$. But that shouldn't alter the validity of the theorem, for if we differentiate we get $\nabla F(\lambda x)\cdot x=F(\lambda x)$ for any $\lambda$ for which the homogeneity equation holds, in particular it holds for $\lambda=1$, which is what we need. OK so all we need is that the equality hold for $\lambda$ in a neighborhood of 1, so we can differentiate. Right? – MickG Sep 08 '15 at 15:05
  • Yes, I think it is enough to have homogeneity for $\lambda$ in any open neighbourhood of $1$. I'm courious if there is any function that actually has this property only for $\lambda$ in some small interval and not in say, positive semiaxis. – Blazej Sep 08 '15 at 22:41