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Legendre transform is defined as follows: Legendre Transform

And the problem I am trying to solve follows: Problem

My thoughts

I searched up to acquired that the matrix concerned is a hessian matrix.

Also, the previous subquestion has the following result: Previous Question

Since this is from implicit function theorem section, I think it will be helpful, but I haven't worked my way around yet.

Any help is sincerely appreciated

Liyang Meng
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    @mathphysicist9283, you should have provided more details, especially the fact that this question is motivated by problem 2 in section 8.5.5 of Zorich's Mathematical Analysis textboook. In particular, he defines the Legendre transform of a function without assuming that the function is convex or concave. It is quite unusual to do this. – Deane Sep 08 '24 at 04:55
  • @mathphysicist9283, I recommend skipping this problem for now. – Deane Sep 08 '24 at 04:57
  • @Deane Thank you for all the replies! And yes I would skip this problem for the moment, thank you. – Liyang Meng Sep 08 '24 at 05:47
  • @Deane I reverted to my first edit and details of Problem 2 a) are provided now. – Liyang Meng Sep 08 '24 at 06:30

1 Answers1

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This is a trivial application of the inverse function theorem; the only thing is you need to understand precisely what the definition is saying. Let $U\subset\Bbb{R}^m$ be open, and let $f:U\to\Bbb{R}$ be a $C^{\ell}$ function ($\ell\geq 2$) with invertible Hessian. Define the two maps

  • $\Phi_f:U\to\Bbb{R}^m$ by $\Phi_f(x):=\left(\frac{\partial f}{\partial x^1}(x),\dots, \frac{\partial f}{\partial x^m}(x)\right)$.
  • $E_f:U\to\Bbb{R}$ by $E_f(x):=\sum\limits_{i=1}^nx^i\frac{\partial f}{\partial x^i}(x) - f(x)=\langle\Phi_f(x),x\rangle-f(x)$.

The map $\Phi_f$ looks simply like the gradient vector of $f$, but I have purposely refrained from calling it that; more generally it is known as the fiber derivative of $f$ (see here, and more generally here). The mapping $E_f$, in the context of classical mechanics, would be called the ‘energy mapping of $f$’ (if you interpret $f$ as the Lagrangian of the system).

In fact, more properly, I should define $\Phi_f:U\to (\Bbb{R}^m)^*$, by saying $\Phi_f(x):=Df_x$; then the definition of $E_f$ reads $E_f(x):=Df_x(x)-f(x)$.

Anyway, the Legendre transform of $f$ is supposed to be a mapping $\lambda_f$ such that $E_f=\lambda_f\circ\Phi_f$ (in the language of classical mechanics, if we call $f$ the Lagrangian, then $\lambda_f$ would be called the Hamiltonian). Does it always exist? No. But if $\Phi_f$ has an inverse on some subset $U_0\subset U$, then we can locally invert to get $\lambda_f=E_f\circ(\Phi_f|_{U_0})^{-1}$. If this local inverse of $\Phi_f$ is smooth, then so is the Legendre-transform.

I hope you see where we’re going with this. We have a mapping $\Phi_f:U\subset\Bbb{R}^m\to\Bbb{R}^m$ which is of class $C^{\ell-1}$ (because $f$ is $C^{\ell}$), so our goal is to apply the inverse function theorem (which we can do because $\ell-1\geq 1$). Fix a point $a\in U$. Then, you can easily see that \begin{align} D(\Phi_f)_a=\text{Hess}_f(a). \end{align} Since the latter is assumed to be invertible, it follows the former is as well. Thus, the inverse function theorem tells us there is an open neighbourhood $U_0$ of $a$ in $U$, an open neighbourhood $V_0$ of $\Phi_f(a)$ in $\Bbb{R}^m$, such that the restriction $\Phi_f|_{U_0}:U_0\to V_0$ is a diffeomorphism of class $C^{\ell-1}$. Therefore, we can define $\lambda_{f,a}:V_0\to\Bbb{R}$ as \begin{align} \lambda_{f,a}:=E_f\circ(\Phi_f|_{U_0})^{-1}, \end{align} and this composition is of class $C^{\ell-1}$. This shows that for smooth enough $f$, having an invertible Hessian locally implies the existence of a Legendre-transform.

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