TITLE: Integral kernel of the Legendre transform
QUESTION [1 upvotes]: First of all, I'm not sure, but I think the Legendre transform can be seen as a linear operator between the functions on a normed space and the functions on its dual. (A functional analysis approach reference would be welcome)
If that's correct, and the following statement of wikipedia article about integral transforms is also correct:
"and in fact if the kernel is allowed to be a generalized function then all linear operators are integral transforms (a properly formulated version of this statement is the Schwartz kernel theorem)."

Then, it is possible to associate an integral transform to the Legendre transform as an operator. My question then would be: What's the kernel of that integral transform which represents the Legendre transform?
This is all related to this question, where there seems to be a relation between the Laplace and the Legendre transform in the macroscopic limit. And I was hoping that the kernel I'm looking for is the same as in the Laplace transform in the limit (logarithms apart). That would be awesome.
I know that I'm not being rigorous about the spaces in which I'm working, but I don't have the knowledge to make this question more rigorous.

REPLY [1 votes]: This is a rather long comment:
The Legendre transform is not an integral operator acting on $L_p$ spaces, but a transformation that acts on proper convex functions. To be more concrete, suppose $X$ is a topological linear space (think of $\mathbb{R}^n$ for example), and let $X^*$ be its dual space (continuous linear functionals on $X$)
Definition: For any function $f:X\rightarrow\overline{\mathbb{R}}$, the function $f^*:X^*\rightarrow\overline{\mathbb{R}}$ given by
\begin{align}
f^*(v)=\sup_{x\in X}\big\{v(x)-f(x)\big\},\qquad v\in X^*\tag{1}\label{frenchel}
\end{align}
is the Frenchel--Legendre transform of $f$.
We have the following result:

For any function $f:X\rightarrow\overline{\mathbb{R}}$,

$f^*$ is convex on $X^*$.
(Frenchel--Young inequality) If $f^{-1}(\{-\infty\})=\emptyset$ and $\{f<\infty\}\neq\emptyset$,
\begin{align}
    f(x)+f^*(v)\geq v(x),\qquad (x,v)\in X\times X^*.
  \end{align}
Under the  $\sigma(X^*,X)$--topology on $X^*$, $f^*$ is lower semicontinuous and
$f^{**}(x):=(f^*)^*(x)\leq f(x)$.
(Frenchel--Legendre duality) Suppose that $X$ is locally convex. If $f$ is proper lower semicontinuous and convex, then $f^*$ is proper convex lower semicontinuous and
\begin{align}
f(x)=f^{**}(x)=\sup\{g(x): \text{g affine},\quad g(y)<f(y)\}.
\end{align}


To give a simple example that shows a connection to the Laplace transform, suppose $\mu$ is a $\sigma$-finite measure on $(\mathbb{R}^n,\mathscr{B}(\mathbb{R}^n))$. Then The set
\begin{align}
\mathbf{\Delta}:=\Big\{\theta\in\mathbb{R}^n: \int e^{\theta\cdot x}\mu(dx)<\infty\Big\}
\end{align}
is a convex set and that the map
$\Lambda:\mathbb{R}^n\mapsto\overline{\mathbb{R}}$ given by
$\Lambda(\theta)=\log\Big(\int e^{\theta\cdot x}\mu(x)\Big)$ is proper, lower semicontinuous and convex with $\operatorname{dom}(\Lambda)=\mathbf{\Delta}$
Define
\begin{align}
\hat{\ell}(x)=\sup_{\theta\in\mathbb{R}^n}\{\theta\cdot x-\Lambda(\theta)\}
\end{align}
Notice that $\hat{\ell}$ is the Frenchel-Legendre transform of $\Lambda$ (here, $\Lambda(\theta):=\infty$ for all $\theta\notin\Delta$)
The relation between $\hat{\ell}$ and $\Lambda$ is given by
\begin{align}
\hat{\ell}(x)&=\sup_{\theta\in\mathbf{\Delta}}\Big(\theta\cdot x-\Lambda(\theta)\Big)\\
\Lambda(\theta)&=\sup_{x\in\mathbb{R}^n}\Big(\theta\cdot x-\hat{\ell}(x)\Big)
\end{align}
This follows from Frenchel-Legendre's duality.

Comment:

In statistics, if $f_\theta(x)=e^{-\Lambda(\theta)}e^{x\cdot\theta}\mu(dx)$ defines an exponential type of distributions, $\ell(x,\theta)=\theta\cdot x-\Lambda(\theta)$ is the maximum likelihood and $\ell(x)$ the super-maximal likelihood. $\Lambda(\theta)$ is the cumulant generating function.

There are other applications related to the study of Large deviations.
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