TITLE: Analogy of Parseval identity for Legendre Transform ? 
QUESTION [24 upvotes]: Parseval's identity states that the sum of squares of coefficients of the Fourier transform of a function equals the integral of the square of the function, or
$$ \sum_{-\infty}^{\infty} |c_n|^2 = (1/2\pi)\int^\pi_{-\pi} |f(x)|^2 dx $$
where the $c_i$ are the Fourier coefficients. 
The Legendre-Fenchel transform can be viewed as a generalization of the Fourier transform. For a given function $f : X \rightarrow R$ over a vector space $X$ which has dual $X^{*}$, the transform $f^* : X^* \rightarrow R$ is defined as:
$$ f^*(p) = \sup_{x \in X}\   \langle x, p\rangle  - f(x) $$
where further $p = f'(x)$ is denoted as $x^*$. 
So my question is:
Is there any natural generalization of Parseval's identity to relate $f^*$ and $f$ ? To be specific, I'm trying to relate quantities like $\|x-y\|$ to $\|p - q\|$ where $p = x^*, q = y^*$

REPLY [15 votes]: I think the identity you want is
$$2\inf_x f(x)=\inf_x(f^\ast(x)+f^\ast(-x))\mbox{.}$$
(I'm skipping a bunch of conditions required of $f$ to make this hold. We'll need convexity at least.)
Let's use $\oplus$ for infimal convolution and let $g(x)=f(-x)$.
By definition $(f\oplus g)(x)=\inf_y(f(x-y)+g(y))$.
Infimal convolution gives us $(f\oplus g)^\ast=f^\ast+g^\ast$.
So
$$\begin{align}
\inf_x(f^\ast(x)+f^\ast(-x)) &=& \inf_x (f^\ast(x)+g^\ast(x))\\
&=& \inf_x (f\oplus g)^\ast(x)\\
&=& \inf_x\inf_y(f(x-y)+f(-y))\\
&=& 2\inf_y f(-y)\\
&=& 2\inf_x f(x)\mbox{.}
\end{align}
$$
That second last equality is because if $f$ takes a minimum value at $-y$ then clearly $f(x-y)\ge f(-y)$ for any $x$ so the double infimum is attained when $f(x-y)$ and $f(-y)$ are "aligned" at $x=0$.
It looks less symmetrical than Parseval's theorem. But if $f$ is real, then Parseval's theorem gives
$$
\int f(x)^2dx=\int\hat f(\omega)\hat f(-\omega)d\omega
$$
Addendum: here's a detailed breakdown of the analogy:
I assumed you wanted Parseval's theorem on the Fourier transform, not the identity for Fourier series. This replaces the infinite sum with an integral.
According to the analogy we replace all integrals with infima. All multiplications with additions. All Fourier transforms with Legendre transforms. Squaring $x$ is multiplication of $x$ by itself, so that becomes adding $x$ to itself resulting in $2x$. Apply all of these replacements to 
$$
\int f(x)^2dx=\int\hat f(\omega)\hat f(-\omega)d\omega
$$
and we get
$$2\inf_x f(x)=\inf_x(f^\ast(x)+f^\ast(-x))\mbox{.}$$
Compare with Fenchel duality when $A=1$ (in the notation of that wikipedia page).