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 $p = x^\*, q = y^\*$



Analogy of Parseval identity for Legendre Transform ? 
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](http://en.wikipedia.org/wiki/Fenchel%27s_duality_theorem) when $A=1$ (in the notation of that wikipedia page).


