Analogy of Parseval identity for Legendre Transform ?  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 $p = x^*, q = y^*$
 A: I'm far from a specialist in this field, but I believe some related stuff can be found in the paper 
Artstein-Avidan, Shiri; Milman, Vitali
The concept of duality in convex analysis, and the characterization of the Legendre transform. 
Ann. of Math. (2) 169 (2009), no. 2, 661–674.
In particular, theorem 14 discusses how the Legendre transform is the 'unique' transform mapping convolution to sum for convex functions. The authors also have a similar paper concerning the classical Fourier transform.
A: 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).
A: Parseval's identity states that Fourier transform is an isometry of $L^2,$ so the right analogue seems to be the duality theorem of convex programming (which states that for convex functions, and various side conditions, Fenchel conjugation preserves  $sup$ norm -- the functions are on a domain and the conjugate domain, but that's not so dissimilar to the Fourier transform). Whether this is the "right" analogy or not is, of course, very much in the eye of the beholder.
A: Why not consider Fourier transform instead Fourier series ?
Parseval's Identity is a direct consequence of the fact that Fourier switches ordinary and convolution products, $\widehat{fg}=\widehat f*\widehat g$. Choosing $g=\bar f$, we have (up to universal constants that cancel at the end)
$$\int_{\mathbb R^n}|f(x)|^2dx=\widehat{|f|^2}(0)=\widehat f*\widehat{\bar f}(0)=\int_{\mathbb R^n}\widehat f(\xi)\widehat{\bar f}(-\xi)d\xi=\int_{\mathbb R^n}\widehat f(\xi)\overline{\widehat{f}(\xi)}d\xi=\int_{\mathbb R^n}|\widehat f(\xi)|^2d\xi.$$
The counterpart of the switching property at the level of the Legendre transform is 
$(f\oplus g)^\ast=f^\ast+g^\ast$
where $ \oplus $ denotes the inf-convolution:
$$f\oplus g(x):=\inf_{y\in\mathbb R^n}(f(y)+g(x-y))$$
Therefore an analogue of Parseval should come from
$$\inf_{y\in\mathbb R^n}(f(y)+g(-y))=f^\ast(0)+ g^\ast(0).$$
But now we should take a special $f$, analogous to $\bar f$ in the case of Fourier analysis. I don't see what is the counterpart of complex conjugation.
