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

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For us philistines: in what way can the Legendre-Fenchel transform be viewed as a generalization of Fourier transform? –  Igor Rivin Feb 2 '11 at 23:38
I think it might be more accurate to say you're looking for an analog of the Plancherel Theorem, see en.wikipedia.org/wiki/Parseval's_identity –  Stopple Feb 2 '11 at 23:39
@Igor, I was hoping you wouldn't ask that :). There are two ways of thinking about this: one is by taking the Fourier transform of a function of the form $f(x) = A(x) exp( i k g(x))$ (see iopscience.iop.org/0305-4470/28/19/008/pdf/ja951908.pdf). Another way is via tropical algebras, which is slightly more involved. But the connection is quite real in terms of properties. Many properties of the L-F transform (see the above link) are akin to those of the Fourier transform. @Stopple you are right: I probably oversimplified in my explanation. –  Suresh Venkat Feb 2 '11 at 23:59
I think the difficulty mainly lies in identifying a place where the Legendre transform is defined so that it has a chance of being its own inverse. Morally, if $f:X \to (-\infty,\infty]$ is lower semicontinuous and convex then the so is its transform. So if $X$ is reflexive the transform interchanges two convex cones and indeed $f^{\ast\ast} = f$. You can't expect the Legendre transform to be continuous wrt the norm, but there is a way to make the transform isometric. Attouch and Wets have a few joint papers on these questions. Further names are Yosida and Moreau, but I'm no expert on that. –  Theo Buehler Feb 3 '11 at 0:44

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).

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Interesting. But I'm not sure I see how this even resembles Parseval's ? –  Suresh Venkat Dec 23 '13 at 20:28
Tell me if you see the resemblance yet. –  Dan Piponi Dec 23 '13 at 20:37
ah I think I get it. Between this and Denis Serre's answer, I think I'm finally seeing it :) –  Suresh Venkat Dec 23 '13 at 20:39
Denis didn't see what the analogue of complex conjugating the Fourier transform was. It's the reflection in $x=0$. –  Dan Piponi Dec 23 '13 at 20:40

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.

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That's an interesting take on this. It might not get me to where I want, but I see how it might be the only "appropriate" analogy here. –  Suresh Venkat Feb 3 '11 at 4:26
+1 and congratulations on 10k. –  quid Sep 24 '11 at 20:48
Thanks! I feel empowered :) –  Igor Rivin Sep 24 '11 at 20:55

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.

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It seems that the editor has troubles with the star used to denote Legendre transform. Does anyone knows how to fix it ? –  Denis Serre Feb 3 '11 at 8:33
thanks Gjergji. –  Denis Serre Feb 3 '11 at 9:57