# Is Fourier transform characterized by its diagonalization properties?

Let us fix the following convention for the Fourier transform in $L^1(\mathbb{R})$ space: $$\hat{f}(\xi)=\int_{-\infty}^\infty f(x)\, e^{-2\pi i x\xi}\, dx.$$ We then have the following properties: \begin{align}\tag{1} \displaystyle \left[\frac{df}{dx}\right]^\hat{}(\xi)=2\pi i\xi\, \hat{f}(\xi); \\ \tag{2} \displaystyle \left[ -2\pi i x\, f\right]^\hat{}(\xi)=\frac{d\hat{f}}{d\xi}(\xi);\\ \tag{3} \displaystyle \hat{f}(0)=\int_{-\infty}^\infty f(x)\, dx. \end{align}

Question Let $K(x, \xi)$ be a bounded function. Suppose that the integral transform $$Tf(\xi)=\int_{-\infty}^{\infty}f(x)K(x, \xi)\, dx,\quad f \in L^1(\mathbb{R})$$ satisfies properties (1), (2) and (3). Is it true that $K(x, \xi)=\exp(-2\pi i x\xi)$?

Motivation for this question comes from the fact that one can evaluate $$\hat{G}(\xi)=\left[ \exp(-\pi x^2)\right]^\hat{}$$ by using only the properties (1), (2) and (3). This is done by Fourier transforming both sides of the differential identity $$\frac{d}{dx}e^{-\pi x^2}=-2\pi x\, e^{-\pi x^2},$$ obtaining the Cauchy problem $$\begin{cases} -2\pi \xi\, \hat{G}(\xi)=\frac{d}{d\xi}\hat{G}(\xi) \\ \hat{G}(0)=1 \end{cases}$$ whose unique solution is $$\hat{G}(\xi)=\exp(-\pi\,\xi^2).$$

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Very nice question. I'd guess the answer is negative, since I'd expect to read such a theorem on books in harmonic analysis. But I have no counter-example... –  Siminore Oct 24 '12 at 14:12

Apart from some quibbling about making sure $K$ is sufficiently integrable or something... this is true. E.g., for precision, take $K$ to be a tempered distribution in two variables. Using the hypothesis that $2\pi ix\cdot T=T\cdot {d\over dx}$ as operators on Schwartz functions, (thinking of "$x$" as multiplication-by-$x$), integration by parts in the integral for $T$ gives ${\partial\over \partial y}K(x,y)=2\pi i x\cdot K(x,y)$ as tempered distribution in two variables. This has obvious classical solutions $C\cdot e^{2\pi ixy}$, as expected. To show that there are no others, among tempered distributions, one way is to divide $K(x,y)$ by $e^{2\pi ixy}$, so the equation becomes ${\partial \over \partial y}K(x,y)=0$. By symmetry, ${\partial\over \partial x}K(x,y)=0$. Integrating, $K(x,y)$ is a translation-invariant tempered distribution in two variables. It is a separate exercise to see that all such are constants.
Edit: in response to @GiuseppeNegro's comment (apart from correcting the sign), the secondary exercise of proving that vanishing first partials implies that a tempered distribution is (integrate-against-) a constant has different solutions depending on one's context, I think. Even in just a single variable, while we can instantly invoke the mean value theorem to prove that a function with pointwise values is constant when it is differentiable and has vanishing derivative, that literal argument does not immediately apply to distributions. In a single variable, integration by parts proves that $u'=0$ for distribution $u$ implies that $u(f')=0\,$ for all test functions $f$, and we can characterize such $f$, namely, that their integrals over the whole line are $0$, from which a small further argument proves that $u$ is a constant. This sort of argument seems to become a little uglier in more than one variable... and, in any case, I tend to favor a slightly different argument that is a special case of proving uniqueness of various group-invariant functionals. E.g., on a real Lie group $G$, there is a unique right $G$-invariant distribution (=functional on test functions), and it is integration-against right Haar measure. The argument is essentially just interchange of the functional and integration against an approximate identity, justified in the context of Gelfand-Pettis (weak) integrals. Probably there are more elementary arguments, but this sort of approach seems clearer and more persuasive in the long run.
Very nice! The fact that all translation-invariant tempered distributions are constants must surely be a consequence of the fact that their derivative is zero, right? (PS: There is a small typo: it is $2\pi i xT=T\frac{d}{dy}$, not $-2\pi i x T=T\frac{d}{dx}$). Thank you very much! –  Giuseppe Negro Oct 24 '12 at 18:07