# Can Fourier transform be seen as a decomposition over a basis in a space of tempered distributions

Fourier series of a function that belongs to $L^2([0,T])$ can be seen as a decomposition of this function over an (orthonormal) basis in the Hilbert space $L^2([0,T])$.

Fourier transform of a function with infinite support—even if the function belongs to $L^2(\mathbb R)$—involves objects outside of $L^2(\mathbb R)$. Complex exponents in the Fourier integration are already outside of $L^2(\mathbb R)$, because they are not square-integrable. Fourier image may as well be outside of $L^2(\mathbb R)$.

Still, as far as i understand, all the objects that can be encountered in direct and inverse Fourier transforms (functions or distributions in real space, Fourier images, and complex exponents) belong to a class of tempered distributions, and tempered distributions—as far as i understand—form a vector space.

So, my question is: can a Fourier transform be seen as a decomposition of a vector from a linear space of tempered distributions over an (orthonormal) basis in this space, which is usually a basis of complex exponents?

• Just a few remarks: 1) The Fourier image of an $L^2$ function lies always in $L^2$, by the Plancherel theorem, contrary to what you state. 2) Yes, the space of tempered distributions forms a vector space. But you cannot talk (at least not without introducing further structure) about orthonormal bases of this space, as no scalar product is given. Note that $<f,g>$ or $f(g)$ do not make sense in general if $f,g$ are tempered distributions, only if $f$ is a tempered distribution and $g$ is a Schwartz function. Commented Dec 30, 2014 at 9:23
• @PhoemueX, perhaps the questioner meant merely that the exponentials themselves are outside $L^2(\mathbb R)$. Commented Dec 30, 2014 at 17:28

Indeed, Fourier inversion can be viewed as a "spectral synthesis" result, where functions (perhaps "generalized") are expressible as superpositions (allowing possibly both sums and integrals...) of eigenfunctions for some natural self-adjoint operator, for example the Laplacian.

The necessity of allowing integrals, and of allowing eigenfunctions outside the initial Hilbert space (e.g., $L^2(\mathbb R)$) arises whenever the operator has more than discrete spectrum (=eigenvalues=point-spectrum). Gelfand et al call eigenfunctions in a larger space "generalized eigenfunctions" (a usage slightly in conflict with Jordan-decomposition uses).
Meanwhile, one should take some care, because Fourier inversion does not directly express a Schwartz function as a superposition of Schwartz functions... so such slogans would be dangerously misleading. Indeed, in all the "explicit" examples I know, the arguments required to demonstrate (and understand) the finer gradations of "spectral synthesis" involve much more than a basic Hilbert-space argument, or even a basic rigged-Hilbert-space (=scales-of-Hilbert-spaces) argument. For example, on the real line, looking also at the Schrodinger/Hamiltonian operator $-\Delta+x^2$ (a.k.a., "with confining potential", a.k.a., "quantum harmonic oscillator") does give a discrete decomposition (related to Hermite polynomials/functions), from which one can prove things about the Schwartz space... and eventually return to Fourier inversion as a corollary of discrete-spectrum set-up. I think N. Wiener took that route to some extent.
An attempt to make sense of "orthogonality" of distinct exponentials on the real line, for example, can be made more purposeful by asking about existence or non-existence of intertwinings between the representations generated by these functions (giving some heft to the barer fact that they have distinct eigenvalues for the Laplacian). That is, the classical integrals involved in spectral decompositions, when convergent, very often are viewable as giving intertwining operators... which often can be proven to be unique even without assuming convergence of the natural integral. Oppositely, re-interpretation of integrals as (unique up to scalars) intertwining operators can give a sense to not-convergent integrals in many useful cases. E.g., this can be thought of as a proof that $\int_{\mathbb R} e^{i\xi x}\;dx=0$ unless $\xi=0$.
• Thank you so much for the reply! Seems that i indeed was under an erroneous impression that Fourier inversion expresses a Schwartz function as a superposition of Schwartz functions. And the simplest counter-example seems to be a decomposition of one of the "basis" functions (a complex exponent) through all the other complex exponents, something like $e^{i \omega t} = \int {\delta(\nu - \omega) e^{i \nu t} } d\nu$, which immediately involves a delta-function, whose "values" do not belong to a complex field. Commented Jan 6, 2015 at 0:29