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If $f$ is an $L^2$ function and $\hat{f}$, its Fourier transform, also in $L^2$, can the Fourier transform and its inverse be written as

$$\hat{f}(\omega)=\int_{-\infty}^\infty f(x) e^{i\omega x}dx$$


$$f(x)=\int_{-\infty}^\infty \hat{f}(\omega) e^{-i\omega x}d\omega$$

respectively, disregarding the constants? When can a Fourier transform and its inverse exist, but the above definitions not hold?

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Hmm, $f \in L^1 \cap L^2$. Otherwise the integrand does not have the be integrable. – Jonas Teuwen Mar 9 '11 at 22:30
@Jonas T: Plancherel helps you to extend the notion of the Fourier transform to all of $L^2$ (like it is conventionally done). See – Fabian Mar 9 '11 at 22:50
@Fabian: Yes, by density. And then by interpolation for $1 < p < 2$. But if you have an explicit integral, doesn't it need to have a meaning then? The question is if it can be written like this. – Jonas Teuwen Mar 9 '11 at 22:52
@Jonas T: I agree if the notation means "integral", it does not have a meaning. If it means "fourier transform" then it does. – Fabian Mar 9 '11 at 23:10

2 Answers 2

up vote 2 down vote accepted

Jonas T already hit on the important point that for the first integral to be defined, $f$ must be in $L^1$. Similarly, for the second integral to be defined, $\hat{f}$ must be in $L^1$.

Fabian has mentioned that the Fourier tranform can be generalized even beyond $L^2$. But if you're only concerned with $L^2$, then one way to define the Fourier transform is to first show that it defines an isometric (with respect to the $L^2$ norm) isomorphism on the Schwartz space, which is dense in $L^2$, and then take the unique extension to all of $L^2$. The fact that the Fourier transform defines an isometry with dense range on a dense subspace of $L^2$, and thus has a unique extension to a unitary operator on $L^2$, is known as Plancherel's theorem. I have only mentioned one approach, which is the one I learned from Chapter 10 of J.B. Conway's A course in functional analysis.

So the answer to your final question, at least restricted to $L^2$, is that every element of $L^2\setminus L^1$ provides an example, strictly speaking. However, you can show that $\frac{1}{\sqrt{2\pi}}\int_{-R}^R f(x)e^{-i\omega x}dx$ converges to $\hat{f}(\omega)$ in $L^2$ norm as $R\to\infty$, and similarly $\frac{1}{\sqrt{2\pi}}\int_{-R}^R \hat{f}(\omega)e^{i\omega x}dx$ converges to $f(x)$ in $L^2$ norm as $R\to\infty$. I don't have a reference to a proof handy, but this and more is summarized in the Springer online encyclopedia's article on the Fourier transform, which includes helpful references and links.

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A good book for non-measure theoretic Fourier analysis is "Fourier Analysis" by Stein & Shakarchi. – Jonas Teuwen Mar 9 '11 at 23:30
I think I understand most of what you say. What exactly does "dense in $L^2$" mean? – user7815 Mar 10 '11 at 1:23
Dense in the sense of One way to think of it in this particular situation is this: A linear subspace $V\subseteq L^2$ is dense if every element of $L^2$ is a limit in the $L^2$ norm of a sequence in $V$. In other words, every element of $L^2$ can be approximated arbitrarily closely by elements of $V$. In my answer, I refer to the fact that the Schwartz space is dense in $L^2$, and in fact the space of smooth functions with compact support is dense in $L^2$, as came up in another question: – Jonas Meyer Mar 10 '11 at 1:29
I also refer to the fact that a continuous linear transformation on a dense subspace has a unique continuous extension to the whole space. More generally, a uniformly continuous function defined on a dense subspace of a metric space has a unique continuous extension to the whole space. – Jonas Meyer Mar 10 '11 at 1:35

There are many cases when the Fourier transform can be defined and the function is not $L^2$. Take a look at tempered distributions for a class of "functions" for which the Fourier transform and its inverse can be defined. The most prominent example is the delta function (distribution).

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