Does the Fourier inversion theorem hold if $f$ is square-integrable?

Question

I know that for $f$ an integrable function, it's Fourier transform may not be integrable, so the inversion theorem does not necessarily hold.

Assume $f$ is square-integrable instead. Does this imply that the inversion theorem holds, i.e. we can recover $f$ by integrating its Fourier transform? If not, what further assumptions are required?

I am getting different information from different notes online. Some say above is enough, others say that we need some continuity properties, and yet others say that the inversion theorem does not hold pointwise but only in $L^2$ sense?

The Placherel Theorem gives the truth of the matter here. – David C. Ullrich Mar 29 '20 at 10:02 — David C. Ullrich, Mar 29 '20 at 10:02

score 3 · Answer 1 · edited Mar 31 '20 at 04:49

3

Here's the story. For $f\in L^1(\Bbb R)$ let $\hat f$ be the usual Fourier transform:

$\newcommand\ft{\frac1{\sqrt{2\pi}}}$

$$\hat f(\xi)=\ft\int e^{-i\xi t}f(t)\,dt.$$

One can state the Inversion Theorem like so:

Inversion Theorem. If $f,\hat f\in L^1$ then $\hat{\hat f}(x)=f(-x)$.

Now for $f\in L^2$ and $A>0$ define $$F_A(f)=\ft\int_{-A}^Ae^{-i\xi t}f(t)\,dt.$$

The main theorem is

Plancherel Theorem There exists an isometric isomorphism $F:L^2\to L^2$ such that $\lim_{A\to\infty}\|Ff-F_Af\|_2=0$ for $f\in L^2$; $F$ is almost its own inverse: $FFf(x)=f(-x)$.

Note that if $f\in L^2\cap L^1$ then it's easy to see that $Ff=\hat f$; this is why $F$ is also called the Fourier transform.

So in particular $F_Af\to Ff$ in norm, not almost everywhere. No. In fact

Theorem (Carleson) If $f\in L^2$ then $F_Af\to Ff$ almost everywhere.

Proof: Very very hard. A huge big deal in Fourier analysis.

Corollary. If $f\in L^2$ then $Ff\in L^2$ and $F_AFf(x)\to f(-x)$ almost everywhere.

It's not clear to me exactly what it means to say the inversion theorem holds almost everywhere, but it seems pretty likely that the corollary answers your question.

edited Mar 31 '20 at 04:49

Community

1

answered Mar 30 '20 at 16:38

David C. Ullrich

92,839

Using Carleson's theorem is kind of overkill, don't you think? – Hugo Mar 30 '20 at 18:46
@Hugocito Huh? There's some question whether "pointwise" means the same to me as to the OP. With my version of "pointwise", Carleson's theorem is the answer to the question! – David C. Ullrich Mar 30 '20 at 18:52
@Hugocito How do you show $F_AFf(x)\to f(-x)$ almost everywhere without using Carleson's theorem? Btw if you can give an easy proof of this you're going to be famous, because Carleson's theorem is a simple consequence. – David C. Ullrich Mar 30 '20 at 18:57
There are two ways of proving it. In both cases you define separately the Fourier transform $\mathcal F$ and the inverse Fourier transform $\mathcal F^{-1}$.
1. Using distribuitions: $\mathcal F \mathcal F^{-1} u = u$ holds for tempered distributions, because it holds for the Schwartz spaces. Now, since both sides of the identity are in $L^2 \subset L_{loc}^1$, the identity also holds in $L_{loc}^1$ as consequence of Lebesgue differentiation theorem. So, the identity of functions $\mathcal F \mathcal F^{-1} u = u$ holds almolst everywhere.
– Hugo Mar 30 '20 at 19:09
You can extend the Fourier transform from from Schwartz space $\mathcal S$ to L^2 by density. You have the maps $\mathcal F$ and $\mathcal F^{-1}$ defined on $\mathcal S$ and you extend both maps to isometries in $L^2$. Now, the identity $\mathcal F \mathcal F^{-1} u = u$ holds for functions in $\mathcal S$, and therefore, by density, it holds in $L^2$. In particular, we have $\mathcal F \mathcal F^{-1} u = u$ almolst everywhere.

Hugo

Mar 30 '20 at 19:13

@Hugocito I didn't ask how to show $\mathcal F\mathcal F^{-1}f=f$ almost everywhere. How do you show $\lim_{A\to\infty}F_AFf(x)=f(-x)$ without Carleson's theorem? – David C. Ullrich Mar 30 '20 at 19:16

Sorry, I don't see why Carleson's theorem is related to the post. What is your point? – Hugo Mar 30 '20 at 19:21

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@Hugocito I may have misinterpreted the question "does the inversion theorem hold almost everywhere" - after all, that's not very precisely formulated, a theorem is true or false, not true almost everywhere. Trying to make sense of the question, i took it to be asking whether the Corollary was true. Carleson's theorem is not "overkill" for the Corollary, Carleson't theorem is trivially equivalent to the corollary. – David C. Ullrich Mar 30 '20 at 19:28

I see. So I think we were just addressing to different interpretations of the question. Thank you for your answer, I didn't know Carleson's theorem also holds for $L^2(\mathbb R^n)$. I just knew it for Fourier series. I understood he was asking if $\mathcal F \mathcal F^{-1} u = u$ holds almost everywhere for $u\in L^2(\mathbb R^n)$. – Hugo Mar 30 '20 at 19:37

score 1 · Answer 2 · answered Mar 29 '20 at 10:30

The Fourier transform is always defined, but you have to work with tempered distributions. In what follows I will use the following definition of Fourier transform $$\mathcal F(\phi)(\xi) := \int_{\mathbb R^n}\phi(x)\exp(-2\pi i x \xi)dx$$ for $\phi \in L^1(\mathbb R^n)$.

The Fourier transform is also defined for the space $\mathcal S'(\mathbb R^n)$ of tempered distributions, and it is indeed a isomorphism. The Fourier transform in this case is given by $$\langle \mathcal F(u),\phi\rangle:= \langle u, \mathcal F(\phi) \rangle,$$ for $\phi \in \mathcal S(\mathbb R^n)$, the Schwartz space of $\mathbb R^n$.

The space $L^2(\mathbb R^n)$ is contained in $\mathcal S'(\mathbb R^n)$, and restricting the Fourier Transform to it we obtain the isomorphism: $$\mathcal F: L^2(\mathbb R^n) \to L^2(\mathbb R^n). $$

Furthermore, this isomorphism is an isometry, that is, $$\Vert \mathcal F(\phi) \Vert = \Vert \mathcal \phi \Vert.$$ This identity is called Placherel theorem.

For a more detailed discussion about this I recommend you to take a look at Folland's book Real Analysis, and at the Fourier Analysis: An Introduction, by Stein and Shakarchi .

There's no need for tempered distributions to give the story for $L^2$. — David C. Ullrich, Mar 30 '20 at 16:25
Yes, I know you don't need tempered distributions to prove the extension to $L^2$. In Folland's book he does the extension using elementary functional analysis and density arguments. Nevertheless, it's much more natural to use distributions here. — Hugo, Mar 30 '20 at 18:43

score 0 · Answer 3 · edited Mar 31 '20 at 04:42

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Your last sentence is correct: the inversion theorem does not hold pointwise but only almost everywhere, which means that you recover the same function in $L^2$ sense (remember that two functions are considered equal in $L^2$ sense if they only differ an a null set). If you want it to hold everywhere you need some continuity properties. Thus all sources are probably correct; they just differ in what they mean with equality.

edited Mar 31 '20 at 04:42

Michael Hardy

1

answered Mar 29 '20 at 12:06

md2perpe

30,042

No, it does hold almost everywhere! That's a hugely deep theorem of Carleson. – David C. Ullrich Mar 30 '20 at 16:24
@DavidC.Ullrich. That's what I wrote. – md2perpe Mar 30 '20 at 16:53
If by pointwise we mean at every point, as opposed to almost everywhere, then yes. But "pointwise" often means almost everywhere, as opposed to just in norm. It's clear from the last sentence of the question that that's what "pointwise" meant in the question, and in that sense the inversion theorem does hold pointwise. – David C. Ullrich Mar 30 '20 at 17:10
@DavidC.Ullrich. I don't think that I've ever seen "pointwise" mean "almost everywhere". And doesn't convergence in $L^2$ norm imply convergence almost everywhere? I can easily find examples that the opposite implication isn't true, but not for this direction. – md2perpe Mar 30 '20 at 17:58
What? Cnvergence in $L^2$ norm certainly does not imply almost everywhere convergence. Describing the example is a little complicated - you can find it in any text on measure theory. – David C. Ullrich Mar 30 '20 at 18:36
Regarding ' I don't think that I've ever seen "pointwise" mean "almost everywhere"': That's perfectly consistent with what I said - I didn't assert that you had ever seen this usage... – David C. Ullrich Mar 30 '20 at 18:38
@DavidC.Ullrich. I found an example: https://math.stackexchange.com/a/138054/168433 – md2perpe Mar 30 '20 at 22:07

Does the Fourier inversion theorem hold if $f$ is square-integrable?

3 Answers3