Having seen the support theorem for the Radon transform, and seeing how much we rely on the Fourier transform as a tool, it is natural to ask an analogous question for the Fourier transform.

## The Easy Pieces: and

For some important spaces, the Fourier transform is very well behaved. In particular

and it is an isomorphism in each case. But what about ? ?

## A Harder Piece:

Say , and specificallyr, say . Then we can write

for all in . It can be extended to :

This makes sense because

So the integral defining converges absolutely for all .

**Claim**

**Proof** We can take any derivative we like and bring it under the integral sign, thanks to the bound we just proved and the compact support of .

In particular we can say

**Claim** is analytic in , i.e. .

**Proof**

This immediately tells us that the situation for compactly supported functions is very different from the situation for Schwartz functions — the Fourier transform of a compactly supported function is analytic, so *it cannot be compactly supported *or it would vanish identically.

This analytic function, is called the *Fourier-Laplace transform* of .

Now we can say quite a bit about how must look

**Claim** If then is an entire analytic function and for all integers there is a constant such that

**Proof** We have already seen that is entire analytic, and because of this it will clearly satisfy these bounds in a given compact set like the unit disk. This means that we only need to prove the inequalities for . For any multi-index we can write

Where the last step follows from integration by parts. But since is Schwartz, this immediately implies that

for all . Taking proves the result.

The main result of this post, the Paley-Wiener theorem, states that these necessary conditions for a function to be in the range of the Fourier transform are in fact sufficient.

**Theorem [Paley-Wiener for smooth functions]**

- If and then extends analytically to and for all non-negative integers there exists a constant such that
- Let be analytic on such that for all non-negative integers there exists a constant such that

for all . Then there exists an with such that For all

**Proof [Paley-Wiener for smooth functions]** We have already shown that (1) is true, now we must prove (2). Given the conditions in (2) we can define a function on by

If we can show that is smooth and supported in , we will be done.

*Step 1: is smooth.*

For an arbitrary multi-index

Which exists thanks to the rapid decay of . So

*Step 2: has compact support.*

We will prove this using contour integration. We can integrate in a large loop, going along the line for fixed , dropping down to the real line, and coming back along the real line. (See figure)

Because of the rapid decay of in the real direction, the contribution of the imaginary segments of this contour will vanish as we extend the loop out to infinity. This lets us see that for all , the integrals along both lines must be equal:

(Yes, this still works in more than one dimension.) This implies

If we take large enough to make integrable, our rapid decrease conditions imply that

was arbitrary. Take where . Then

for *all* ! If this does not tell us much, but if $latex |x| > A$ then we can make the right hand side vanishingly small. must be zero when . This completes the proof.

## Extending to Distributions

For with support in the ball of radius about the origin, we can still define the Fourier transform and its analytic continuation:

This function is analytic because the same arguments used before show that

Furthermore

For some . (I’m sorry, I did not quite follow this. If someone can clarify, please let me know and I’ll add a bit of explanation.) Again, we find that this condition characterizes the range for the Fourier transform.

**Theorem [Paley-Wiener for distributions]**

- If and then has an analytic extension to andfor some
- If latex \mathcal{U}(z)$ is analytic andfor some then there is a distribution such thatand .

**Proof** We ahve already proven (1) above. To prove (2), approximate the distribution with a function. Choose a bump function with and . Define

We know that the Fourier transform of has an analytic extension of exponential type thanks to the Paley-Wiener theorem for smooth functions. Define

Then the rapid decrease in the real direction of imples that

for all integers . Thus is the Fourier transform of a smooth function supported in the ball .

This implies that for all test functions supported outside , vanishes , hence . Letting proves the result.

Hello Rolfe,

I’m really thankful for your work in posting these course notes. Your explanations are very clear and the intuitive descriptions are helpful (even though it takes me forever to figure out the details sometimes!)

Here’s my guess for the bound on in part (1) of the theorem for distributions:

It looks like N is the order of the distribution u, which would imply the first inequality. The second inequality would follow from .