# Inverse Problems Course Notes — The Paley-Wiener Theorem

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: $L^2$ and $\mathcal{S}$

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

$\mathfrak{F}: \mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}(\mathbb{R}^n)$

$\mathfrak{F}: \mathcal{S}^{\prime}(\mathbb{R}^n) \rightarrow \mathcal{S}^{\prime}(\mathbb{R}^n)$

$\mathfrak{F}: L^2(\mathbb{R}^n) \rightarrow L^2(\mathbb{R}^n)$

and it is an isomorphism in each case.  But what about $C_0^{\infty}(\mathbb{R}^n)$$\mathcal{E}^{\prime}(\mathbb{R}^n)$?

## A Harder Piece: $f \in C_0^{\infty}(\mathbb{R}^n)$

Say $f \in C_0^{\infty}(\mathbb{R}^n)$, and specificallyr,  say $A \leq |x| \implies f(x) = 0$.  Then we can write

$\hat{f}(\xi) = \int_{|x| < A} e^{-ix\cdot\xi}f(x)dx$

for all $\xi$ in $\mathbb{R}^n$.  It can be extended to $\mathbb{C}^n$:

$\hat{f}(z) = \int_{|x| < A} e^{-ix\cdot z}f(x)dx$

This makes sense because

$|e^{-ix\cdot z}| = |e^{x\cdot \Im z}e^{-ix\cdot \Re z}| \leq e^{A\Im z}$

So the integral defining $f(z)$ converges absolutely for all $z$.

Claim $\hat{f} \in C^{\infty}(\mathbb{C}^n)$

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 $f$.

In particular we can say

Claim $\hat{f}$ is analytic in $\mathbb{C}^n$, i.e. $\overline{\partial}\hat{f} = \sum \frac{\partial}{\partial \overline{z}_i}\hat{f} d\overline{z}_i = 0$.

Proof

$\frac{\partial}{\partial \overline{z}_i}\hat{f} = \int_{|x| < A}(\frac{\partial}{\partial \overline{z}_i}e^{-ix\cdot z})f(x)dx = 0$

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, $\hat{f}(z)$ is called the Fourier-Laplace transform of $f$.

Now we can say quite a bit about how $\hat{f}$ must look

Claim If $\text{supp} f \subset B_A(0)$ then $\hat{f}(z)$ is an entire analytic function and for all integers $N \geq 0$ there is a constant $C_N > 0$ such that

$|\hat{f}(z)| \leq C_N e^{A|\Im z|}(1 + |z|)^{-N}$

Proof We have already seen that $\hat{f}$ 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 $|z| > 1$.  For any multi-index $\alpha$ we can write

$\hat{f}(z) = \int_{\mathbb{R}^n} e^{-ix\cdot z}f(x)dx$

$= \int_{\mathbb{R}^n} \frac{1}{(-iz)^{\alpha}}\frac{\partial^{\alpha}}{\partial x^{\alpha}}e^{-ix\cdot z}f(x)dx$

$= \frac{1}{(-iz)^{\alpha}}\int_{\mathbb{R}^n}e^{-ix\cdot z}\frac{\partial^{\alpha}f}{\partial x^{\alpha}}(x)dx$

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

$|\hat{f}(z)| \leq C_\alpha \frac{e^{A|\Im z|}}{|z|^{|\alpha|}}$

for all $\alpha$.  Taking $C_N = sup_{|\alpha| = N} C_\alpha$ 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]

1. If $f \in C_0^{\infty}(\mathbb{R}^n)$ and $\text{supp} f \subset B_A(0)$ then $\hat{f}$ extends analytically to $\mathbb{C}^n$ and for all non-negative integers $N$ there exists a constant $C_n > 0$ such that$|\hat{f}(z)| \leq C_N(1 + |z|)^{-N}e^{A|\Im z|}$
2. Let $\mathcal{U}(z)$ be analytic on $\mathbb{C}^n$ such that for all non-negative integers $N$ there exists a constant $C_n > 0$ such that$|\mathcal{U}(z)| \leq C_N(1 + |z|)^{-N}e^{A|\Im z|}$
for all $z \in \mathbb{C}^n$.  Then there exists an $f \in C_)^{\infty}(\mathbb{R}^n)$ with $\text{supp} f \subset B_A(0)$ such that $\mathcal{U}(\xi) = \hat{f}(\xi)$For all $\xi \in \mathbb{R}^n$

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 $u$ on $\mathbb{R}^n$  by

$\hat{u}(\xi) \doteq \mathcal{U}(\xi)$

$u(x) = \frac{1}{2\pi}\int_{\mathbb{R}^n}e^{ix\cdot \xi}\mathcal{U}(\xi)d\xi$

If we can show that $u$ is smooth and supported in $B_A(0)$, we will be done.

Step 1: $u$ is smooth.

For an arbitrary multi-index $\alpha$

$\partial^{\alpha}_x u = \frac{1}{2\pi}\int_{\mathbb{R}^n} \partial^{\alpha}_x(e^{ix\cdot \xi})\mathcal{U}(\xi)d\xi$

$= \frac{1}{2\pi}\int_{\mathbb{R}^n}(i\xi)^{\alpha}\mathcal{U}(\xi)d\xi < \infty$

Which exists thanks to the rapid decay of $\mathcal{U}$.  So $u \in C^{\infty}(\mathbb{R}^n)$

Step 2: $u$ has compact support.

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

Because of the rapid decay of $\mathcal{U}$ 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 $\eta$, the integrals along both lines must be equal:

$u(x) = \frac{1}{(2\pi)^n} \int_{\mathbb{R}^n} e^{ix\cdot\xi} \;\mathcal{U}(\xi) d\xi$

$= \frac{1}{(2\pi)^n} \int_{\mathbb{R}^n} e^{ix\cdot(\xi + i\eta)} \;\mathcal{U}(\xi + i \eta) d\xi$

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

$|u(x)| \leq e^{-x\cdot\eta}\int_{\mathbb{R}^n}|\mathcal{U}(\xi + i\eta)|d\xi$

If we take $N$ large enough to make $(1 + |\xi| + |\eta|)^{-N}$ integrable, our rapid decrease conditions imply that

$|u(x)| \leq C_Ne^{-x\cdot \eta + A|\eta|}\int_{\mathbb{R}^n}\frac{d\xi}{(1 + |\eta| + |\xi|)^N} = \tilde{C}_Ne^{-x\cdot \eta + A|\eta|}$

$\eta$ was arbitrary.  Take $\eta = t\frac{x}{|x|}$ where $t > 0$. Then

$|u(x)| \leq C_Ne^{t(A -|x|)}$

for all $t > 0$!  If $|x| \leq A$ this does not tell us much, but if $latex |x| > A$ then we can make the right hand side vanishingly small.  $u(x)$ must be zero when $|x| > A$.  This completes the proof.

## Extending to Distributions

For $u \in \mathcal{E}^{\prime}(\mathbb{R}^n)$ with support in the ball of radius $A$ about the origin, we can still define the Fourier transform and its analytic continuation:

$\hat{u}(z) = \langle u, e^{-i\langle \cdot, z\rangle}\rangle$

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

$\frac{\partial}{\partial \overline{z}_j}\hat{u}(z) = \langle u, \frac{\partial}{\partial \overline{z}_j}e^{-i\langle \cdot, z\rangle}\rangle = 0$

Furthermore

$|\hat{u}(z)| = |\langle u, e^{-i\langle \cdot, z\rangle}\rangle| \leq C_N\sum_{|\alpha| \leq N} \sup_{B_A(0)}|\partial^{\alpha} e^{-i\langle \cdot, z\rangle}|$

$\leq C_Ne^{A|\Im z|}(1 + |z|)^N$

For some $N$.  (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]

1. If $u \in \mathcal{E}^{\prime}(\mathbb{R}^n)$ and $\text{supp } u \subset B_A(0)$then $\hat{u}$ has an analytic extension to $\mathbb{C}^n$ and$|\hat{u}(z)| \leq Ce^{A|Im z|}(1 + |z|)^N$for some $N$
2. If latex \mathcal{U}(z)\$ is analytic and$|\mathcal{U}(z)| \leq Ce^{A|\Im z|}(1 + |z|)^N$for some $N > 0$ then there is a distribution $u \in \mathcal{E}^{\prime}(\mathbb{R}^n)$ such that$\mathcal{U}(z) = \hat{u}(z)$and $\text{supp } u \subset B_A(0)$.

Proof We ahve already proven (1) above.  To prove (2), approximate the distribution with a function.  Choose a bump function $\varphi$ with $\text{supp }\varphi \subset B_1(0)$ and $\int \varphi = 1$.  Define

$\varphi_\epsilon(x) = \frac{1}{\epsilon^n}\varphi(\frac{x}{\epsilon})$

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

$\mathcal{U}_\epsilon(z) = \mathcal{U}(z)\hat{\varphi}_{\epsilon}(z)$

Then the rapid decrease in the real direction of $\hat{\varphi}_{\epsilon}$ imples that

$|\mathcal{U}_{\epsilon}(z)| \leq Ce^{(a+\epsilon)||Im z|}(1 + |z|)^{-N}$

for all integers $N \geq 0$.  Thus $\mathcal{U}_{\epsilon}$ is the Fourier transform of a smooth function $u_\epsilon = u * \varphi_\epsilon$ supported in the ball $B_{A+\epsilon}(0)$.

This implies that for all test functions $\psi$  supported outside $B_{A+\epsilon}(0)$, $\langle u, \psi\rangle$ vanishes , hence $\text{supp } u \subset B_{A + \epsilon}(0)$.  Letting $\epsilon \rightarrow 0$ proves the result.

Here’s my guess for the bound on $\hat u$ 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 $|\partial^\alpha e^{-i\langle\cdot,z\rangle}| = |(-i)^{|\alpha|} z^\alpha e^{-i\langle\cdot,z\rangle}| \leq |z|^{|\alpha|} e^{\Re (-i\langle x,z\rangle)}\leq |z|^{|\alpha|} e^{A|\Im z|}$.