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.


\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)

Contour of integration used in the proof of the Paley-Wiener Theorem


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


|\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|)^Nfor some N
  2. If latex \mathcal{U}(z)$ is analytic and|\mathcal{U}(z)| \leq Ce^{A|\Im z|}(1 + |z|)^Nfor 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.


One thought on “Inverse Problems Course Notes — The Paley-Wiener Theorem

  1. 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 \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|}.

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