Rolfe's Lecture Notes

A great tool

Rolfe Schmidt — Wed, 30 Dec 2009 21:57:43 +0000

I’ve started using latex2wp to convert my lecture notes to wordpress format, and it is great. I spend far less time futzing with wordpress, and I get nicer looking posts. I also get to write in standard LaTeX, and can produce pdfs from the same document I use to produce blog posts. In that mythical future when I have free time, I may start seeing if I can tweak it or contribute a bit.

In the mean time, I hope to rewrite all of the earlier posts using this tool to get a more consistent look and an easier to manage set of source documents. In other news: next week brings a new quarter, and a set of lectures on electrical impedance tomography.

Inverse Problems Course Notes — Numerical Methods for X-Ray Transform Reconstruction

Rolfe Schmidt — Wed, 30 Dec 2009 21:51:28 +0000

These notes are based on Gunther Uhlmann’s lectures for MATH 581 taught at the University of Washington in Autumn 2009.

An index to all of the notes is available here.

The X-Ray transform is particularly interesting because we can use it for tomography. We can send X-Rays through a patient, over many different lines, and measure the accumulated attentuation. Once this is done, we know the X-Ray transform of our object, so we can simply pass this in to the inversion formula and reconstruct a three-dimensional model. There are two substantial problems with this:

We cannot collect data for every line, we can only sample a finite number of lines.
Our measurements will have error.

There are in fact even more complications — beam hardening comes to mind — but these two must be dealt with first. Loosely speaking, the solution to the first problem relies on the stability estimates for the X-Ray Transform, and the solution to the second problem depends on the range characterization. In this post we will discuss the first problem, finite sample size, in detail. The second problem is manageable too: the range characterization of the X-Ray Transform lets us find the image function most likely to have produced the error-filled data. It turns out that the filtered backprojection algorithm we introduce below is an important step in this process. The rest of this post is organized as follows: we will start by showing that for an infinite, discrete set of samples we have unique reconstruction. Then we will consider finite sets of samples and see that reconstruction is hopeless without a priori assumptions. From there we will introduce the Filtered Backprojection algorithm and state and Natterer’s theorem about filtered backprojection error estimates.

1. Unique reconstruction for infinite, semi-discrete sample sets

In a practical tomography problem, we are not given the entire X-Ray Transform of a function, instead we just have

for some and , , . Our first result is a bit of good news: even with discrete angular samples we can get a unique reconstruction (at least for distributions with compact support).

Theorem 1 For , if

for an infinite number of , then .

Proof:

for all and for all . The set has an accumulation point, call it , where we must have in . For we can also find a sequence such that . Since , all derivatives in the direction are zero:

But is identically zero on too, so all of its directional derivatives vanish at . Since has compact support, Paley-Wiener implies that is analytic. An analytic function in several variables whose derivatives all vanish must be identically zero. So , and hence , is zero.

2. Reconstruction is impossible for finite sample sets

If the result in the last section gave us some hope, the theorem we see in this section should disillusion us. It turns out that when we have only a finite number of samples, not only is the reconstruction non-unique (which is clear), but it can be horrendously bad. Behavior on the margins of arbitrary data of interest can cause all of our observed values to vanish.

Theorem 2 (Finite number of X-rays) Let have compact closure and a nonempty interior. Let and Consider a finite number of directions, given by where . Then there exists an and such that on the interior of and

for all .

Proof: We will start with a direct proof of the case of a single line, . We want to construct an that has . Using the Fourier Slice Theorem, we see that need to construct with

To make sure we have this property, we will construct so that

Let’s start by looking at a concrete case, and take . Now we want

But this happens exactly when

where . This suggests that if we can solve the differential equation

on the interior of then we will be done. (Exercise: why can we ignore the constant term?) To do this, let with on the interior of . Then we want to solve the equation

which can be solved by integrating along the direction. Now we can just choose

to complete the proof. For general , use the same approach to solve

And take

General case, . The extension to larger is straightforward. Now we want to find an such that

To do this, we solve

Which we can do by integrating times. Now take

to complete the proof.

So we cannot reconstruct a function from a finite sample of its X-Ray Transform without a priori assumptions. So let’s make those assumptions and reconstruct.

3. Filtered Back-projection

We want to reconstruct a function from a discrete set of samples of its X-Ray Transform. We will just look at the case of practical interest: . Here the X-Ray Transform is the same as the Radon Transform, and we want to compute

given data . We will do this in two steps. First we will cut off the high frequencies, then we will discretize the integrals.

3.1. Step 1: Cut off high frequencies

We start by observing that for any distribution we have the identity

where is the delta function. We can choose a band-limited approximation to the identity, where

for all

For example, we could take . Now we will modify our goals. High frequency components cannot be sampled properly by a finite set of observations. Roughly, we cannot hope to measure angular frequencies greater than given samples. So instead of trying to reconstruct precisely, we will try to reconstruct, a band-limited approximation of . Observe

Proposition 3

Proof: The proof uses the Fourier Transform

Proposition 4

Proof: Again, we use the Fourier Transform.

In other words, since the Fourier Transform diagonalizes all of these operators, they commute.

Putting these facts together, we see that we want to reconstruct from our samples of , and

Where

3.2. Step 2: Discretize

If we can compute the quantity in Equation 1 then we are done. But

which is an integral that must be approximated. There are many ways to do this, but for our purposes the trapezoid rule will suffice. Specifically, assume that the support of is a subset of the cube . Then the support of is a subset of the cube . Let where is our discretization step size. Then we will use the following approximation

Which we will call the discretized convolution. We also need to discretize

To do this, pick and define

Putting these together, we can precisely define our reconstruction formula.

Definition 5 (Filtered Backprojection)

3.3. Bounding reconstruction errors: Natterer’s Theorem

Variants of this formula are the standard reconstruction method used in practical tomography. Now we have one pressing question: how good is it? Without making a priori assumptions about , we know we can’t hope for much. In fact Natterer has shown that under reasonable conditions we can get a lot.

Theorem 6 (Natterer) Let have support in the cube . Assume that we have observed samples as indicated above. Then

where

That is, the first error comes from discretization of the convolution and the second comes from the discretization of . Let , , and . Then we have pointwise estimates
0 \end{array} ' title='\displaystyle \begin{array}{rcl} |e_1(x)| &\leq& C \sup_{\theta \in S^1} \int_{|\sigma| \geq b}|\sigma||\hat{f}(\sigma\theta)|d\sigma \quad \forall x, |x_i| \leq 1 \\ |e_2(x)| &\leq& C(\alpha) e^{-D(\alpha)N}\|f\|_{\infty} \qquad D(\alpha) > 0 \end{array} ' class='latex' />

Note that the estimate on is related to control of the Sobolev norm of . We will not prove this theorem here, look at Natterer’s book for details. Instead, we will spend the remaining lectures exploring helical tomography, local tomography, and other related problems. (A pdf version of these notes is available here.)

Inverse Problems Course Notes — The Range of the X-Ray Transform

Rolfe Schmidt — Fri, 04 Dec 2009 21:07:08 +0000

These notes are based on Gunther Uhlmann’s lectures for MATH 581 taught at the University of Washington in Autumn 2009.

An index to all of the notes is available here.

Now we are prepared to study the range of the X-Ray Transform. We will will look at the image of in detail — once we understand this we can easily use our understanding of the Radon transform to characterize the image of for smooth, compactly supported functions.

So let . Recall that

It will be convenient to extend to all of as follows

So is said to be positive homogeneous of degree -1 in .

Claim 1 For , The partial differential equations

hold for all .

To show this, check the following

which is symmetric in and .

These equations are called John’s Equations, named for Fritz John who was the first to point this fact out. The main result is that these are the only conditions we need to characterize the range of

Theorem 1 (Fritz John) If , define its extension to as

Then is in the range of if and only if and

for all

This is nice: we characterize the range with only a finite number of conditions. Contrast this with the range characterization of the Radon Transform, which required an infinite number of conditions and could not be effectively checked.

Proof: We want to solve

By the Fourier Slice Theorem for the X-Ray transform, this is solved if we take such that

But for any fixed , there are many such that , so at first glance this definition looks problematic. To make it work we must show that, for fixed , for all orthogonal to .

We will do this by showing that the directional derivatives of are zero in all directions orthogonal to . It is easier to work on than on the manifold , so we will study an extension of to and show that the -gradient of this extension is parallel to , thus the directional derivatives in all directions orthogonal to will be zero.

Defining this extension, we need to make sure that it does not vary as we scale , or the -gradient will have a component in the direction and will not be parallel to . With that in mind, define the extension as follows

using the homogeneity in the definition of . Applying the inverse Fourier Transform gives

When . For general we have

Because , and hence , is well behaved, we can differentiate this expression under the integral sign.

at points where . The general expression is quite complicated, but many of the terms are multiples of and . It is a worthwhile exercise to verify this expression.

Now we can use John’s equation to say that, when

But this is an (inverse) Fourier Transform of a function and can only vanish if the transformed function vanishes ( is injective). So

The expression on the left hand side is nothing but the coordinate of the wedge product . So this wedge product vanishes. That can only happen when the two (co)vectors are parallel, so there is a scalar valued function such that

If you aren’t comfortable with wedge products, it is just a short hand way of making the following calcuation.

which clearly implies that and are parallel.

This is exactly what we were looking for: is an extension of , and its derivatives vanish in all directions orthogonal to . So when we look at a small change in , as long as that change remains perpendicular to :

So for fixed , is locally constant on . This set is connected, so is constant on and our definition of is sound.

The rest of the proof is straightforward.

(A pdf version of these notes is available here.)

Inverse Problems Coruse Notes — Stability Estimates for the X-Ray Transform

Rolfe Schmidt — Mon, 30 Nov 2009 19:17:27 +0000

These notes are based on Gunther Uhlmann’s lectures for MATH 581 taught at the University of Washington in Autumn 2009.

An index to all of the notes is available here.

With an inversion formula in hand, we are ready to state and prove the basic stability estimates for the X-Ray Transform. They are almost identical to the estimates for the Radon transform, the only difference coming from the fact that smoths by of a derivative, where smooths by derivatives.

Theorem 1 For all , compactly supported, and , the following estimates hold

The first inequality holds even when does not have compact support.

In other words X-Ray inversion is stable, and so is the X-Ray transform itself, as long as you measure errors with the right norms.

Proof: Step 1. Let’s start by proving the first inequality — stability, or boundedness, of X-Ray inversion — in the case (). The argument goes as follows: use the inversion formula to rewrite the formula for the norm of , then use Plancherel’s Theorem and the Fourier Slice Theorem to move everything to the frequency domain. Here are the details.

Now we can extend this to general by replacing with . Then

Proving the first inequality. Here we relied on the following result, an easy generalization of the intertwining results for , and we used before.

Claim 1

The proof is left as an exercise.

Step 2. Again, we will start with the case.

To go forward we need a result that relates the measure to standard Lebesgue measure. As varies over , clearly covers all of , but it covers it more than once and points close to the origin are covered “more densely” than points far away. The following claim makes this intuition precise, and the proof can be found in the appendix of Natterer’s book.

Lemma 2

This lets us continue, writing

We have not used the compact support of yet, but now it appears in a move that should seem familiar. We will split the integral into a sum of two integrals, one over the low fequencies, and the other over the high frequencies. Define

Then

But

And

We can repeat this argument for any with the usual modifications (pick smooth compactly supported on and use ).

(A pdf version of these notes is available here.)

Inverse Problems Course Notes — The X-Ray Transform for Distributions

Rolfe Schmidt — Mon, 23 Nov 2009 19:47:56 +0000

These notes are based on Gunther Uhlmann’s lectures for MATH 581 taught at the University of Washington in Autumn 2009.

An index to all of the notes is available here.

We have studied the X-Ray transform on very restricted domains — and — and found an inversion formula there. Now we want to move to our next basic question: is the inversion stable?

As with the Radon transform, we will see that it is stable: small errors in the X-Ray transform lead to small errors in the reconstructed function, and vice versa, when errors are measured in an appropriate norms.

Sobolev spaces provide one family of norms that will work, but to use these we need to extend the domain of to distributions, and prove the inversion formula on the broader domain.

1. The X-Ray Transform of Distributions

As usual, we define on distributions by duality, and most of the results are tautologies.

Definition 1 For , define by its action on test functions

Proposition 2 is linear and continuous.

Proof: Exercise.

We define similarly.

Definition 3 For , define by its action on test functions

Proposition 4 is linear and continuous.

We can also define and for tempered distributions but we will see that these domains are more natural for the problem at hand because we want to study domains where we can use our inversion formula. But the inversion formula involves a fractional power of the Laplacian, which is not tempered!

2. Powers of the Laplacian, Sobolev Spaces

For define powers of the negative Laplacian using the fourier Transform as follows.

for -n}' title='{\alpha > -n}' class='latex' />, so that the right hand side is locally integrable.

Proposition 5 for and -\frac{n}{2}}' title='{\alpha > -\frac{n}{2}}' class='latex' />.

Proof:

If then , so

proving the result. Now we need to consider the case . We will proceed by splitting the integral into a high-frequency and low-frequency part.

1}|\xi|^{2\alpha}|\hat{u}(\xi)|^2(1 + |\xi|^2)^{s-\alpha} d\xi \\ &\leq& \int_{|\xi| \leq 1}|\xi|^{2\alpha}|\hat{u}(\xi)|^2(1 + |\xi|^2)^{s-\alpha} d\xi + \|u\|_{H^s(\mathbb{R}^n} \\ &\leq& \sup_{|\xi| \leq 1} |\hat{u}(\xi)|\int_{|\xi| \leq 1} |\xi|^{2\alpha} + \|u\|_{H^s(\mathbb{R}^n} \end{array} ' title='\displaystyle \begin{array}{rcl} \|(-\bigtriangleup)^{\frac{\alpha}{2}}u\|_{H^{s-\alpha}(\mathbb{R}^n)} &=& \int_{|\xi| \leq 1}|\xi|^{2\alpha}|\hat{u}(\xi)|^2(1 + |\xi|^2)^{s-\alpha} d\xi \\ & & \quad + \int_{|\xi| > 1}|\xi|^{2\alpha}|\hat{u}(\xi)|^2(1 + |\xi|^2)^{s-\alpha} d\xi \\ &\leq& \int_{|\xi| \leq 1}|\xi|^{2\alpha}|\hat{u}(\xi)|^2(1 + |\xi|^2)^{s-\alpha} d\xi + \|u\|_{H^s(\mathbb{R}^n} \\ &\leq& \sup_{|\xi| \leq 1} |\hat{u}(\xi)|\int_{|\xi| \leq 1} |\xi|^{2\alpha} + \|u\|_{H^s(\mathbb{R}^n} \end{array} ' class='latex' />

But

Where is some smooth function — your choice — that is identically 1 on a neighborhood of . But then

If you chose well, this shows that , and thus (with a slightly larger constant)

completing the proof.

So now we have defined on and defined on . You might wonder why we did not define these on tempered distributions; the definitions are straightforward after all.

The answer is that these operators appear alongside in the inversion formula and this operator is not tempered. For some ,try applying to a test function :

But this is not well defined because . For now, we will restrict our attention to domains where the inversion formula makes sense.

We’ll close this post with the obvious theorem. The Proof is left as an exercise.

Theorem 6 (X-Ray Inversion I and II For Distributions) The following identities hold

(A pdf version of these notes is available here.)

Inverse Problems Course Notes — An Alternative Development of the X-Ray Transform

Rolfe Schmidt — Thu, 19 Nov 2009 22:50:00 +0000

These notes are based on Gunther Uhlmann’s lectures for MATH 581 taught at the University of Washington in Autumn 2009.

An index to all of the notes is available here.

Now that we’ve sketched the basic facts about the X-Ray transform, let’s look at it from a different perspective. When we developed the theory of the Radon transform, we made extensive use of the Fourier Slice Theorem — something we did not mention at all for the X-Ray transform. We also notice that our inversion formula looked different from the one we developed for .

In this post we will see that these differences are superficial. We will find the analog for the Fourier Slice Theorem and use it to redevelop our X-Ray transform results, mimicking our theory of the Radon transform.

1. The Fourier Slice Theorem

When analyzing the Radon Transform, we noticed that the “wave functions” used to define the Fourier Transform are constant on hyperplanes, so the Fourier Transform of a function could be represented as a phase-weighted sum of integrals over these constant hyperplanes, i.e. as a weighted integral of the Radon Transform of the function. Specifically

This was the Fourier Slice Theorem. If the waves are constant over hyperplanes, they are certainly constant over lines, so we can find a similar formula for the X-Ray Transform. There are many ways we can foliate a hyperplane into a family of parallel lines, so we should expect to get many formulas. This is exactly what happens. (I try to give a bit more intuition in a separate post.)

Given , select an arbitrary orthogonal unit vector . Then

As varies over and varies over , varies over and is standard Lebesgue measure. In this case, notice that because so

As promised, this gives us a different formula for each choice of . Let’s summarize the result in a theorem

Theorem 1 (Fourier Slice Theorem for the X-Ray Transform) For all and with

For the Radon transform, the slice theorem immediately allowed us to prove that the Radon Transform intertwines (powers of) the Laplacian on with on . We can do the same thing for the X-Ray transform.

Theorem 2 (Intertwining) For and -(n-1)}' title='{\alpha > -(n-1)}' class='latex' />

Where is defined by .

The condition that -(n-1)}' title='{\alpha > -(n-1)}' class='latex' /> insures that the operator is defined by an absolutely convergent integral on .

Proof: The proof is exactly the same as it was for the Radon Transform. We use the slice Theorem for :

We can do the same thing for the transpose.

Theorem 3 (Intertwining) For and -(n-1)}' title='{\alpha > -(n-1)}' class='latex' />

Proof: As often happens with dual results, the proof is just a formal manuipulation. First note that is its own transpose:

A similar argument works for on , so we can write

This result immediately lets us rewrite the inversion formula in a style that looks more like our original Radon inversion formula

Theorem 4 (X-Ray Inversion II) For all

(A pdf version of these notes is available here.)

The Geometry of the Slice Theorems

Rolfe Schmidt — Thu, 19 Nov 2009 18:27:51 +0000

When studying the Radon transform, we saw that we could reconstruct the Fourier transform of a function from its Radon Transform:

The Fourier Transform integrates the product of a function with “waves” that are constant on hyperplanes.
The Radon Transform computes the integral of a function over these hyperplanes.
So the Fourier Transform of a function is a sort of phase-weighted sum of integrals over hyperplanes, i.e. a phase-weighted sum of the Radon transform.

When evaluating the Fourier Transform of a function at a point , the hyperplane appears with “weight” . So we can write the Fourier transform of in terms of the Radon Transform as follows:

The same arguments apply to the X-Ray transform. In fact, we can compute the Radon transform from the X-Ray transform at a hyperplane by integrating over a set of parallel lines that covers the hyperplane (there is more than one way to do this!).

To compute the Fourier Transform directly from the X-Ray transform, consider the following. Say we want to compute the Fourier Transform of at a frequency vector . We can pick any orthogonal , and have . Now the “wave” with frequency will be constant in direction , so to compute the Fourier Transform of at we just need to add up the integrals of along the lines in direction — — weighted it by the value of the wave on these lines. In other words

for . Notice that our choice of was arbitrary, so this really gives us a continuum of formulas, one for each .

Inverse Problems Course Notes — The X-Ray Transform

Rolfe Schmidt — Mon, 16 Nov 2009 01:06:11 +0000

We motivated the study of the Radon transform with a tomographic problem: given the change in intensity of X-rays along all lines through a region, can we reconstruct the attenuation (think of this as density) in the region?

For our purposes in this problem, light travels along straight lines and in two dimensions, those lines are hyperplanes. So inverting the Radon transform — which sums functions over hyperplanes — solves our problem in 2-D. But in higher dimensions this is not what we need. We need to integrate along lines, not hyperplanes.

So we introduce the X-ray transform.

Definition Given a function , the X-ray transform of is defined as

where .

The first thing to notice is that there is some redundancy here. For all we will have

So the natural domain of is actually smaller, we can pick just one representative from the line without losing any information. There is a distinguished point on this line: the point is the unique point on the line that is perpendicular to , .

Since for all , it is natural to think of as a function and , the hyperplane normal to . Formally, the domain of is

Where is the hyperplane normal to . This is a manifold of dimension .

Now , and we will use standard Lebesgue measure on this space, denoting it by . As with the hyperplanes in the Radon transform, we will keep using the fact that

is standard Lebesgue measure on $mathbb{R}^{n}$.

Function Spaces on

To proceed we need to define the standard function spaces on this domain. There is nothing surprising here. Let’s start with

Here is a tangential derivative to . For example, if

then take

For any other hyperplane, we can rotate it into this position and use these operators to define .

Now we will define. The sphere is compact, so we only need to worry about compact support in the directions.

(Sorry — having wordpress-latex trouble. Will try to get this fixed.) The Schwartz space is similar:

As with the spaces defined for the radon transform, we could choose to ignore the derivatives in the angular variables without impacting the subsequent theory, but we will not do that.

In the same way, we can define , etc. Topologies on these spaces are also defined in the natural way.

The Transpose of

It is easy to see that is injective: the Radon transform is injective and can be computed from by integrating the value of over all the lines that make up the hyperplanes in the definition of . But we will do better and produce an explicit inversion formula.

First we need to find the formal transpose of , i.e. the operator such that .

Let

As varies over and t varies over in this integral, the variable varies over all of and

is standard Lebesgue measure. If we perform this change of variables, we must remember that is only defined on , but when , we have . So

So we will define

Proposition The maps

are linear and continuous.

Proof Exercise.

The Normal Operator

We will use a different approach to develop the inversion formula for the X-ray transform. We could parallel our development for the Radon transform, and we could rederive the Radon inversion formula using this approach. It is good to see more than one way to look at a problem.

In linear algebra, it is often convenient to work with an operator . It is a normal operator and is usually “much better behaved” than .

Because in the integral defining , we catch the integral over twice: once for , and once for . This can be reinterpreted as an integral in polar coordinates. Let and . Then

up to constants. This is a singular integral, but not badly behaved. We do not need to regularize it.

To invert this convolution, we use the Fourier transform.

for .

Claim

We will prove this claim (in fact, we’ll prove a more general result) in a supporting post. This result immediately lets us invert the normal operator. If

then we must have

In other words,

is a power of the negative Laplacian. Now we can state the main result

Theorem [The X-Ray Inversion Formula] For all

So we gain half a derivative by applying . Notice that this formula does not depend on the dimension, because the dimension of the manifolds we are integrating over to define do not change. Also note that the inversion operator is never local.

We will see that for 2' title='n > 2' class='latex' /> this has too many lines, and the formula is not very useful.

Parallel Development for the Radon Transform

We could develop the inversion formula for the Radon transform using its normal operator in a completely analogous way.

So the same manipulations show that

Theorem [Radon Inversion Formula II] For

If is even, we get a square root and the operator is not local. If is odd, the inversion operator is local.

At first glance, this looks different, but it is really the same as what we had before. The proof is left as an exercise. It relies on the observation that the Radon transform intertwines (functions of the) Laplacian on with those on :

A Quick Look Ahead

In the coming lectures, we will

Generalize these results to distributions
Rewrite the inversion formula to make it look more like the formula for .
Make stability estimates
Characterize the range of

The range characterization is perhaps the most interesting part. depends on variables and is overdetermined. We can extend it to as follows

and immediately see that we have a compatibility condition:

We will see the Fritz John’s theorem that states that this system of PDEs gives necessary and sufficient conditions for a function to be in the range of .

Inverse Problems Course Notes — The Paley-Wiener Theorem

Rolfe Schmidt — Fri, 13 Nov 2009 21:42:09 +0000

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 0' title='C_N > 0' class='latex' /> 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 1' title='|z| > 1' class='latex' />. 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 0' title='C_n > 0' class='latex' /> such that
Let be analytic on such that for all non-negative integers there exists a constant 0' title='C_n > 0' class='latex' /> 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 0' title='t > 0' class='latex' />. Then

for all 0' title='t > 0' class='latex' />! 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 A' title='|x| > A' class='latex' />. 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 0' title='N > 0' class='latex' /> 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.

Inverse Problems Course Notes — Motivating the Range Characterization for the Radon Transform

Rolfe Schmidt — Thu, 12 Nov 2009 23:18:21 +0000

With the proofs of the range characterization and support theorems for the radon transform complete, let’s take a step back and look at the intuition behind these results again. In particular, let’s try to understand the moment conditions (It might be good to read this post before reading the full proofs). It was easy to directly verify that the moment conditions were necessary, so

We need to show it is sufficient — that . Take $g \ in \mathcal{S}_H(\mathbb{R}\times S^{n-1})$, we need to find with

Define

Because is even, is well defined away from the origin. It is also easy to see that for , is smooth and rapidly decreasing. To prove that is Schwartz, we need to show that it

is smooth at the origin. The problem is that is not smooth at the origin, so we need to get rid of it.

How can we do this?

Write

and expand the power series

Thanks to the moment conditions, we can write this as

Where is a homogeneous polynomial of degree . But then the terms cancel out:

Which certainly has derivatives of order 0, 1, and 2 at the origin. Of course we can continue this process. This should convince you that it is worth trying to prove the result rigorously, but shouldn’t quite convince you that the result is true — that is what the rigorous proof is for.