1. Definition and main properties.
For , the Fourier transform of is the function
Here denotes the inner product of and :
Observe that this inner product in is compatible with the Euclidean norm since . It is easy to see that the integral above converges for every and that the Fourier transform of an function is a uniformly continuous function.
(i) The Fourier transform is linear and for any .
(ii) The function is uniformly continuous.
(iii) The operator is bounded operator from to and
(iv) (Riemann-Lebesgue) We have that
Proof: The properties (i), (ii) and (iii) are easy to establish and are left as an exercise. There are several ways to see (iv) based on the idea that it is enough to establish this property for a dense subspace of . For example, observe that if is the indicator function of an interval of the real line, , then we can calculate explicitly to show that
Tensoring this one dimensional result one easily shows that whenever is the indicator function of an -dimensional interval of the form . Obviously the same is true for finite linear combinations of -dimensional intervals since the Fourier transform is linear.
Now let be any function in and and consider a simple function which is a finite linear combination of -dimensional intervals, such that . Let also be large enough so that whenever . Using (iii) and the linearity of the Fourier transform we have that
whenever , which finishes the proof.
In view of (ii) and (iv) we immediately get the following.
Exercise 1 Show the properties (ii) and (iii) in the previous Theorem.
The discussion above and especially Corollary 2 shows that a necessary condition for a function to be a Fourier transform of some function in is . However, this condition is far from being sufficient as there are functions which are not Fourier transforms of functions. See Exercise 8.
Let us now see two important examples of Fourier transforms that will be very useful in what follows.
Proof: Observe that in one dimension we have
where the third equality is a consequence of Cauchy’s theorem from complex analysis. The -dimensional case is now immediate by tensoring the one dimensional result.
Remark 1 Replacing in the previous example we see that is its own Fourier transform.
which is a simple consequence of the identities
Using (1) we can write
by the definition of the -function.
Exercise 2 This exercise gives a first (qualitative) instance of the uncertainty principle. Prove that there does not exist a non-zero integrable function on such that both and have compact support.
Hint: Observe that the function
extends to an entire function (why ?).
The definition of the Fourier transform extends without difficulty to finite Borel measures on . Let us denote by this class of finite Borel measures and let . We define the Fourier transform of to be the function
We have the analogues of (i), (ii) and (iii) of Theorem 1 if we replace the norm by the total variation of the measure. However property (iv) fails as can be seen by consider the Fourier transform of a Dirac mass at the point . Indeed observe that
which is a constant function.
The Fourier transform interacts very nicely with convolutions of functions, turning them to products. This turns out to be quite important when considering translation invariant operators as we shall see later on in the course.
Exercise 3 Prove Proposition 3.
Another important property of the Fourier transform is the multiplication formula.
We will now describe some easily verified symmetries of the Fourier transform. We introduce the following basic operations on functions:
Dilation operator: .
(iii) , where .
Exercise 4 Prove the symmetries in Proposition 5 above. Also, let be an invertible linear transformation, that is, . Define the general dilation operator
where is the (real) adjoint of , that is the matrix for which we have for all .
We now come to one of the most interesting properties of the Fourier transform, the way it commutes with derivatives.
(b) We will say that a function has a partial derivative in the norm with respect to if there exists a function such that
where is a non-zero vector along the -th coordinate axis. If has a partial derivative with respect to in the -norm, then
Exercise 5 Prove Proposition 6.
A similar result that involves the classical derivatives of a function is the following:
Exercise 6 Prove Proposition 7.
where here we abuse notation and denote by the operator of multiplication by . Thus the Fourier transform turns derivatives to multiplication by the corresponding variable, and vice versa, it turns multiplication by the coordinate variable to a partial derivative, whenever this is technically justified. This is a manifestation of the heuristic principle that smoothness of a function translates to decay of the Fourier transform and on the other hand, decay of a function at infinity translates to smoothness of the Fourier transform.
A second remark is that these commutation relations generalize, in an obvious way, to higher derivatives. To make this more precise, let be a polynomial on :
Slightly abusing notation again we write for the differential operator
We then have that the following commutation relations are true
2. Inverting the Fourier transform
On of the most important problems in the theory of Fourier transforms is that of the inversion of the Fourier transform. That is, given the Fourier transform of an function, when can we recover the original function from ? We begin with a simple case where the recovery is quite easy.
for almost every .
Proof: The proof is based on the following calculation. For we have that
where in the last equality we have used Example 1. We can thus write
Since as and , Lebesgue’s dominated convergence theorem shows that is almost everywhere equal to the -limit of the sequence of functions
as (technically speaking we need to consider a sequence ). On the other hand since , another application of Lebesgue’s dominated theorem shows that the -limit of the functions is also equal to . This completes the proof of the proposition.
An immediate corollary is that the Fourier transform is a one-to-one operator:
Corollary 9 Let and suppose that for all . The we have that for almost every .
The proof is an obvious application of Proposition 8.
Exercise 7 (i) Suppose that . Show that
Conclude that whenever , we have that
(ii) Show that maps the Schwartz space onto .
(i) Show that is a proper subset of .
Hint: While there are different ways to do that, a possible approach is the following. For simplicity we just consider the case :
(a) Show that for all where is a numerical constant that does not depend on .
(b) Suppose that is such that is an odd function. Use (a) to show that for every we have that
for some numerical constant which does not depend on .
(c) Construct a function which is not the Fourier transform of an function. To do this note that it is enough to find a function which does not satisfy the condition in (b).
(ii) Show that where the closure is taken in the topology.
Hint: Observe that is dense in , in the topology of the supremum norm.
It is convenient to define the formal inverse of the Fourier transform in the following way. For we set
Here we denote by the reflection of a function , that is, . Observe that is the conjugate of the Fourier transform. Thus the operator is very closely connected to the operator and enjoys essentially the same symmetries and properties.
As we shall see later on, it is also the adjoint of the Fourier transform with respect to the inner product
Although we haven’t yet defined the Fourier transform on we can calculate for that
Proposition 8 claims that is also the inverse of the Fourier transform in the sense that
The proof of Proposition 8 is quite interesting in the following ways. First of all observe that we have actually showed that whenever , is equal (a.e.) to the limit of the functions
as . This does not require any additional hypothesis and actually provides us with a method of inverting the Fourier transform of any function, at least in the sense. The second remark is that the proof of Proposition 8 can be generalized to different methods of summability. Indeed, let be such that and . For we consider the integrals
The following more general version of Proposition 8 is true.
converge to in , as .
Proof: The proof is just a consequence of formula (3). Indeed, is an approximation to the identity since and and thus converges to in the norm as .
Proposition 8 says that the inversion formula is true whenever . This however is not the most natural assumption since the Fourier transform of an function need not be integrable. The idea behind Proposition 10 is to `force’ in by multiplying it by the function . Thus, we artificially impose some decay on . This is equivalent to smoothing out the function itself by convolving it with a smooth function . Although no smoothness is explicitly assumed in Proposition 10, there is a hidden smoothness hypothesis in the requirement . Indeed, we could have replaced this assumption by directly assuming that is (say) a smooth function with compact support and taking ; then the conclusion would follow automatically. The trick of multiplying the Fourier transform of a general function with an appropriate function in or, equivalently, smoothing out the function itself allows us then to invert the Fourier transform, at least in the -sense. This process is usually referred to as a summability method.
As we shall see now, the inversion of a Fourier transform by means of a summability method is also valid in a pointwise sense. Because of formula (3), in order to understand the pointwise convergence of the -means of we have to examine the pointwise convergence of the convolution to , whenever is an approximation to the identity.
Definition 11 Let . The Lebesgue set of is the set of points such that
The Lebesgue set of a locally integrable function is closely related to the set where the integral of is differentiable:
Definition 12 Let . The set of points where the integral of is differentiable is the set of points such that
where is the volume of the unit ball in . In other words, we say that the integral of is differentiable at some point if the average of with respect to Euclidean balls centered at the value of at the point .
We shall come back to these notions a bit later in the course when we will introduce the maximal function of which is just the maximal average of around every point. For now we will use as a black box the following theorem:
While postponing the proof of this theorem for later on in the course, we can already see the following simple proposition connecting the Lebesgue set of to to the set of points where the integral of is differentiable. In particular we see that almost every point in is Lebesgue point of .
Proof: For any rational number we have that the function is locally integrable. Theorem 13 then implies that
for almost every . Thus the set where the previous statement is not true has measure zero and so does the set . Now let . Indeed, let and be such that . We then have
The first summand converges to as since while the second summand is smaller than . This shows that the Lebesgue set of is contained in and thus that almost every point in is a Lebesgue point of .
We can now give the following pointwise convergence result for approximations to the identity.
whenever is a Lebesgue point for . If in addition then the -means of ,
converge to as for almost every .
Proof: Let be a Lebesgue point of and fix . By Corollary 14 there exists such that
We can estimate as usual
We claim that
First of all observe that is radially decreasing. We will abuse notation and write . For every we have that
Now since , the left hand side in the previous estimate tends to when and when we get the claim.
We write (4) in polar coordinates to get
Setting we can rewrite the previous estimate in the form
whenever . We now estimate as follows
At this point the proof simplifies a bit if we assume that is differentiable. In this case we have that and we can estimate the last integral by
The argument actually goes through without the assumption that is differentiable by a clever use of the Riemann-Stieljes integral. Note that the function is decreasing thus almost everywhere differentiable. This shows that .
For we estimate as follows
For the second summand we have that
as , since .
On the other hand, we have
Now since is decreasing we have
We have showed that
whenever is a Lebesgue point of . Since was arbitrary this completes the proof of the theorem.
Remark 2 The previous theorem is true in the case that is a radially decreasing function in or, in general, a function that satisfies a bound of the form for some .
We conclude the discussion on the inversion of the Fourier transform with a useful corollary.
for almost every . In particular,
Proof: By identity (3) we have that
for all . Observe that the functions on both sides of this identity are continuous functions of . Now let satisfy the conditions of Theorem 15. Assume furthermore that is non-negative and continuous at . For example we can consider the function . Now since the point is a point of continuity of , it certainly belongs to the Lebesgue set of . Thus we have that which gives
Since is positive, we can use Fatou’s lemma to write
so . Thus the inversion formula holds true for and we get
for almost every . However
2.1. Two special summability methods
The Gauss-Weierstrass summability method. By dilating the function we get
The function is called the Gauss kernel and it gives rise to the Gauss-Weierstrass method of summability. The Fourier transform of is
It is also clear that
for all . The discussion in the previous sections applies to the Gauss-Weierstrass summability method and we have that the means
convergence to in and also in the pointwise sense, for every in the Lebesgue set of . One of the aspects of Gauss-Weierstrass summability is that the function defined above satisfies the heat equation:
To see that the Gauss-Weierstrass means of satisfy the Heat equation with initial data , one can use the formula for and calculate everything explicitly. However it is easier to consider the Fourier transform of the solution of the Heat equation in the variable and show that it must agree with the Fourier transform of , again in the variable. Observe that under suitable assumptions on the initial data we get that the solution converges to the initial data as `time’ .
Exercise 9 Let , . Using the properties of the Fourier transform show that the function satisfies the initial value problem
Solve the initial value problem to give an alternative proof of the fact that . Observe that the differential equation above is invariant under the Fourier transform.
The Abel summability method. We consider the function where . By dilating the function we have
The function is called the Poisson kernel (for the upper half plane) and it gives rise to the Abel method of summability. The Fourier transform of is
This is just a consequence of the calculation in Example 2, the inversion formula and the easily verified fact that . It is also clear by a direct calculation or through the previous Fourier transform relation that
for all . Everything we have discussed in these notes applies to the Abel summability method. In particular we have that whenever , the means
converge to in as and also in the pointwise sense for all in the Lebesgue set of . The function is also called the Poisson integral or extension of . It is not difficult to see that it satisfies the Dirichlet problem
Here we denote by the upper half plane . Thus, if we are given an function on the `boundary’ , the Poisson integral of provides us with a harmonic function in the upper half plane which has boundary value in the sense that converges to as both in the sense as well as almost everywhere.
Remark 3 The Poisson extension of ,
is harmonic in , that is, that it satisfies the Laplace equation:
This is essentially a consequence of the fact that for .
In general, we can ask for a harmonic function in which has boundary value where the limit is taken -sense. In the case this extension is uniquely given by the Poisson integral of . Also, the same is true if and . On the other hand, if we ask for a function which is harmonic in , continuous in and has boundary function , then no assumption on can guarantee that this extension is unique. Take for example and , . The solution of the Dirichlet problem becomes unique though if we require in addition that the harmonic extension is a bounded function in . See [SW] for more information.
Exercise 10 Prove the subordination identity:
For this, first prove the identities
The second identity above is obvious. In order to prove the first, use the theory of residues for the function
[Update 11 Mar 2011: Exercise 10 added.]
[Update 11 Mar 2011: Statement of Theorem 15 completed with the corollary about the convergence of the means of .]