DMat0101, Notes 4_ the Fourier Transform of the Schwartz Class and Tempered Distributions _ I Woke Up in a Strange Place

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DMat0101, Notes 4: The Fourier transform of the Schwartz class and tempered distributionsPosted on March 20, 2011 by ioannis pariss is

In this section we go back to the space of Schwartz functions and we define the Fourier transform in this

set up. This will turn out to be extremely useful and flexible. The reason for this is the fact that Schwartz

functions are much `nicer’ than functions that are just integrable. On the other hand, Schwartz functions are

dense in all spaces, , so many statements established initially for Schwartz functions go through in the

more general setup of spaces. A third reason is the dual of the space , the space of tempered

distributions, is rich enough to allow us to define the Fourier transform of much rougher objects than

integrable functions

1. The space of Schwartz functions as a Fréchet space

We recall that the space of Schwartz functions consists of all smooth (i.e. infinitely differentiable)

functions such that the function itself together with all its derivatives decay faster than any

polynomial at infinity. To make this more precise it is useful to introduce the seminorms defined for any

non-negative integer as

where are multi-indices and as usual we write . Thus if and only if

and for .

It is clear that is a vector space. We have already seen that a basic example of a function in is the

Gaussian and it is not hard to check that the more general Gaussian function , where

is a positive definite real matrix, is also in . Furthermore, the product of two Schwartz functions is again a

Schwartz function and the space is closed under taking partial derivatives or multiplying by complex

polynomials of any degree. As we have already seen (and it’s obvious by the definitions) the space of infinitely

differentiable functions with compact support is contained in , , and each one of

these spaces is a dense subspace of for any and also in , in the corresponding topologies.

The seminorms defined above define a topology in . In order to study this topology we need the following

definition:

Definition 1 A Fréchet space is a locally convex topological vector space which is induced by a

complete invariant metric.

A translation invariant metric on .It is not hard to actually define a metric on which induces

the topology. Indeed for two functions we set

The function is translation invariant, symmetric and that it separates the elements of

. The metric induces a topology in ; a set is open if and only if there exists exists and

such that

I Woke Up In A Strange Place A day in the life of a mathematician.

From an analysis point of view.



Convergence in . By definition, a sequence converges to if as . A more

handy description of converging sequences in is given by the following lemma.

Lemma 2 A sequence converges to if and only if

for all .

Proof: First assume that as . Then, since

converges to zero as and all summands are positive, we conclude that for every we have that

as . However, this easily implies that as , for every .

Assume now that as for every and let . We choose a positive integer such that

.

Thus,

Now, every term in the finite sum of the first summand converges to as and we get that as

.

is a topological vector space. The topology induced by turns into a topological vector space.

To see this we need to check that addition of elements in and multiplication by complex constants are

continuous with respect to . This is very easy to check.

Local convexity. For and consider the family of sets

We claim that is a neighborhood basis of the point for the topology induced by . Indeed, the

system defines a neighborhood basis of . On the other hand it is implicit in the proof of Lemma 2 that

for every there is some and some such that . This proves the claim.

Now, in order to show that endowed with the topology induced by is locally convex it suffices (by

translation invariance) to show that the point has a neighborhood basis which consists of convex sets. This is

clear for the neighborhood basis defined above since the seminorms are positive homogeneous. Observehowever that the balls are not convex.

Exercise 1 Show that the balls , , are not convex sets.



Completeness. The space is a complete topological vector space with the topology induced by . If is

a Cauchy sequence in then for every , the sequence

is a Cauchy sequence in the space , with the topology induced by the supremum norm. Since this space is

complete we conclude that converges uniformly to some . A standard uniform convergence

argument shows now that .

Remark 1 In general, a sequence in a topological vector space is called a Cauchy sequence if

for every open neighborhood of zero , there exists some positive integer so that for

all . If the topology is induced by a translation invariant metric, this definitions coincides

with the more familiar one, that is: for every there exists such that

whenever .

The discussion above gives the following:

Theorem 3 The space , endowed with the metric and the topology induced by , is aFréchet space.

We now give a general Lemma that describes continuity of linear operators acting on by giving a simple

description of continuity of linear transformations.

Lemma 4 (i) Let be a Banach space and be a linear operator. Then is

continuous if and only if there exists and such that

for all .

(ii) Let be a linear operator. Then is continuous if and only if for each

there exists and such that

for all .

Proof: For (i) it is clear that is continuous if (1) holds. On the other hand, suppose that iscontinuous and let be the open ball of center and radius in . Then is a neighborhood of

in and hence it contains some . Thus implies that . Now we have that

Similarly, if is continuous then for every there is so that

This implies (2) using the same trick we used to deduce (1).

It is obvious that for every , . Let us show however that this embedding is also

continuous:



Proposition 5 Let . Then the identity map is continuous, that is,

there exists so that

for all .

Proof: Let . For and we have that

If observe that so there is nothing to prove.

2. The Fourier transform on the Schwartz class

Since there is no difficulty in defining the Fourier transform on by means of the formula

All the properties of that we have seen in the previous week’s notes are of course valid for the Fourier

transform on . As we shall now see, there is much more we can say for the Fourier transform on .

For and every polynomial we have that . Using the commutation relations

we see that . Furthermore, since we can use the inversion formula to write

This shows that is onto and of course it is a one to one operator as we have already seen.

Finally let us see that it is also a continuous map. To see this observe that

for every , by Proposition 5. However, so we get that

for every which shows that is continuous.

We have thus proved the following:

Theorem 6 The Fourier transform is a homeomorphism of onto itself. The operator



is the continuous inverse of on :

on .

We immediately get Plancherel’s identities:

Corollary 7 Let . We have that

In particular, for every we have that

Proof: The multiplication formula of the previous week’s notes reads

for and thus for . Now let and apply this formula to the functions

where . Observing that we get the first of the identities in the corollary. Applying this identity to the

functions and we also get the second.

We also get an nice proof of the fact that convolution of Schwartz functions is again a Schwartz function.

Corollary 8 Let . Then .

Proof: For we have that . Since we conclude that and thus that

.

3. The Fourier transform on

We have already seen that the Fourier transform is defined for functions by means of the formula

While this integral converges absolutely for , this is not the case in general for . However,

Corollary 7 says that the Fourier transform is a bounded linear operator on which is a dense subset of

and in fact we have that

for every . As we have seen several times already, this means that the Fourier transform has a unique

bounded extension, which we will still denote by , throughout . In fact the Fourier transform is an

isometry on as identity (3) shows.

Definition 9 A linear operator which is an isometry and maps onto is

called a unitary operator.



Corollary 10 The Fourier transform is a unitary operator on .

The definition of the Fourier transform on given above suggest that given , one should find a

sequence such that in and define

This, however, is a bit too abstract. The following lemma gives us an alternative way to calculate the Fouriertransform on .

Lemma 11 Let . The following formulas are valid

where the notation above means that the limits are considered in the norm.

Proof: Given let us define the functions

Then on the one hand we have that in . On the other hand the functions belong to

for all so we can write

Since the Fourier transform is an isometry on we also have that as in . The proof of

the second formula is similar.

4. The Fourier transform on and Hausdorff-Young

Having defined the Fourier transform on and on we are now in position to interpolate between

these two spaces. Indeed, we have established that

and that is of strong type and of strong type both with norm . We have also seen that it is well

defined on the simple functions with finite measure support and on the Schwartz class, both dense subsets of

all spaces for . Setting we get where is the dual exponent of . This shows that

. The Riesz-Thorin interpolation theorem now applies to show the following:

Theorem 12 (Hausdorff-Young Theorem) For the Fourier transform extends to

bounded linear operator

of norm at most , that is we have



Remark 2 This is one instance where the Riesz-Thorin interpolation theorem fails to give the

sharp norm, although the endpoint norms are sharp. Indeed, the actual norm of the Fourier

transform is

This is a deep theorem that has been proved firstly by K.I. Babenko in the special case that is an

even integer and then by W. Beckner in the general case.

Exercise 2 Let be a general Gaussian function of the form

for some positive definite real matrix . Show that

Observe that this gives a lower bound on the norm .

Hint: Write as a composition of translations, modulations and generalized dilations of the

basic Gaussian function .

Remark 3 The inversion problem for , has a similar solution as the case. One can

easily see that the means of converge to in as well as for every Lebesgue point of if is

appropriately chose. In particular this is the case for the Abel or Gauss means of .

We also have the following extension on the action of the Fourier transform on convolutions.

Proposition 13 Let and for some . Then, as we know, the function

belongs to . We have that

for almost every .

We close this section by discussing the possibility of other mapping properties of the Fourier transform,

besides the ones given by the Hausdorff-Young theorem. In particular we have seen that the Fourier transformis of strong type for all . But are there any other pairs for which the Fourier transform is of

strong, or even weak type ?

The easiest thing to see is that whenever is of type we must have that .

Exercise 3 Suppose that is of weak type . Show that we must necessarily have .

Hint: Exploit the scale invariance of the Fourier transform; in particular remember the

symmetry .

The previous exercise thus shows that the only possible type for is of the form . The Hausdorff-Young

theorem shows that this is actually true whenever . It turns out however that the bound fails for

. The following exercise describes one way to prove this.





an integrable function. We shall see that the fact that is closed under taking partial derivatives,

multiplying by polynomials and by taking the Fourier transform of its elements, its dual space is also closed

under the corresponding operations.

In what follows we will many times write for the dual and for the pairing .

Definition 14 A linear functional will be called a tempered distribution if it is

continuous on with respect to the topology on described in the previous sections.

That is, the linear functional is a tempered distribution if and only if there exists some and

such that

for all .

We equip the space with the weak-* topology ; a sequence of tempered distributions converges to a

limit if one has for all . This is the weakest topology such that for each the

functional

is continuous. The space equipped with this topology will also be denoted by .

In what follows we will also use the notation for whenever and . Be careful

not to confuse this pairing with .

6. Examples of tempered distributions

We now describe several examples of classes of tempered distributions. We begin by showing how we can

identify some known function classes with tempered distributions.

(i) Any element , can be identified with an element by means of the formula

and the map is continuous. We will say in this case that the tempered distribution is an

function.

It is clear that is linear. Furthermore we have that

for some non-negative integer , by Proposition 5, which shows that by Lemma 4. Furthermore, the

mapping is continuous. Indeed, if in we set . We need to show that in the

weak-* topology, that is, that for every . However this is a consequence of the previous

estimate.

(ii) Any element can be identified with an element by means of the formula



and the map is continuous. We will say in this case that the tempered distribution is an

Schwartz function. The proof is very similar to that of (i).

(iii) If be a finite Borel measure. Then can be identified with a tempered distribution by

means of the formula

and the map is continuous. We will say in this case that the tempered distribution is a (finite

Borel) measure. The proof is the same as that of the preceding cases.

(iv) Let . A measurable function such that for some non-negative integer is

called a tempered function. Again the functional is an element of . For such a function is often

called a slowly increasing function. Similarly a Borel measure such that

is called a tempered Borel measure and it defines an element of by setting

We will say that the tempered distribution is a tempered Borel measure.

Exercise 5 Show that if is a tempered Borel measure then and the map is

continuous. Conclude the corresponding statement if is a tempered function. Observe that

defines a tempered measure.

Exercise 6 Show that a Borel measure is a tempered measure if and only if it is of polynomial

growth: for every we have that

for some positive integer and all . In particular, is locally finite.

Remark 4 From the previous definitions one gets the impression that the term `tempered’ is

closely connected to `of at most polynomial growth’. This is in some sense correct since all

functions or measure of at most polynomial growth define tempered distribution. On the otherhand, the opposite claim is not true. Indeed, observe that the function is a slowly increasing

function (actually it is bounded) and thus defines a tempered distribution. Thus, the derivative of

this function, is also a tempered distribution although it grows exponentially fast.

All the previous examples identify functions and measures (of moderate growth) with tempered distributions

and the embeddings are continuous. However the space also contains `rougher’ objects which are

neither functions nor measures.

Exercise 7 Show that the functional for all is a tempered distribution

which does not arise from a tempered measure (and thus it does not arise from a temperedfunction either).

Example 1 (The principal value distribution) We define the functional as



Then . To see that let us fix some and and write

Now observe that thus the limit of the first summand as exists and

Moreover we have that

Furthermore this distribution does not arise from any locally finite measure. It is also easy to see

that this tempered distribution cannot arise from any locally finite Borel measure. For this

consider a Schwartz function adopted to an interval of the form for .

Exercise 8 (The principal value distribution in many dimensions) Let be a

homogeneous function of degree . This means that

(i) Show that there exists a function such that where .

(ii) Assume that . For we define

Show that the limit in the previous definition exists and that defines a tempered

distribution.

7. Basic operations on the space of tempered distributions

We have already seen that the space is closed under several basic operations: differentiation, multiplying

by polynomials, multiplication between elements of the Schwartz space and, most notably, the Fourier

transform. The space of tempered distributions has very similar properties:

Derivatives in : We begin the discussion by considering and writing down the integration

by parts formula

According to the previous definitions we can rewrite the previous formula as

or



The right hand side of the previous identity though makes sense for any in the place of whenever

. Also, for the mapping is continuous since is continuous and the map is

continuous. We thus define the partial derivative of any by means of

The previous discussion implies that .

Example 2 Let be the tempered function defined as

The function is many times called the Heaviside step function. Clearly defines a tempered

distribution in the usual way

For every we then have

That is

Remark 5 The fact that the distributional derivative of the Heaviside step function is the Dirac

mass at is intuitively obvious. The function is differentiable everywhere except at and

whenever . On the other hand there is a jump discontinuity of weight equal to at

which roughly speaking requires an infinite derivative to be realized. In general, a jump

discontinuity of weight at a point has a distributional derivative which coincides with Dirac

mass of weight at the point .

Example 3 Let be a Dirac mass at . We then have

This also explains the minus sign in Exercise 7.

Exercise 9 In dimension show that:

(i) The distributional derivative of the signum function is .

(ii) The distributional derivative of the locally integrable function is equal to .

(iii) The distributional derivative of the locally integrable function is equal to .

Translations, Modulations, Dilations and reflections in : We have see that the translation

operator maps a measurable function to the function , where . A trivial change of variables

shows that whenever we have that

Now assume that is a tempered function (say). In the language of distributions we can rewrite the previous



identity as

for all . Again, the write hand side of this identity is well defined for any and we define the

translation of any distribution as

It is easy to see that .

Similarly we define for and the tempered distributions

Convolution in : Let . Then it is an easy application of Fubini’s theorem that

where is the reflection of . In the language of distributions the previous identity reads

Now the right hand side of the previous identity is well defined whenever while in order to define

the distribution we need to have that . Now assume that is a function such that for

all . This is obviously the case if . Thus we can define the convolution of any with a

function by means of the formula

It is easy to see that the function is continuous as a composition of the continuous maps and

thus for every and .

Exercise 10 Actually, the condition is a bit too much to ask if one just wants to define

the convolution . As we have observed, the only requirement is that whenever

. Suppose that is a rapidly decreasing function, that is for all

. Show the convolution of and can be defined and that is again an element

of .

It turns out that the convolution of a tempered distribution with a Schwartz function is a function:

Theorem 15 Let and . Then the convolution is the function given by the

formula

Moreover, and for all multi-indices the function is slowly increasing.

For the proof of this theorem see [SW].



The Fourier transform on : We now come to the definition and action of the Fourier transform of

tempered distribution. As in all the other definitions, first we investigate what happens in the case the

tempered distribution is a Schwartz function. So, letting the multiplication formula implies that

In the language of tempered distributions we have that

Observing once more that the right hand side is well defined for all and that the map

is well defined and continuous we define the Fourier transform of any tempered distribution as

We have that whenever . It is also trivial to define the inverse Fourier transform of a

tempered distribution as

and to show that is a homeomorphism of onto itself. Also the operator satisfies all

the symmetry properties that the classical Fourier transform satisfies and commutes with derivatives in the

same way.

Example 4 (The Fourier transform of in ) We consider the function

Note that is locally integrable in and it decays at infinity thus it can be identified with a

tempered distribution which we will still call . On the other hand is not in any space so we

can’t consider its Fourier transform in the classical sense. We claim that the Fourier transform of

in the sense of distributions is given as

First of all observe that it suffices to show that

for all . Here it is convenient to express the function as an average of functions with known

Fourier transforms. Indeed, this can be done my means of the identity

which can be proved by simple integration by parts. Now fix a function . We have that



by an application of Fubini’s theorem since the function is an integrable function on . The

inner integral can be calculated now by using the multiplication formula and the (known) Fourier transform of

a Gaussian. Indeed we have

Putting the last two identities together we get

Now observe that by changing variables we have

and thus

since is locally integrable in and . A second application of Fubini’s theorem then gives (4) and

proves the claim.

Exercise 11 (i) Let be a smooth function such that for all multi-indices the partial derivatives

have at most polynomial growth: for some . Then the product of a

tempered distribution with is well defined by means of the formula

and .

(ii) If and then show that

Remark 6 The definition of the Fourier transform on implies that whenever ,

we have that

Thus the Fourier transform on tempered distributions is an extension of the classical definition of

the Fourier transform. If on the other hand for some then is a tempered

function and thus is a tempered distribution. This allows us to define the Fourier transform of

by looking at as a tempered distribution. The discussion the followed the Hausdorff-Young

theorem however suggests that will not be a function in general.

Exercise 12 (Poisson summation formula) For we define

Note that can be identified with the sum of a unit masses positioned on every point of the

integer lattice



Show that and that .

Hints: (a) First prove the case of dimension by proving the following intermediate

statements.

(i) Show that satisfies the invariances and .

(ii) Consider a Schwartz function with support in the interval and . If

has compact support show that the function

is a smooth function with compact support.

(iii) Let be a tempered distribution which satisfies the invariances and .

Show that

whenever are as in step (ii). Conclude that

for some , whenever is a Schwartz function with compact support. Extend this equality to

all by a density argument.

(iv) Step (iii) essentially shows that any tempered distribution that has the symmetries in must

agree with up to a multiplicative constant. Observe that satisfies the same invariances.

Conclude that by step (i). Determine the numerical constant by testing against the

Schwartz function . This concludes the proof for the one dimensional case.

(b) For general use Fubini’s theorem to show that

where denotes the (one-dimensional) Fourier transform in the direction. Thus step (a)

implies that

for every . Conclude the proof by iterating this identity.

Exercise 13 (Equivalent form of Poisson summation formula) If and

then we have that

8. Translation invariant operators



Let be vector spaces of functions on and suppose that is an operator that maps into . We will say

that commuted with translations or that is translation invariant if for all . To see an

example of such an operator, consider and define for all , . We

have seen that is well defined and furthermore that

that is, is of strong type . We have seen that the convolution commutes with translations which implies

that commutes with translations. It is quite interesting that, in some sense, all translation invariant

operators are given by a convolution with an appropriate `kernel’ (which might not be a function).

Theorem 16 Let , , be a bounded linear operator that commutes

with translations. Then there exists a unique tempered distribution such that

Thus bounded linear operators of strong type are in a one to one correspondence with the subclass of

tempered distributions such that

for all In this case we will slightly abuse language and say that the tempered distribution is of type

. It would be desirable to characterize this class of tempered distribution for all but such a

characterization is not known in general and probably does not exist. Here we gather some partial results in

this direction:

Proposition 17 (`The high exponents are on the left’) Suppose that is a linear operator

which is translation invariant and of strong type . Then we must have that . In particularthe class of tempered distributions of type is empty whenever .

Exercise 14 Prove Proposition 17 above.

Hint: Suppose that a that is translation invariant and of strong type with . Let

and consider the function

for some large positive integer and points that will be chosen appropriately. Show that

by choosing the points to be far apart from each other (how far depends only on ) we

have that while the left hand side will be of the order for large.

However, if is of strong type this is only possible if .

We also have a characterization of translation invariant operators in the following two special cases.

Theorem 18 ( ) A distribution is of type if and only if there exists such

that . In this case, the norm of the operator

defined on as



is equal to . Moreover, .

Theorem 19 ( ) A distribution is of type if and only if it is a finite Borel measure.

In this case, the norm of the operator

defined on as

is equal to the total variation of the measure .

For the proofs of these theorems and more details see [SW].

In this course we will not actually need that every translation invariant operator is a convolution operator since

we will mostly consider specific examples where this is obvious. We will focus instead on the following case.

8.1. Multiplier Operators

Let . For we define

We will say that is amultiplier operator associated to the multiplier .

Observe that is a well defined linear operator on and in fact it is bounded. Rather than relying on

Theorem 18 let us see this directly:

In fact it is not hard to check that the opposite inequality is true so that .

Exercise 15 If is a multiplier operator associated to the multiplier show that

Thus is a linear operator of type . If extends to a linear operator of type , that is if there is an

estimate of the form

for all , then we will say that is multiplier on .

Remark 7 The previous discussion and in particular Theorem 18 shows that is in fact given in

the form



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About ioannis parissis

I'm a postdoc researcher at the Center for mathematical analysis, geometry and dynamical systems at IST, Lisbon, Portugal.

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for some . In fact will be the inverse Fourier transform of in the sense of

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DMat0101, Notes 4_ the Fourier Transform of the Schwartz Class and Tempered Distributions _ I Woke Up in a Strange Place