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 1A 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

**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 2A 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. Observe however that the balls are not convex.

Exercise 1Show that the balls , , arenotconvex 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 1In general, a sequence in a topological vector space is called aCauchy sequenceif 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 3The space , endowed with the metric and the topology induced by , is a Fréchet 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 is continuous 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 5Let . 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 6The Fourier transform is ahomeomorphismof onto itself. The operator

is the continuous inverse of on :

on .

We immediately get Plancherel’s identities:

Corollary 7Let . 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 8Let . 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 9A linear operator which is an isometry and maps onto is called aunitary operator.

Corollary 10The 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 Fourier transform on .

Lemma 11Let . 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 2This 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 2Let 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 3The 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 13Let 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 transform is 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 3Suppose 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.

Exercise 4Suppose that cannot be of strong type when . (i) Let be a large positive integer and . For consider the function

Show that

(ii) For any show that

if and are large enough. For this show first the endpoint bounds for and . This will also give you the intermediate upper bounds by log-convexity. For the lower bounds, consider the values of close to integer multiples of .

(iii) The previous steps show that

which allows you to conclude the proof.

**5. The space of tempered distributions **

The purpose of this paragraph is to introduce a space of `generalized functions’ that is much larger than all the spaces we have seen so far, namely *the space of tempered distributions*. Let us begin with an informal discussion, drawing some analogies with some more classical (though not so classical) function spaces.

We have seen already that at whenever and the underlying measure is -finite, then the space can be identified with the dual , by means of the pairing:

This is already quite interesting. A function in is already a generalized object in the sense that it is only defined up to sets of measure zero; so in fact it represents and equivalent class. Furthermore, it can be identified with a linear functional acting on another function space.

We have see that the space is contained in every space and furthermore that it is dense in for all . Restricting our attention to a smaller class of function, the space , we get a larger dual space:

We thus obtain a space of generalized functions that contains the `classical’ spaces. As we shall see, this space is much bigger and in particular it allows us to differentiate (in the appropriate sense) and remain in this class of generalized functions and, most notably, consider the Fourier transform of these objects and still remain in the class. These operation many times are not even available on spaces; for example we cannot even define the Fourier transform on for . Furthermore, even when there is a way to define these operations on functions we don’t necessarily stay in the given class of functions. For example, while it is perfectly legitimate to define the Fourier transform of an function, the resulting function is not in general 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 14A linear functional will be called atempered distributionif 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 5Show 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 6Show that a Borel measure is a tempered measure if and only if it is ofpolynomial growth: for every we have that

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

Remark 4From 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 other hand, 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 7Show 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 tempered function 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 2Let be the tempered function defined as

The function is many times called theHeaviside step function. Clearly defines a tempered distribution in the usual way

For every we then have

That is

Remark 5The 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 3Let be a Dirac mass at . We then have

This also explains the minus sign in Exercise 7.

Exercise 9In 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 10Actually, 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 15Let and . Then the convolution is thefunctiongiven 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 theproductof a tempered distribution with is well defined by means of the formula

and .

(ii) If and then show that

Remark 6The 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 16Let , , 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 particular the class of tempered distributions of type is empty whenever .

Exercise 14Prove 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 a *multiplier 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 15If 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 7The previous discussion and in particular Theorem 18 shows that is in fact given in the form

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

In “For the proof of this theorem see [SW],” which book/paper is SW referring to? I would be very interested in taking a look at it. BTW, great post!!!

Alex, The references are defined in the introductory post here (clicky). The book [SW] is “Stein and Weiss, An introduction to Fourier Analysis on Euclidean spaces”.

Thank you for such a quick response!!! Just noticed that email notifications come from “do not reply” email address.

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I stumbled across this while looking for information on band-limited signals. Perhaps this question is fodder for an article.

If I have a function in t, f(t), that has a Fourier transform, F(w), and the Fourier transform is band-limited (i.e there is a finite value, w0, such that F(w)=0 for all |w| > w0), is the signal, f(t), continuous?

My intuitive sense is that the “sharpest edge” that f(t) can have (the suprenum of the derivative of f(t)) is the maximum of the slope available in the maximum frequency component of f(t); i.e. F(w0).

This frequency component would be e^{2 \pi i t * w0} and this maximum slope would be w0*F(w0).

Is there a way to make this intuition rigorous?

Dear John

It´s quite important to clarify in what sense you take the Fourier transform of some function . For example, in the simplest case that say and has compact support, then clearly and the inversion formula is true. Thus is equal to the inverse Fourier transform of everywhere, and thus it is continuous (in fact it is much smoother than that because of the Paley-Wiener theorem).

One can answer this question in much greater generality, that is, without the apriori assumption that (you will still however need some mild assumption on to guarantee that you can embed it in the class of tempered distributions and take the Fourier transform in that weaker sense). Then all the Fourier transforms involved should be considered in the sense of distributions and your assumption would be something like “ is a distribution of compact support”. This case is also covered by a version of the Paley-Wiener theorem.

Hope this helps.

yannis

It may. I thank you for the new direction (the Schwartz-Paley-Wiener Theorem). My Fourier transform is of an almost periodic function from number theory. The transform is thus limited to a spectrum of +-1, but consists of Dirac delta “functions” located at the rational points within the interval. I am not sure the transform is well behaved enough for Schwartz-Paley-Wiener, but perhaps it is sufficient for the Hormander generalization mentioned in the Wikipedia article. It is fortunate for me that the only property I desire to be proved for the underlying signal is that is be continuous for all t greater than some interval about zero. Hopefully the modesty of this goal is enough to overcome the difficulties presented by a Fourier transform that is a distribution and not continuous.

Thank you very much for your consideration and on this matter. After all a new direction is a new adventure in analysis, eh?

I am sorry that last bit should have been the inverse transform of spectrum, wF(w), which is the Fourier transform of the derivative of f(t). But that very statement, “transform of the derivative of f(t)”, already makes presumptions on the continuity of f(t) I was hoping to avoid with the band lmited question. Are there spectral only conditions of F(w) which are sufficient to demonstrate f(t) is continuous?

The Fourier transform of a finite Borel measure is, by definition,

I can’t see the connection with its transform viewed as a tempered distribution. Should it be the same or the definition is made by analogy to functions thinking as the measure’s density?

Great post btw.

Dear Pablo, thanks for your comment.

You’re right, the Fourier transform of a finite Borel measure can be defined directly, without appealing to the theory of distributions. As you commented, one thinks of an function as the density of the Borel measure . Replace this by and you have a perfectly meaningful definition of the Fourier transform on the class of finite Borel measures. One can quite easily check that this definition coincides with the definition given by distribution theory. Indeed we have (by definition)

Thank you so much for your quick answer and your time!