Let . Then the Fourier transform of is defined as follows:
We recall that is integrable is integrable.
Now we're ready to prove the next theorem:
Theorem 8.2: The Fourier transform of an integrable is well-defined.
Proof: Since is integrable, lemma 8.2 tells us that is integrable. But
, and therefore is integrable. But then, is integrable, which is why
has a unique complex value, by definition of integrability.
Theorem 8.3: Let . Then the Fourier transform of , , is bounded.
Once we have calculated the Fourier transform of a function , we can easily find the Fourier transforms of some functions similar to . The following calculation rules show examples how you can do this. But just before we state the calculation rules, we recall a definition from chapter 2, namely the power of a vector to a multiindex, because it is needed in the last calculation rule.
For a vector and a -dimensional multiindex we define , to the power of , as follows:
Now we write down the calculation rules, using the following notation:
to mean the sentence 'the function is the Fourier transform of the function '.
Let be the Fourier transform of . Then the following calculation rules hold:
If additionally is the Fourier transform of , we have
Proof: To prove the first rule, we only need one of the rules for the exponential function (and the symmetry of the standard dot product):
For the next two rules, we apply the general integration by substitution rule, using the diffeomorphisms and , which are bijections from to itself.
In order to proceed with further rules for the Fourier transform which involve Schwartz functions, we first need some further properties of Schwartz functions.
Let be a Schwartz function and let . Then the function
is a Schwartz function as well.
Let . Due to the general product rule, we have:
We note that for all and , equals to to some multiindex power. Since is a Schwartz function, there exist constants such that:
Hence, the triangle inequality for implies:
Every Schwartz function is integrable.
We use that if the absolute value of a function is almost everywhere smaller than the value of an integrable function, then the first function is integrable.
Let be a Schwartz function. Then there exist such that for all :
The latter function is integrable, and integrability of follows.
Now we can prove all three of the following rules for the Fourier transform involving Schwartz functions.
If is the Fourier transform of a function in the Schwartz space , in addition to the rules in theorem 8.4, also the following rules hold:
If additionally is the Fourier transform of a , then
For the first rule, we use induction over .
It is clear that the claim is true for (then the rule states that the Fourier transform of is the Fourier transform of ).
We proceed to the induction step: Let , and assume that the claim is true for all such that . Let such that . We show that the claim is also true for .
Remember that we have . We choose such that (this is possible since otherwise ), define
by Schwarz' theorem, which implies that one may interchange the order of partial derivation arbitrarily.
Let be an arbitrary positive real number. From Fubini's theorem and integration by parts, we obtain:
Due to the dominated convergence theorem (with dominating function ), the integral on the left hand side of this equation converges to
as . Further, since is a Schwartz function, there are such that:
Hence, the function within the large parentheses in the right hand sinde of the last line of the last equation is dominated by the function
and hence, by the dominated convergence theorem, the integral over that function converges, as , to:
From the uniqueness of limits of real sequences we obtain 1.
We use again induction on , note that the claim is trivially true for , assume that the claim is true for all such that , choose such that and such that and define .
Theorems 8.6 and 8.7 imply that
for all , and
for all , .
exists for all .
Hence, Leibniz' integral rule implies:
Let , and let be the Fourier transform of . Then .
Let be two arbitrary -dimensional multiindices, and let . By theorem 8.6 is a Schwartz function as well. Theorem 8.8 implies:
By theorem 8.3, is bounded. Since were arbitrary, this shows that .
We define the Fourier transform on the Schwartz space to be the function
Theorem 8.9 assures that this function really maps to . Furthermore, we define the inverse Fourier transform on the Schwartz space to be the function
This function maps to since .
Both the Fourier transform and the inverse Fourier transform are sequentially continuous:
Let and let be a sequence of Schwartz functions such that . Then and , both in the sense of Schwartz function convergence as defined in definition 3.11.
1. We prove .
Let . Due to theorem 8.8 1. and 2. and the linearity of derivatives, integrals and multiplication, we have
As in the proof of theorem 8.3, we hence obtain
Due to the multi-dimensional product rule,
Let now be arbitrary. Since as defined in definition 3.11, for each we may choose such that
Further, we may choose such that
Hence follows for :
Since was arbitrary, we obtain .
2. From 1., we deduce .
If in the sense of Schwartz functions, then also in the sense of Schwartz functions, where we define
Therefore, by 1. and integration by substitution using the diffeomorphism , .
In the next theorem, we prove that is the inverse function of the Fourier transform. But for the proof of that theorem (which will be a bit long, and hence to read it will be a very good exercise), we need another two lemmas:
If we define the function
then and .
By the product rule, we have for all
Due to 1. of theorem 8.8, we have
from 2. of theorem 8.8 we further obtain
Hence, is constant. Further,
By substitution using the diffeomorphism ,
For the next lemma, we need example 3.4 again, which is why we restate it:
Example 3.4: The standard mollifier, given by
, where , is a bump function (see exercise 3.2).
Let , and for each define . Then in the sense of Schwartz functions.
Let be arbitrary. Due to the generalised product rule,
By the triangle inequality, we may hence deduce
Since both and are Schwartz functions (see exercise 3.2 and theorem 3.9), for each we may choose such that
Further, for each , we may choose such that
Let now be arbitrary. We choose such that for all
Further, we choose such that
This is possible since
due to our choice of .
Then we choose such that for all
Inserting all this in the above equation gives for . Since , and were arbitrary, this proves in the sense of Schwartz functions.
Let . Then and .
1. We prove that if is a Schwartz function vanishing at the origin (i. e. ), then .
So let be a Schwartz function vanishing at the origin. By the fundamental theorem of calculus, the multi-dimensional chain rule and the linearity of the integral, we have
and multiplying both sides of the above equation by , we obtain
Since by repeated application of Leibniz' integral rule for all
all the are bump functions (due to theorem 4.15 and exercise 3.?), and hence Schwartz functions (theorem 3.9). Hence, by theorem 8.8 and the linearity of the Fourier transform (which follows from the linearity of the integral),
Let . By Fubini's theorem, the fundamental theorem of calculus and since is a bump function, we have
If we let , theorem 8.11 and lemma 8.13 give the claim.
2. We deduce from 1. that if is an arbitrary Schwartz function, then .
As in lemma 8.12, we define
Let now be any Schwartz function. Then is also a Schwartz function (see exercise 3.?). Further, since , it vanishes at the origin. Hence, by 1.,
Further, due to lemma 8.12 and the linearity of the Fourier transform,
3. We deduce from 2. that if is a Schwartz function and is arbitrary, then (i. e. .
Let and be arbitrary. Due to the definition of ,
Further, if we define ,
Hence, by 2.,
4. We deduce from 3. that for any Schwartz function we have .
Let and be arbitrary. Then we have
The Fourier transform of tempered distributionsEdit
Let be a tempered distribution. We define
is a tempered distribution.
1. Sequential continuity follows from the sequential continuity of and (theorem 8.11) and that the composition of two sequentially continuous functions is sequentially continuous again.
2. Linearity follows from the linearity of and and that the composition of two linear functions is linear again.