Math 362: Real and Abstract AnalysisCollege of the Holy Cross, Spring 2013

Fourier Transforms

Given a function f : R → R, its Fourier transform is the function

f(ξ) =

∫Rf(x)e−ixξ dx

and its inverse Fourier transform is the function

f(x) =1

2π

∫Rf(x)eixξ dξ

Thought of as an operator, the Fourier transform is denoted by F and the inverse Fouriertransform by F−1. That is, F(f) = f and F−1(f) = f . It should be noted that it is not atall obvious that the second formula really is the inverse of the first. Before proving this, wewill look at some of the basic properties of the Fourier transform. It is helpful to first workwithin a special class of functions called the Schwartz class.

Schwartz Class

Definition 1. A function f is said to be rapidly decreasing if for every integer N ≥ 0there exists a constant CN such that

|x|N |f(x)| ≤ CN

for all x ∈ R.

Definition 2. The Schwartz class S is the set of all functions f ∈ C∞(R) such that f andall of its derivatives are rapidly decreasing.

It is easy to see that the Schwartz class is closed under differentiation and multiplicationby polynomials. Also, since functions in S are bounded and decay faster than any polynomialas |x| → ∞, it follows that Schwartz class functions are integrable, and therefore it makessense to take their Fourier transform.

Example 1. For a > 0, f(x) = e−a|x|2 is in S. Its Fourier transform is given by

f(ξ) =

∫ ∞

−∞e−ax2

e−ixξ dx

=

∫ ∞

−∞e−a[(x+iξ/2a)2+ξ2/4a2] dx

= e−ξ2/4a

∫ ∞

−∞e−a(x+iξ/2a)2 dx

In the remaining integral, we substitute z = x+ iξ/2a to get∫ ∞

−∞e−a(x+iξ/2a)2 dx =

∫ ∞+iξ/2a

−∞+iξ/2a

e−az2 dz

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This integral is actually a line integral over the line z = x + iξ/2a in the complex plane.Using methods from complex analysis, one can show that it is equivalent to the integral overthe real axis: ∫ ∞

−∞e−az2 dz.

Finally, making the change of variable z = y/√a this becomes

1√a

∫ ∞

−∞e−y2 dy =

√π

a,

so

f(ξ) =

√π

ae−ξ2/4a.

The following theorem is the most important algebraic property of Fourier transforms.

Theorem 1. If f ∈ S then f ∈ S and

df

dx= iξf

xf = idf

dξ

for 1 ≤ k ≤ n.

So differentiation of f corresponds to multiplication of f by a polynomial, and converselymultiplication of f by a polynomial corresponds to differentiation of f .

Proof. By the integrability of Schwartz class functions, the following calculations are justi-fied. Integration by parts in x gives

df

dx(ξ) =

∫R

df

dxe−ixξ dx =

∫Riξf(x)e−ixξ dx = iξf(ξ)

while differentiation with respect to ξ gives

df

dξ(ξ) =

∫R−ixf(x)e−ix·ξ dx = −ixf(ξ)

This proves the two formulas.To prove that f ∈ S implies f ∈ S, first notice that f ∈ S implies f is bounded, since f

is integrable. Next, sinceξf = −ifx

and since fx ∈ S it follows that ξf is also bounded. By induction it follows that

|ξ|N |f(ξ)|

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is bounded for any positive integer N , and hence f is rapidly decreasing. So we have shownthat f ∈ S implies f is rapidly decreasing. Finally, since

fξ = −ixf

and since xf ∈ S it follows from what was just proven that fξ is rapidly decreasing. By

induction it follows that all derivatives of f exist and are rapidly decreasing.

Theorem 2. (Fourier Inversion Theorem) Let f ∈ S. Then ˇf = f .

Proof. For n ∈ N, define

In(x) =1

2π

∫ ∞

−∞f(ξ)e−ξ2/n2+ixξ dξ

Since the integrand converges pointwise to f(ξ)eixξ as n → ∞ and since

|f(ξ)e−ξ2/n2+ixξ| ≤ |f(ξ)|

and |f(ξ)| is integrable since f ∈ S, it follows from the Dominated Convergence Theoremthat

limn→∞

In(x) =1

2π

∫ ∞

−∞f(ξ)eixξ dξ =

ˇf(x).

On the other hand,

In(x) =1

2π

∫ ∞

−∞

∫ ∞

−∞f(y)e−i(y−x)ξe−ξ2/n2

dydξ

Since |f(y)|e−ξ2/n2is integrable over R× R, we can apply Fubini’s theorem to get

In(x) =1

2π

∫ ∞

−∞f(y)

(∫ ∞

−∞e−i(y−x)ξe−ξ2/n2

dξ

)dy

The term in parentheses is the Fourier transform of e−ξ2/n2evaluated at y − x, so applying

Example 1 with a = ϵ, it equals √n2πe−n2(y−x)2/4,

and thus

In(x) =

∫ ∞

−∞f(y)

n√4π

e−n2(y−x)2/4dy.

Setting

gn(x) =n√4π

e−n2x2/4 =n

2ϕ(nx/2)

where ϕ(x) = 1√πe−x2

, it follows that gn is an approximate identity and therefore

In = gn ∗ f → f

pointwise as n → ∞. Thusˇf(x) = lim

n→∞In(x) = f(x).

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Fourier Transforms and Convolution

Theorem 3. Let f, g ∈ S. Then

F(f ∗ g) = F(f)F(g)

F(fg) =1

2πF(f) ∗ F(g)

Proof. The first formula was assigned for homework. For the second formula, we first notethat F−1f(y) = 1

2πFf(−y). So for any u, v ∈ S, we have

F−1(u ∗ v)(ξ) = 1

2πF(u ∗ v)(−ξ)

=1

2πFu(−ξ)Fv(−ξ)

= 2πF−1u(ξ)F−1v(ξ).

Letting u = Ff and v = Fg and using the Fourier inversion theorem, this becomes

F−1(Ff ∗ Fg)(ξ) = 2πf(ξ)g(ξ)

Applying the inversion theorem once more gives

F(fg) =1

2πFf ∗ Fg,

as desired.

Here is an application of some of these results.

Example 2. Consider the heat equation in one dimension,

ut = kuxx

u(x, 0) = ϕ(x)

Take the Fourier transform of both sides with respect to x, and use Theorem 1:

ut = kuxx = k(iξ)(iξ)u = −kξ2u

For each fixed ξ, solve the differential equation ut = −kξ2u to get

u(ξ, t) = C(ξ)e−kξ2t

To determine C(ξ), take the Fourier transform of the initial condition:

u(ξ, 0) = ϕ(ξ)

Thus u(ξ, t) = ϕ(ξ)e−kξ2t. Now using the formula in Example 1, with a = 1/4kt gives

e−x2/4kt =√4πkte−kξ2t =⇒ e−kξ2t =

1

4πkte−x2/4kt

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Thus if we define

S(x, t) =1

4πkte−x2/4kt

thenu = ϕS = ϕ ∗ S

so the Fourier inversion theorem implies

u = ϕ ∗ S

So the solution of the heat equation is simply the initial data ϕ convolved with the heatkernel S. (See the handout on convolution.)

L2 Properties

The Fourier transform behaves very nicely with respect to L2.

Lemma 1. For f, g ∈ S, ⟨f , g⟩ = 2π ⟨f, g⟩.

Proof. By direct calculation

⟨f , g⟩ =∫Rf(ξ)g(ξ) dξ

=

∫R

∫Rf(x)g(ξ)e−ixξ dx dξ

=

∫R

∫Rf(x)g(ξ)e−ixξ dξ dx

=

∫Rf(x)

∫Rg(ξ)eix·ξ dξ dx

=

∫Rf(x)2πg(x) dx

= 2π ⟨f, g⟩

where once again the integrability of f and g justifies changing the order of integration.

Theorem 4. (Plancherel’s Formula) Let f, g ∈ S. Then

⟨f , g⟩ = 2π⟨f, g⟩

∥f∥L2 = 2π∥f∥L2

Proof. The second formula follows from the first by setting g = f , while the first followsfrom the previous lemma by replacing g with g and using the inversion theorem.

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Remark. Plancherel’s formula seems to indicate that the Fourier transform is bounded asan operator from L2 to L2. However, one can find functions in f ∈ L2 for which the integralwhich defines f is not an L2 function. One must therefore be careful when thinking of F asan operator on L2.

Extension to L2

Plancherel’s formula allows us to extend the Fourier transform to an operator F : L2 →L2. The proof relies on the following facts:

1. L2 is complete.

2. S is dense in L2.

To extend F to all of L2, use the density of S to approximate elements of L2 by elementsof S, and take a limit. More precisely, let f ∈ L2. Since S is dense, there exists a sequencefk in S such that

limk→∞

∥fk − f∥L2 = 0

Since fk is convergent, it is a Cauchy sequence. By Plancherel’s formula it follows that thesequence F(fk) is also a Cauchy sequence. By completeness of L2 there exists a limit. Callthis limit F(f). Then

∥F(f)∥L2 = limk→∞

∥F(fk)∥L2 = limk→∞

2π∥fk∥L2 = 2π∥f∥L2

so Plancherel’s formula still holds for the extension.

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