Lecture 11.Fourier transform

1 Definition and main resultsLet f ∈ L2(R). The Fourier transform of a function f is a function

f̃(α) =

∫R

f(x)t−iαxdx (1)

The normalized Fourier transform of f is a function

f̂ =1√2π

f̃ (2)

It is not obvious that the Fourier transform exists for any function f ∈ L2R. This isproved below.

The main results of this lecture are:

Theorem 1 (Plancherel equality) f ∈ L2(R) ⇒ f̂ ∈ L2(R), and ||f || = ||f̂ ||.

Theorem 2 [Inversion Formula] ∀f ∈ L2(R),

f(x) =1

2π

∫R

f̃(α)eiαxdα

We will prove these theorems for the case when f is a C2 function with a finite support,and then extend it to the whole of L2(R).

The Fourier transform is a continuous analog of the Fourier series. The Fourier seriesare defined on a circle, and the Fourier transform is defined on the real line. The latter isobtained from the first one by passing to the limit when the circle length tends to infinity.

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2 Decreasing of the Fourier transform for a smooth func-tion

Theorem 3 Fourier transform of a Cm function with a compact support decreases noslower than |x|−m.

The proof is quite the same as for the Fourier series.Proof The Fourier transform brings the derivation to the multiplication by iα:

f̃ ′(α) = iαf̃(α). (3)

This formula is proved by the integration by parts. Denoting the Fourier transform byF , we have:

F(f ′)(α) =

∫f ′(x)e−iαxdx =

∫e−iαxdf(x) = −

∫f(x)de−iαxdx = iα

∫f(x)e−iαxdx = iαFf(α).

This implies

Ff(α) =1

iαF(f ′)(α).

By induction in M we get:

Ff(α) =1

(iα)mF(f (m))(α).

But the function F(f (m)) is bounded because f (m) is continuous with the finite support. ¤

Corollary 1 If f ∈ C2,0, then there exists a C such that:

|f̃(α)| <C

1 + α2. (4)

3 Fourier series on a “long” circleThe circle may be represented by a segment with the endpoints identified. Another repre-sentation: the functions on a circle of “length T ” are represented by T -periodic functionson the real line. We will use this latter representation.

Consider a 2πl-periodic function f : R → R and decompose it to the Fourier series withrespect to a system that we will now construct. Consider a 2π-periodic function g(x) = f(xl)and decompose it into a classical Fourier series:

g(x) = Σk∈Zgkeikx. (5)

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The factors ik in the exponential ar called the wave numbers. For the classical Fourier seriesthe set of wave numbers coincides with Z.

By definition of g,

f(x) = g(x

l

)= Σgke

ik xl = Σα∈ Z

lcae

iαx (6)

The wave numbers in the latter series range over the set Zl. For the large l this set is much

thicker than Z. Somewhat loosely we may say that as l → ∞ this set tends to R.Equality (6) implies that the vectors eiαx, a ∈ Z

lform a basis in L2([−πl, πl]). Their

norms equal√

2πl. Hence, for α ∈ Zl,

cα =1

2πl

∫ πl

−πl

f(x)e−iαxdx

Note that for α ∈ Zl,

cα =1

2πlF(fχ[−πl,πl])(α) (7)

The Plancherel equality for f has the form

||f ||2 = 2πl Σα∈ Zl|cα|2. (8)

The Fourier decomposition of f is given by (6), (7).

4 Passing to the limit: heuristic proofs of the Plancherelequality and Inversion Formula

Fix a function f ∈ C2,0 with the support supp f ⊂ [−πl0, πl0], and for any l > l0 con-sider the restriction of f to the segment [−πl, πl].Formally, these restrictions are differentfunctions, but we will denote them all by f .

Let us prove the Plancherel equality:

||f || = ||f̂ ||.

It is sufficient to prove that

||f ||2 =1

2π||f̃ ||2 (9)

Let us extend the function f 2πl-periodically, end decompose the resulting function to theFourier series. By (8) and (7),

||f ||2 =1

2πlΣα∈ Z

l|f̃(α)|2 := Σl. (10)

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The expression Σl is an integral sum for the integral

1

2π||f̃ ||2 =

1

2π

∫|f̃(α)|2dα := I

This sum corresponds to the partition of the line by the segments of length 1l

with theendpoints in the set Z

l. On one hand, this sum tends to the integral (this should be proved!);

on the other hand, the sequence Σl is stationary (does not depend on l). This “proves” (9).In the same way, the Inversion Formula may be proved. By (6), for |x| < l,

f(x) =1

2πlΣα∈ Z

lf̃(α)eiαx := Sl.

Sl is an integral sum for the integral

I(x) =1

2π

∫f̃(α)eiαxdα.

Passing to the limit as above (this taking of the limit should be justified), we obtain theInversion Formula:

f(x) =1

2π

∫R

f̃(α)eiαxdα (11)

5 Formal proof

Lemma 1 Sl −−−→l→∞

∫R |f̃(α)|2dα provided that: f ∈ C2,0.

Lemma 2 If C2,0, then Σl(x) → I(x) for l → ∞.

The previous arguments and Lemma 1 imply the Plancherel equality, and Lemma 2implies the Inversion Formula.Proof [of Lemma 1]. If the integral I have been proper, the integral sums would tend tothe integral by the classical theorem about Riemann integrals. We need to overcome animproper integral. This is done by the estimates below.

By Corollary 1, there exists C > 0 : |f̃(α)| < C(1 + α2)−1. Take ε > 0 and such N that

1

lΣα∈ Z

l\[−N,N ]|f̃(α)|2 <

1

lΣα∈ Z

l\[−N,N ]

C

(1 + α2)2 < C

∫|α|≥N− 1

l

dα

(1 + α2)2 ≤ ε

3.

Then ∫|α|≥N

|f(α)|2dα <ε

3.

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Take l so large that ∣∣∣∣1l Σα∈ Zl|, α|≤N |f̃(α)|2 −

∫|α|≤N

|f̃(a)|2dx

∣∣∣∣ <ε

3.

Then∣∣∣Sl − ||f̃ ||2

∣∣∣ < ε. ¤

Lemma 2 is proved in the same way.Conclusion. We have proved Theorems 1 and 2: the Plancherel equality and Inversion

Formula for smooth functions f with the compact support.The goal: prove the same for f ∈ L2.

6 Operators and their extensionsDefinition 1 A map A : H → H is a linear operator, provided that A(αξ+βη) = αA(ξ)+βA(η).

Definition 2 A map A is an isometry provided that ||Aξ|| = ||ξ|| ∀ξ ∈ H; by default, Ais linear.

Theorem 4 Let E be a dense subset of H, A : E → E ′ ⊂ H be an isometry. Then A maybe extended to H up to an isometry A : H → H.

Proof Let x ∈ H, (xr) ⊂ E, xn → x for n → ∞. Letyn = Axn. Then (xn) is a Cauchysequence ⇒ (yn) is a Cauchy sequence as well. Let y = limn→∞yn . Set A(x) = y.

Exercise 1 The map A is well defined: y depends on x only, and not on xn → x.

The map A is an isometry because ||yn|| = ||xn|| ⇒ ||y|| = ||x||. ¤Therefore, the Fourier transform may be extended to the whole of L2(R) as an isometry;

that is, on the whole os this space the Plancherel equality holds.The Inversion Formula is extended to the space L2(R) in the same way. Namely, let,

S : f(x) 7→ f(−x) be an operator of the reversion of the indeterminate. The InversionFormula is equivalent to the following one:

1

2πF2 = S.

The operators in both sides of this equation are the isometries. Their coincidence on adense set C2,0 implies the same on the whole space L2(R).

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7 Case of many variablesIn case of many variables the formula for the Fourier transform has the same form (1) asfor one variable, with the use of the multiindex notations. The formula for the normalizedFourier transform is:

f̂ =1√

(2π)nf̃ (12)

The Plancherel equality for f ∈ L2(Rn) is quite the same as for n = 1. The inversionformula obtains a different factor

f(x) =1

(2π)n

∫Rn

f̃(α)eiαxdα.

All this is proved in the very same way as for n = 1. We will use these facts, but willnot repeat the proofs.

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