Lecture Note Supplements. Math 671 Fourier

Analysis

Teck-Cheong LimDepartment of Mathematical Sciences

George Mason University4400, University DriveFairfax, VA 22030

U.S.A.e-mail address: tlim@gmu.edu

1. Another formula for Fejer kernel is

FN (x) =

N−1∑n=−(N−1)

(1− |n|

N

)einx

= 1 + 2

N−1∑n=1

(1− n

N

)cos(nx)

Proof.Write ω = eix. Recall that NFN (x) = D0(x) + · · · + DN−1(x) whereDn(x) =

∑nj=−n ω

j is the Dirichlet kernel. Then

NFN (x)

=

N−1∑n=0

n∑j=−n

ωj

=

N−1∑j=−(N−1)

N−1∑n=|j|

1 · ωj

=

N−1∑j=−(N−1)

ωjN−1∑n=|j|

1

=

N−1∑j=−(N−1)

ωj(N − |j|).

Using 2 cos y = eiy + e−iy, we get the second part.

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2. (Extension of Theorem 2.1) Let f be an integrable function on the circle.

Suppose at a point t0, f(t−0 ), f(t+0 ), limt→0+f(t0+t)−f(t+0 )

t and limt→0−f(t0+t)−f(t−0 )

t

all exist (and are finite). Then SN (f)(t0)→ 12 (f(t−0 ) + f(t+0 )) as N tends

to infinity.Proof.Define

F (t) =

{f(t0−t)−f(t+0 )

t −π ≤ t < 0f(t0−t)−f(t−0 )

t 0 < t ≤ π

(We may define F (0) to be any number.)It follows from our assumption that F (t) is integrable.[In Riemann sense,if we define Dg to be the set of points of discontinuity of g, then F isbounded and DF ⊂ Df ∪ {t0}. Since Df is a set of measure 0, so is DF .Thus F is Riemann integrable. In Lebesgue sense, F is bounded on asmall interval I = (t0 − ε, t0 + ε), and outside of I, 1/|t| is bounded. SoF = FχI + FχI′ is integrable since both FχI and FχI′ are.] Recall that

1

2π

∫ 0

−πDN (t) dt =

1

2π

∫ π

0

DN (t) dt =1

2.

It follows that

SN (f)(t0)− 1

2(f(t+0 ) + f(t−0 ))

=1

2π

∫ π

−πf(t0 − t)DN (t) dt− 1

2(f(t+0 ) + f(t−0 ))

=1

2π

∫ 0

−π(f(t0 − t)− f(t+0 ))DN (t) dt+

1

2π

∫ π

0

(f(t0 − t)− f(t+0 ))DN (t) dt

=1

2π

∫ π

−πtF (t)DN (t) dt

=1

2π

∫ π

−π

t

sin(t/2)F (t) sin(N +

1

2)t dt

which approaches to 0 as N → ∞ by Riemann-Lebesgue lemma (see theproof of Theorem 2.1). Q.E.D.

3. Corollary of Lemma 1.5, Chapter 3, p. 81:

1

2π

∫ π

−πf(x)g(x)e−inx dx = f · g(n) = (f ∗ g)(n) =

∞∑k=−∞

f(k)g(n− k).

Proof.Apply Lemma 1.5 to the function G(x) = g(x)einx.

4. Prove Corollary 1.10 p.142:The Fourier transform is a bijective mapping on the Schwartz space.

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Proof.If F(f) = 0, then by Theorem 1.19 (Fourier Inversion theorem) f = 0.Hence the (linear) map F is one to one. Suppose now g is a function inS(R). Let

f(x) = F∗(g)(x) =

∫ ∞−∞

g(ξ)e2πiξx dξ.

Note that f(x) = F(g)(−x) = g(−x). Therefore

F(f)(ξ) =

∫ ∞−∞

g(−x)e−2πiξx dx

=

∫ ∞−∞

g(x)e2πiξx dx

= g(ξ)

where we made a change of variable and used Fourier inversion theorem.This proves that the Fourier transform is onto. This proof also shows thatF∗ is the inverse of F .

5. Prove formula (4) on p. 297:Let f be a continuous function defined on Sd−1 and R a rotation of Rd.Prove that ∫

Sd−1

f(R(γ))dσ(γ) =

∫Sd−1

f(γ)dσ(γ).

Proof.Define g on Rd by

g(x) =

{f(x/|x|)|x|e−|x|d+1/(d+1) for x 6= 00 for x = 0

g is continuous and of moderate decrease. Thus∫Rd

g(R(x)) dx =

∫Rd

g(x) dx

by the equation on p, 297 preceding equation (4). Now∫Rd

g(R(x)) dx =

∫Sd−1

∫ ∞0

g(rR(γ))rd−1 dr dσ(γ)

=

∫Sd−1

∫ ∞0

f(R(γ))rde−rd+1/(d+1) dr dσ(γ)

=

∫Sd−1

f(R(γ)) dσ(γ)

because ∫ ∞0

rde−rd+1/(d+1) dr = 1.

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Similarly ∫Rd

g(x) dx =

∫Sd−1

f(γ) dσ(γ).

Note that we could have let g(x) = h(r)f(γ), where h(r) is any func-tion of moderate decrease such that rd−1h(r) is integrable on (0,∞) and∫∞0rd−1h(r) dr 6= 0.

6. Prove supx |x|k|g(x− y)| ≤ Ak(1 + |y|)k in the proof of Proposition 1.11,Chapter 5.Proof.By change of variable x = y + t, we need to bound the expression(

|y + t|1 + |y|

)k|g(t)|.

Since|y + t|1 + |y|

≤ |y|+ |t|1 + |y|

≤ 1 + |t|,

and(1 + |t|)k

1 + t2k

and(1 + t2k)|g(t)|

are bounded, say by Bk and Ck, we find that(|y + t|1 + |y|

)k|g(t)| ≤ (1 + |t|)k|g(t)| ≤ BkCk.

7. Prove (f ∗ g)(ξ) = f(ξ)g(ξ) in Proposition 1.11.Proof.

(f ∗ g)(ξ)

=

∫ ∞−∞

(f ∗ g)(x)e−2πixξ dx

=

∫ ∞−∞

∫ ∞−∞

f(y)g(x− y)e−2πixξ dy dx

=

∫ ∞−∞

∫ ∞−∞

f(y)g(x− y)e−2πixξ dx dy( by Fubini’s theorem)

=

∫ ∞−∞

f(y)g(ξ)e−2πiyξ dy( by Proposition 1.2 (ii))

= g(ξ)

∫ ∞−∞

f(y)e−2πiyξ dy

= g(ξ)f(ξ)

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8. We may add the following to the list of Fourier transforms in Proposition1.2, Chapter 5.

f(−x) −→ f(−ξ)f(x) −→ f(−ξ)f(−x) −→ f(ξ)

f(x) −→ f(−ξ)f(x) −→ f(ξ)

f(−x) −→ f(ξ)

f(h− x) −→ f(ξ)e−2πihξ

9. We may add the following to the list of Proposition 1.11, Chapter 5, p.142.

(iv) f · g(ξ) = (f ∗ g)(ξ).Proof. Apply the above formula g(h− x) −→ g(ξ)e−2πihξ to Proposition1.8 (multiplication formula).

10. Prove Theorem 1.12 (Plancherel), Chapter 5.Proof. Let g(x) = f(−x). Then∫ ∞

−∞|f(ξ)|2 dξ

=

∫ ∞−∞

f(ξ)f(ξ) dξ

=

∫ ∞−∞

f(ξ)g(ξ) dξ( by item above )

=

∫ ∞−∞

(f ∗ g)(ξ) dξ( by Proposition 1.11 (iii))

= f ∗ g(0)( by Fourier inversion theorem with x = 0)

=

∫ ∞−∞

f(x)g(0− x) dx

=

∫ ∞−∞

f(x)f(x) dx

=

∫ ∞−∞|f(x)|2 dx

11. If f, g ∈ S(R), then (f, g) = (f , g). (This generalizes Theorem 1.12(Plancherel).)Proof. Apply the formula g(x) −→ g(ξ) to Proposition 1.8 (multiplica-tion formula).

12. (A variant of Poisson Summation Formula) Suppose f ∈ S(R), and f beits Fourier transform. Then

∞∑n=−∞

f(x+ n) =

∞∑n=−∞

f(n)e−2πinx.

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Proof.We first prove that the Fourier transform of f(x) is f(−ξ):∫ ∞

−∞f(x)e−2πiµx dx =

∫ ∞−∞

f(ξ)e2πi(−µ)ξ dξ

= f(−µ)( by the Fourier inversion formula )

So by Poisson summation formula, we have

∞∑n=−∞

f(x+ n) =

∞∑n=−∞

f(−n)e2πinx =

∞∑n=−∞

f(n)e−2πinx.

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