The Poisson Summation Formula

3.1 Fourier Coefficients

3.1.5 The Poisson Summation Formula

(5) For all k∈Zⁿwe have cf g(k) =

∑

m∈Zⁿ

bf(m)bg(k−m) =

∑

m∈Zⁿ

bf(k−m)bg(m).

Proof. (1) and (2) follow from the corresponding statements in Proposition 3.1.15.

Parseval’s relation (3) follows from polarization. First replace f by f+g in (1) and expand the squares. We obtain that the real parts of the expressions in (3) are equal. Next replace f by f+igin (1) and expand the squares. We obtain that the imaginary parts of the expressions in (3) are equal. Thus (3) holds. Next we prove (4). We already know that the map f 7→ {bf(m)}m∈Zⁿ is an injective isometry. It remains to show that it is onto. Given a square summable sequence{a_m}_m∈Zn of complex numbers, define

f_N(t) =

∑

|m|≤N

a_me^2πim·t.

Observe that f_N is a Cauchy sequence inL²(Tⁿ)and it therefore converges to some f ∈L²(Tⁿ). Then we have bf(m) =a_mfor allm∈Zⁿ. Finally, (5) is a consequence

of (3) and Proposition 3.1.2 (6) and (3).

Proof. Since bf is integrable onRⁿ, inversion holds and f can be identified with a continuous function. Define a 1-periodic function onTⁿby setting

F(x) =

∑

m∈Zⁿ

f(x+m).

It is straightforward to verify thatF∈L¹(Tⁿ). The calculation F(m) =b

Z

Tⁿ

F(x)e^−2πim·xdx=

∑

k∈Zⁿ Z

[−¹₂,¹₂]ⁿ−k

f(x)e^−2πim·xdx=bf(m) gives that the sequence of the Fourier coefficients ofFcoincides with the restriction of the Fourier transform of f onZⁿ. Since we have that

m∈Z

∑

ⁿ

|F(m)|b =

∑

m∈Zⁿ

|bf(m)| ≤C

∑

m∈Zⁿ

1

(1+|m|)^n+δ <∞,

Proposition 3.1.14 implies conclusion (3.1.19).

Example 3.1.18.We have seen earlier (see Exercise 2.2.11) that the following iden- tity gives the Fourier transform of the Poisson kernel inRⁿ:

(e^−2π|x|)b(ξ) =Γ(ⁿ⁺¹₂ ) πⁿ⁺¹²

1 (1+|ξ|²)ⁿ⁺¹²

.

The Poisson summation formula yields the identity Γ(ⁿ⁺¹₂ )

π

n+1

2

∑

k∈Zⁿ

ε (ε²+|k+x|²)ⁿ⁺¹²

=

∑

k∈Zⁿ

e^{−2π ε|k|}e^−2πik·x. (3.1.20) It follows that

k∈Z

∑

ⁿ\{0}

1 (ε²+|k|²)ⁿ⁺¹²

=1 ε

π

n+1 2

Γ(ⁿ⁺¹₂ )

∑

k∈Zⁿ

e^{−2π ε|k|}− 1 εⁿ

,

from which we obtain the identity

∑

k∈Zⁿ\{0}

1

|k|ⁿ⁺¹=lim

ε→0

1 ε

πⁿ⁺¹² Γ(ⁿ⁺¹₂ )

∑

k∈Zⁿ

e^{−2π ε|k|}− 1 εⁿ

. (3.1.21)

The limit in (3.1.21) can be calculated easily in dimension 1, since the sum inside the parentheses in (3.1.21) is a geometric series. Carrying out the calculation, we obtain

k6=0

∑

1 k²=π²

3 .

Example 3.1.19.Letη andg be as in Example 2.4.9. Let 0<Reα <n and let x∈[−¹₂,¹₂)ⁿ. The Poisson summation formula gives

∑

m∈Zⁿ\{0}

e^2πim·x

|m|^α =

∑

m∈Zⁿ

η(m)e^2πim·x

|m|^α =g(x) +

∑

m∈Zⁿ\{0}

g(x+m).

It was shown in Example 2.4.9 thatg(ξ)decays faster than the reciprocal of any polynomial at infinity and is equal toπ^α−n2Γ(^n−α₂ )Γ(^α₂)⁻¹|ξ|^α−n+h(ξ), whereh is a smooth function onRⁿ. Then, forx∈[−¹₂,¹₂)ⁿ, the function

m∈Z

∑

ⁿ\{0}

g(x+m) is also smooth, and we conclude that

∑

m∈Zⁿ\{0}

e^2πim·x

|m|^α =π^α−n2Γ(^n−α₂ )

Γ(^α₂) |x|^α−n+h1(x), whereh₁(x)is aC^∞function on[−¹₂,¹₂)ⁿ.

For other applications of the Poisson summation formula related to lattice points, see Exercises 3.1.12 and 3.1.13.

Exercises

3.1.1.LetPbe a trigonometric polynomial onTⁿ. (a) Prove thatP(x) =∑P(m)eb ^2πim·x.

(b) Let f be inL¹(Tⁿ). Prove that(f∗P)(x) =∑P(m)b bf(m)e^2πim·x.

3.1.2.OnT¹letPbe a trigonometric polynomial of degreeN>0. Show thatPhas at most 2Nzeros. Construct a trigonometric polynomial with exactly 2Nzeros.

3.1.3.Prove the identities (3.1.15), (3.1.16), and (3.1.17) about the Fej´er kernel F(n,N)onTⁿ. Deduce from them that the family{F_N}_Nis an approximate identity asN→∞.

Hint:Express the functions sin²(πx)and sin²(π(N+1)x)in terms of exponentials.

3.1.4.(de la Vall´ee Poussin kernel). OnT¹define V_N(x) =2F2N+1(x)−F_N(x).

(a) Show that the sequenceVN is an approximate identity.

(b) Prove thatVcN(m) =1 when|m| ≤N+1, andVcN(m) =0 when|m| ≥2N+2.

3.1.5.(Hausdorff–Young inequality) Prove that when f ∈L^p, 1≤p≤2, the se- quence of Fourier coefficients of f is inl^p0 and

m∈Z

∑

ⁿ

|bf(m)|^p01/p⁰

≤ f

L^p.

Also observe that 1 is the best constant in the preceding inequality.

3.1.6.Use without proof that there exists a constantC>0 such that for allt∈R we have

N k=2

∑

e^iklogke^ikt

≤C√

N, N=2,3,4, . . . , to construct an example of a continuous functiongonT¹with

m∈Z

∑

|g(m)|b ^q=∞

for allq<2. Thus the Hausdorff–Young inequality of Exercise 3.1.5 fails forp>2.

Hint:Considerg(x) =∑^∞_k=2 ^e

iklogk

k^1/2(logk)²e^2πikx. For a proof of the previous estimate, see Zygmund [303, Theorem (4.7) p. 199].

3.1.7.The Poisson kernel onTⁿis the function P_r1,...,r_n(x) =

∑

m∈Zⁿ

r₁^|m1|· · ·r^|mn^n|e^2πim·x

and is defined for 0<r₁, . . . ,r_n<1. Prove thatP_r1,...,r_ncan be written as P_r1,...,r_n(x₁, . . . ,x_n) =

n

∏

j=1

Re

1+r_je^2πixj 1−r_je^2πixj

=

n

∏

j=1

1−r²_j

1−2rjcos(2πx_j) +r²_j , and conclude thatPr,...,r(x)is an approximate identity asr↑1.

3.1.8.LetDN=D(1,N)be the Dirichlet kernel onT¹. Prove that 4

π²

N k=1

∑

1 k ≤

D_N

L¹≤2+π 4 + 4

π²

N k=1

∑

1 k. Conclude that the numbers

D_N

L¹ grow logarithmically asN→∞and therefore the family {D_N}_N is not an approximate identity on T¹. The numbers

D_{N L}1, N=1,2, . . ., are called theLebesgue constants.

Hint:Use that ¹

sin(πx)−_πx¹

≤^π₄when|x| ≤¹₂.

3.1.9.LetD_Nbe the Dirichlet kernel onT¹. Prove that for all 1<p<∞there exist two constantsC_p,c_p>0 such that

c_p(2N+1)^1/p0≤ D_N

L^p≤C_p(2N+1)^1/p0.

Hint:Consider the two closest zeros ofD_N near the origin and split the integral into the intervals thus obtained.

3.1.10.(S. Bernstein) LetP(x)be a trigonometric polynomial of degreeNonT¹. Prove that

P⁰

_L∞≤4πN P

_L∞.

Hint:Prove first thatP⁰(x)/2πiNis equal to (e^−2πiN(·)P)∗FN−1

(x)e^2πiNx− (e^2πiN(·)P)∗FN−1

(x)e^−2πiNx and then takeL^∞norms.

3.1.11.(Fej´er and F. Riesz) LetP(ξ) =∑^Nk=−Na_ke^2πikξ be a trigonometric poly- nomial onT¹of degreeN such that P(ξ)>0 for all ξ. Prove that there exists a trigonometric polynomialQ(ξ)of the form∑^N_k=0b_ke^2πikξ such thatP(ξ) =|Q(ξ)|². Hint:Note thatNzeros of the polynomialR(z) =∑^Nk=−Na_kz^k+N lie inside the unit circle and the otherNlie outside.

3.1.12.(Landau [167]) Points inZⁿare calledlattice points. Follow the following steps to obtain the number of lattice pointsN(R)inside a closed ball of radiusR inRⁿ. LetBbe the closed unit ball inRⁿ,χBits characteristic function, andv_nits volume.

(a) Using the results in Appendices B.6 and B.7, observe that there is a constantC_n such that for allξ ∈Rⁿwe have

|χcB(ξ)| ≤C_n(1+|ξ|)⁻ⁿ⁺¹² . (b) For 0<ε< ₁₀¹ letΦ^ε =χ₍₁₋ε

2)B∗ζε, whereζε(x)¹

εⁿζ(^x

ε)andζ is a smooth function that is supported in|x| ≤ ¹₂ and has integral equal to 1. Also letΨ^ε = χ₍₁₊ε

2)B∗ζε. Prove that

Φ^ε(x) =1 when|x| ≤1−ε and Φ^ε(x) =0 when|x| ≥1, Ψ^ε(x) =1 when|x| ≤1 and Ψ^ε(x) =0 when|x| ≥1+ε, and also that

cΦ^ε(ξ)

+ cΨ^ε(ξ)

≤C_n,N(1+|ξ|)⁻ⁿ⁺¹² (1+ε|ξ|)^−N for everyξ ∈RⁿandNa large positive number.

(c) Use the result in (b) and the Poisson summation formula to obtain

m∈Z

∑

ⁿ

χ_B(^m_R)≥

∑

m∈Zⁿ

Φ^ε(^m_R) =RⁿΦc^ε(0) +

∑

m∈Zⁿ\{0}

RⁿΦc^ε(Rm)

≥ v_n(1−ε)ⁿ−C_n,N

∑

m∈Zⁿ\{0}

Rⁿ(1+R|m|)⁻ⁿ⁺¹² (1+εR|m|)^−N.

Now use(1−ε)ⁿ≥1−nεand pickεsuch thatεRⁿ=ε⁻

n−1

2 to deduce the estimate N(R)≥v_nRⁿ+O(R^nn−1n+1)asR→∞. Argue similarly withΨ^ε to obtain the identity

N(R) =v_nRⁿ+O(R^nn−1n+1), asR→∞.

3.1.13.(Minkowski) LetSbe an open convex symmetric set inRⁿand assume that the Fourier transform of its characteristic function satisfies the decay estimate

|χbS(ξ)| ≤C(1+|ξ|)⁻ⁿ⁺¹² .

(This is the case if the boundary ofShas nonzero Gaussian curvature.) Assume that

|S|>2ⁿ. Prove thatScontains at least one lattice point other than the origin.

Hint:Assume the contrary, set f =χ1 2S∗χ1

2S, and apply the Poisson summation formula to f to prove that f(0)≥ bf(0).