Elements of Graduate Analysis

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Elements of Graduate Analysis

Nicolas Lerner

Universit´ e Pierre et Marie Curie (Paris 6)

October 2, 2014

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Chapter 1 Fourier Analysis

1.1 Preliminaries

The Fourier transform of a function u∈L¹(Rⁿ) can be defined as ˆ

u(ξ) = Z

Rⁿ

u(x)e^−2iπx·ξdx. (1.1.1)

Lemma 1.1.1 (Riemann-Lebesgue Lemma). Let u be in L¹(Rⁿ). Then we have u(ξ)b −→

|ξ|→∞0.

Moreover the function uˆ is uniformly continuous on Rⁿ.

Proof. We note first that (1.1.1) is meaningful as the integral of anL¹ function and we have also

sup

ξ∈Rⁿ

|u(ξ)| ≤ kukb _L¹₍_Rⁿ₎. (1.1.2) Letϕ∈C_c^∞(Rⁿ). With α= (α₁, . . . , α_n)∈Nⁿ, we define

D^α =D₁^α¹. . . D_n^αⁿ, Dj = 1 2iπ

∂

∂x_j, ξ^α=ξ₁^α¹. . . ξ_n^αⁿ. (1.1.3) We find the identities

ξ₁ϕ(ξ) =b Dd₁ϕ(ξ), D[^αϕ(ξ) =ξ^αϕ(ξ),b (1.1.4) entailing 1 +|ξ|²

ϕ(ξ) = Fourierb

ϕ+P

1≤j≤nD²_jϕ

.We find thus 1 +|ξ|²

|ϕ(ξ)| ≤ kϕb + X

1≤j≤n

D_j²ϕk_L¹_(Rⁿ₎,

which implies lim|ξ|→+∞ϕ(ξ) = 0.b Foru∈L¹(Rⁿ), we have

|u(ξ)| ≤ |b (u\−ϕ)(ξ)|+|ϕ(ξ)| ≤ kub −ϕk_L¹₍_Rⁿ₎+|ϕ(ξ)|,b 5

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so that for all ϕ∈C_c^∞(Rⁿ), lim sup

|ξ|→∞

|bu(ξ)| ≤ ku−ϕk_L¹₍_Rⁿ₎ =⇒lim sup

|ξ|→∞

|u(ξ)| ≤b inf

ϕ∈C_c^∞(Rⁿ)ku−ϕk_L¹₍_Rⁿ₎ = 0.

We have also u(ξb +η)−u(ξ) =b R

Rⁿe^−2iπx·ξ e^−2iπx·η−1

u(x)dx, so that

|bu(ξ+η)−bu(ξ)| ≤ Z

Rⁿ

|u(x)| |e^−2iπx·η −1|

| {z }

≤2

dx,

and Lebesgue’s dominated convergence Theorem shows that, for all ξ∈Rⁿ,

η→0lim|bu(ξ+η)−bu(ξ)|= 0, proving continuity.We have also for R >1,|η| ≤1,

|u(ξb +η)−u(ξ)| ≤b sup

|ξ|≤R

|bu(ξ+η)−bu(ξ)|+ 2 sup

|ξ|≥R−1

|ˆu(ξ)|

so that for 0< ε <1, ifωρ is a modulus of continuity¹ of the continuous function ˆu on the compact set {|x| ≤ρ}

sup

|η|≤ε,ξ∈R^m

|u(ξb +η)−u(ξ)| ≤b ω_R+1(ε) + 2 sup

|ξ|≥R−1

|ˆu(ξ)|, proving that the lim sup of the lhs when ε goes to 0 is smaller than

2 sup

|ξ|≥R−1

|ˆu(ξ)|, for all R > 1.

Since that quantity is already proven to go to 0 when R goes to +∞, we obtain the uniform continuity of ˆu.

We need to extend this transformation to various other situations and it turns out that L. Schwartz’ point of view to define the Fourier transformation on the very large space of tempered distributions is the simplest. However, the cost of the distribution point of view is that we have to define these objects, which is not a completely elementary matter. We have chosen here to limit our presentation to the tempered distributions, topological dual of the so-called Schwartz space of rapidly decreasing functions; this space is a Fr´echet space, so its topology is defined by a countable family of semi-norms and is much less difficult to understand than the space of test functions with compact support on an open set. Proving the Fourier inversion formula on the Schwartz space is a truly elementary matter, which yields almost immediately the most general case for tempered distributions, by a duality abstract nonsense argument. This chapter may also serve to the reader as a motivation to the explore the more difficult local theory of distributions.

1For a continuous functionv defined on a compact subsetK ofR^m, the modulus of continuity ω is defined onR⁺ byω(ρ) = sup_x,y∈K

|x−y|≤ρ

|v(x)−v(y)|.We have lim_ρ→0₊ω(ρ) = 0.

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1.2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 7

1.2 Fourier Transform of tempered distributions

The Fourier transformation on S(Rⁿ)

Definition 1.2.1. Let n ≥ 1 be an integer. The Schwartz space S(Rⁿ) is defined as the vector space ofC^∞ functions ufrom Rⁿ toC such that, for all multi-indices.

α, β ∈Nⁿ,

sup

x∈Rⁿ

|x^α∂_x^βu(x)|<+∞.

Here we have used the multi-index notation: forα = (α1, . . . , αn)∈Nⁿ we define x^α=x^α₁¹. . . x^α_nⁿ, ∂_x^α =∂_x^α₁¹. . . ∂_x^α_nⁿ, |α|= X

1≤j≤n

α_j. (1.2.1) A simple example of such a function is e^−|x|², (|x| is the Euclidean norm of x) and more generally, if A is a symmetric positive definiten×n matrix, the function

v_A(x) =e^−πhAx,xi (1.2.2)

belongs to the Schwartz class. The space S(Rⁿ) is a Fr´echet space equipped with the countable family of semi-norms (p_k)k∈N

p_k(u) = sup

x∈Rⁿ

|α|,|β|≤k

|x^α∂_x^βu(x)|. (1.2.3)

Lemma 1.2.2. The Fourier transform sends continuously S(Rⁿ) into itself.

Proof. Just notice that ξ^α∂_ξ^βu(ξ) =ˆ

Z

e^−2iπxξ∂_x^α(x^βu)(x)dx(2iπ)^|β|−|α|(−1)^|β|, and since sup_x∈_Rn(1 +|x|)ⁿ⁺¹|∂_x^α(x^βu)(x)|<+∞, we get the result.

Lemma 1.2.3. For a symmetric positive definite n×n matrix A, we have

vc_A(ξ) = (detA)^−1/2e^−πhA⁻¹^ξ,ξi, (1.2.4) where v_A is given by (1.2.2).

Proof. In fact, diagonalizing the symmetric matrix A, it is enough to prove the one-dimensional version of (1.2.4), i.e. to check

Z

e^−2iπxξe^−πx²dx= Z

e^−π(x+iξ)²dxe^−πξ² =e^−πξ², where the second equality is obtained by taking the ξ-derivative of R

e^−π(x+iξ)²dx : we have indeed

d dξ

Z

e^−π(x+iξ)²dx

= Z

e^−π(x+iξ)²(−2iπ)(x+iξ)dx

=i Z d

dx e^−π(x+iξ)²

dx= 0.

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For a >0, we obtain R

Re^−2iπxξe^−πax²dx= a^−1/2e^−πa⁻¹^ξ², which is the sought result in one dimension. If n ≥ 2, and A is a positive definite symmetric matrix, there exists an orthogonal n×n matrix P (i.e. ^tP P = Id) such that

D=^tP AP, D= diag(λ₁, . . . , λ_n), all λ_j >0.

As a consequence, we have, since |detP|= 1, Z

Rⁿ

e^−2iπx·ξe^−πhAx,xidx= Z

Rⁿ

e−2iπ(P y)·ξ

e−πhAP y,P yi

dy

= Z

Rⁿ

e^−2iπy·(^t^{P ξ)}e^−πhDy,yidy (with η=^tP ξ) = Y

1≤j≤n

Z

R

e^−2iπy^j^η^je^−πλ^j^y²^jdy_j = Y

1≤j≤n

λ^−1/2_j e^−πλ⁻¹^j ^η²^j

= (detA)^−1/2e^−πhD⁻¹^η,ηi = (detA)^−1/2e^−πh^t^{P A}⁻¹^P ^t^{P ξ,}^t^{P ξi}

= (detA)^−1/2e^−πhA⁻¹^ξ,ξi.

Proposition 1.2.4. The Fourier transformation is an isomorphism of the Schwartz class and for u∈S(Rⁿ), we have

u(x) = Z

e^2iπxξu(ξ)dξ.ˆ (1.2.5)

Proof. Using (1.2.4) we calculate for u∈S(Rⁿ) and >0, dealing with absolutely converging integrals,

u(x) = Z

e^2iπxξu(ξ)eˆ ^−π²^|ξ|²dξ

= Z Z

e^2iπxξe^−π²^|ξ|²u(y)e^−2iπyξdydξ

= Z

u(y)e^−π⁻²^|x−y|²⁻ⁿdy

= Z

u(x+y)−u(x)

| {z }

with absolute value≤|y|ku⁰k_L∞

e^−π|y|²dy+u(x).

Taking the limit when goes to zero, we get the Fourier inversion formula u(x) =

Z

e^2iπxξu(ξ)dξ.ˆ (1.2.6)

We have also proven for u∈S(Rⁿ) and ˇu(x) =u(−x)

u=u.ˇˆˆ (1.2.7)

Since u 7→uˆ and u 7→ uˇ are continuous homomorphisms of S(Rⁿ), this completes the proof of the proposition.

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Proposition 1.2.5. Using the notation D_x_j = 1

2iπ

∂

∂xj

, D^α_x =

n

Y

j=1

D_x^α^j

j with α= (α₁, . . . , α_n)∈Nⁿ, (1.2.8) we have, for u∈S(Rⁿ)

Dd^α_xu(ξ) =ξ^αu(ξ),ˆ (D^α_ξu)(ξ) = (−1)ˆ ^|α|x\^αu(x)(ξ) (1.2.9) Proof. We have foru∈S(Rⁿ), ˆu(ξ) = R

e^−2iπx·ξu(x)dx and thus (D^α_ξu)(ξ) = (−1)ˆ ^|α|

Z

e^−2iπx·ξx^αu(x)dx, ξ^αu(ξ) =ˆ

Z

(−2iπ)^−|α|∂_x^α e^−2iπx·ξ

u(x)dx = Z

e^−2iπx·ξ(2iπ)^−|α|(∂_x^αu)(x)dx, proving both formulas.

N.B. The normalization factor _2iπ¹ leads to a simplification in Formula (1.2.9), but the most important aspect of these formulas is certainly that the Fourier transformation exchanges the operation of derivation with the operation of multiplication.

For instance with

P(D) = X

|α|≤m

a_αD_x^α,

we have for u∈S(Rⁿ), P u(ξ) =c P

|α|≤maαξ^αu(ξ) =ˆ P(ξ)ˆu(ξ),and thus (P u)(x) =

Z

Rⁿ

e^2iπx·ξP(ξ)ˆu(ξ)dξ. (1.2.10)

Proposition 1.2.6. Let φ, ψ be functions in S(Rⁿ). Then the convolution φ∗ψ belongs to the Schwartz space and the mapping

S(Rⁿ)×S(Rⁿ)3(φ, ψ)7→φ∗ψ ∈S(Rⁿ) is continuous. Moreover we have

φ[∗ψ = ˆφψ.ˆ (1.2.11)

Proof. The mapping (x, y)7→F(x, y) =φ(x−y)ψ(y) belongs to S(R²ⁿ) since x, y derivatives of the smooth functionF are linear combinations of products

(∂^αφ)(x−y)(∂^βψ)(y) and moreover

(1 +|x|+|y|)^N|(∂^αφ)(x−y)(∂^βψ)(y)|

≤(1 +|x−y|)^N|(∂^αφ)(x−y)|(1 + 2|y|)^N|(∂^βψ)(y)| ≤p(φ)q(ψ),

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wherep, qare semi-norms onS(Rⁿ). This proves that the bilinear mapping (φ, ψ)7→

F(φ, ψ) is continuous from S(Rⁿ)×S(Rⁿ) into S(R²ⁿ). We have now directly

∂_x^α(φ∗ψ) = (∂_x^αφ)∗ψ and (1 +|x|)^N|∂_x^α(φ∗ψ)| ≤

Z

|F(∂^αφ, ψ)(x, y)|(1 +|x|)^Ndy

≤ Z

|F(∂^αφ, ψ)(x, y)|(1 +|x|)^N(1 +|y|)ⁿ⁺¹

| {z }

≤p(F)

(1 +|y|)⁻ⁿ⁻¹dy,

where p is a semi-norm of F (thus bounded by a product of semi-norms of φ and ψ), proving the continuity property. Also we obtain from Fubini’s Theorem

(φ[∗ψ)(ξ) = Z Z

e−2iπ(x−y)·ξ

e^−2iπy·ξφ(x−y)ψ(y)dydx = ˆφ(ξ) ˆψ(ξ), completing the proof of the proposition.

The Fourier transformation on S⁰(Rⁿ)

Definition 1.2.7. Letnbe an integer≥1. We define the spaceS⁰(Rⁿ) as the topological dual of the Fr´echet space S(Rⁿ): this space is called the space of tempered distributions onRⁿ.

We note that the mapping

S(Rⁿ)3φ7→ ∂φ

∂x_j ∈S(Rⁿ),

is continuous since for all k ∈ N, p_k(∂φ/∂x_j) ≤ p_k+1(φ), where the semi-norms p_k are defined in (1.2.3). This property allows us to define by duality the derivative of a tempered distribution.

Definition 1.2.8. Letu∈S⁰(Rⁿ). We define∂u/∂x_j as an element ofS⁰(Rⁿ) by h∂u

∂x_j, φi_S⁰,S =−hu, ∂φ

∂x_ji_S⁰,S. (1.2.12) The mapping u 7→∂u/∂x_j is a well-defined endomorphism of S⁰(Rⁿ) since the estimates

∀φ∈S(Rⁿ), |h∂u

∂x_j, φi| ≤Cupku(∂φ

∂x_j)≤Cupku+1(φ), ensure the continuity on S(Rⁿ) of the linear form ∂u/∂xj.

Definition 1.2.9. Let u ∈ S⁰(Rⁿ) and let P be a polynomial in n variables with complex coefficients. We define the product P uas an element of S⁰(Rⁿ) by

hP u, φi_S⁰_,_S =hu, P φi_S⁰_,_S. (1.2.13)

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The mapping u 7→ P u is a well-defined endomorphism of S⁰(Rⁿ) since the estimates

∀φ∈S(Rⁿ), |hP u, φi| ≤C_up_k_u(P φ)≤C_up_k_u_+D(φ),

whereD is the degree of P, ensure the continuity on S(Rⁿ) of the linear form P u.

Lemma 1.2.10. Let Ω be an open subset of Rⁿ, f ∈ L¹_loc(Ω) such that, for all ϕ∈C_c^∞(Ω), R

f(x)ϕ(x)dx= 0. Then we have f = 0.

Proof. LetK be a compact subset of Ω and letχ∈C_c^∞(Ω) equal to 1 on a neighborhood ofK (see e.g. Exercise 2.8.7 in [11]). With ρ∈C_c^∞(Rⁿ) such thatR

ρ(t)dt= 1, and for >0, ρ(x) =ρ(x/)⁻ⁿ, we get that

→0lim+

ρ∗(χf) =χf inL¹(Rⁿ), since for w∈L¹(Rⁿ),

kρ∗w−wk_L¹₍_Rⁿ₎= Z

Rⁿ

Z

Rⁿ

ρ(y) w(x−y)−w(x) dy

dx

≤ Z Z

|ρ(z)||w(x−z)−w(x)|dzdx= Z

|ρ(z)|kτ_zw−wk_L¹₍_Rⁿ₎dz.

We know² that limh→0kτ_hw−wk_L¹₍_Rⁿ₎ = 0 and kτ_hw−wk_L¹₍_Rⁿ₎ ≤ 2kwk_L¹₍_Rⁿ₎ so that Lebesgue’s dominated convergence theorem provides

lim→0kρ∗w−wk_L¹_(Rⁿ₎ = 0.

We have ρ∗(χf) (x) =

Z

f(y)χ(y)ρ (x−y)⁻¹ ⁻ⁿ

| {z }

=ϕx(y)

dy, with suppϕx ⊂ suppχ, ϕx ∈ C_c^∞(Ω), and from the assumption of the lemma, we obtain ρ∗(χf)

(x) = 0 for all x, implying χf = 0 from the convergence result and thus f = 0, a.e. onK; the conclusion of the lemma follows since Ω is a countable union of compact sets (see e.g. Exercise 2.8.10 in [11]).

Definition 1.2.11 (support of a distribution). For u ∈ S⁰(Rⁿ), we define the support of u and we note suppu the closed subset of Rⁿ defined by

(suppu)^c={x∈Rⁿ,∃V open∈Vx, u|V = 0}, (1.2.14) where Vx stands for the set of neighborhoods of x and u|V = 0 means that for all φ∈C_c^∞(V), hu, φi= 0.

2Forφ∈C_c⁰(Rⁿ), we havekτhw−wk_L1(Rⁿ)≤ kτhw−τhφk_L1(Rⁿ)+kτhφ−φk_L1(Rⁿ)+kφ−wk_L1(Rⁿ), so that for|h| ≤1,

kτhw−wk_L1(Rⁿ)≤2kφ−wk_L1(Rⁿ)+ Z

|φ(x−h)−φ(x)|dx

≤2kφ−wk_L1(Rⁿ)+|suppφ+Bⁿ|sup|φ(x−h)−φ(x)|

which implies that lim sup_h→0kτ_hw−wk ≤2 inf_φ∈C0

c(Rⁿ)kφ−wk_L1(Rⁿ)= 0.

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Proposition 1.2.12.

(1) We have S⁰(Rⁿ)⊃ ∪1≤p≤+∞L^p(Rⁿ),with a continuous injection of each L^p(Rⁿ) into S⁰(Rⁿ). As a consequence S⁰(Rⁿ) contains as well all the derivatives in the sense (1.2.12) of all the functions in some L^p(Rⁿ).

(2) For u∈C¹(Rⁿ) such that

|u(x)|+|du(x)|

(1 +|x|)^−N ∈L¹(Rⁿ), (1.2.15) for some non-negative N, the derivative in the sense (1.2.12) coincides with the ordinary derivative.

Proof. (1) For u∈L^p(Rⁿ) and φ∈S(Rⁿ), we can define hu, φi_S⁰_,_S =

Z

Rⁿ

u(x)φ(x)dx, (1.2.16)

which is a continuous linear form on S(Rⁿ):

|hu, φi_S⁰_,_S| ≤ kuk_L^p₍_Rⁿ₎kφk_Lp0

(Rⁿ), kφk_Lp0

(Rⁿ) ≤ sup

x∈Rⁿ

(1 +|x|)ⁿ⁺¹^p⁰ |φ(x)|

C_n,p≤C_n,pp_k(φ), for k ≥k_n,p= n+ 1 p⁰ , with p_k given by (1.2.3) (when p = 1, we can take k = 0). We indeed have a continuous injection ofL^p(Rⁿ) intoS⁰(Rⁿ): in the first place the mapping described by (1.2.16) is well-defined and continuous from the estimate

|hu, φi| ≤ kuk_L^pC_n,pp_k_n,p(φ).

Moreover, this mapping is linear and injective from Lemma 1.2.10.

(2) We have forφ ∈S(Rⁿ), χ0 ∈C_c^∞(Rⁿ), χ0 = 1 near the origin, A=h∂u

∂x_j, φi_S⁰_,_S =−hu, ∂φ

∂x_ji_S⁰_,_S =− Z

Rⁿ

u(x)∂φ

∂x_j(x)dx so that, using Lebesgue’s dominated convergence theorem, we find

A =− lim

→0₊

Z

Rⁿ

u(x)∂φ

∂x_j(x)χ₀(x)dx.

Performing an integration by parts on C¹ functions with compact support, we get A= lim

→0+

nZ

Rⁿ

(∂_ju)(x)φ(x)χ₀(x)dx+ Z

Rⁿ

u(x)φ(x)(∂_jχ₀)(x)dxo , with ∂ju standing for the ordinary derivative. We have also

Z

Rⁿ

|u(x)φ(x)(∂jχ0)(x)|dx≤ k∂jχ0)kL^∞(Rⁿ)

Z

|u(x)|(1 +|x|)^−Ndx pN(φ)<+∞, so that

h∂u

∂x_j, φi_S⁰_,_S = lim

→0+

Z

Rⁿ

(∂_ju)(x)φ(x)χ₀(x)dx.

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Since the lhs is a continuous linear form onS(Rⁿ) so is the rhs. On the other hand forφ∈C_c^∞(Rⁿ), the rhs isR

Rⁿ(∂_ju)(x)φ(x)dx.SinceC_c^∞(Rⁿ) is dense inS(Rⁿ) (cf.

Exercise 1.5.3), we find that h∂u

∂x_j, φi_S⁰_,_S = Z

Rⁿ

(∂_ju)(x)φ(x)dx, since the mapping φ 7→ R

Rⁿ(∂_ju)(x)φ(x)dx belongs to S⁰(Rⁿ), thanks to the assumption ondu in (1.2.15). This proves that _∂x^∂u

j =∂_ju.

The Fourier transformation can be extended to S⁰(Rⁿ). We start with noticing that for T, φ in the Schwartz class we have, using Fubini Theorem,

Z

Tˆ(ξ)φ(ξ)dξ = Z Z

T(x)φ(ξ)e^−2iπx·ξdxdξ = Z

T(x) ˆφ(x)dx, and we can use the latter formula as a definition.

Definition 1.2.13. Let T be a tempered distribution ; the Fourier transform ˆT of T is the tempered distribution defined by the formula

hT , ϕiˆ _S⁰_,_S =hT,ϕiˆ _S⁰_,_S. (1.2.17) The linear form ˆT is obviously a tempered distribution since the Fourier transformation is continuous onS. Thanks to Lemma1.2.10, ifT ∈S, the present definition of ˆT and (1.1.1) coincide.

This definition gives that, withδ₀standing as the Dirac mass at 0,hδ₀, φi_S⁰_,_S =φ(0) (obviously a tempered distribution), we have

δb0 = 1, (1.2.18)

since hδb₀, ϕi=hδ₀,ϕib =ϕ(0) =b R

ϕ(x)dx=h1, ϕi.

Theorem 1.2.14. The Fourier transformation is an isomorphism of S⁰(Rⁿ). Let T be a tempered distribution. Then we have³

T =T,ˇˆˆ Tˇˆ=T .ˆˇ (1.2.19) With obvious notations, we have the following extensions of (1.2.9),

D[^α_xT(ξ) = ξ^αTˆ(ξ), (D_ξ^αTˆ)(ξ) = (−1)^|α|x\^αT(x)(ξ). (1.2.20) Proof. We have forT ∈S⁰

hT, ϕiˇˆˆ _S⁰_,_S =hT ,ˆˆ ϕiˇ _S⁰_,_S =hT ,ˆ ϕiˆˇ _S⁰_,_S =hT,ϕiˆˆˇ _S⁰_,_S =hT, ϕi_S⁰_,_S,

3We define ˇT as the distribution given byhT , ϕiˇ =hT,ϕiˇ and ifT ∈S⁰, ˇT is also a tempered distribution sinceϕ7→ϕˇ is an involutive isomorphism ofS.

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where the last equality is due to the fact that ϕ7→ ϕˇ commutes⁴ with the Fourier transform and (1.2.6) means

ˇˆ ˆ ϕ=ϕ,

a formula also proven true on S⁰ by the previous line of equality. Formula (1.2.9) is true as well for T ∈S⁰ since, with ϕ∈S and ϕ_α(ξ) =ξ^αϕ(ξ), we have

hD[^αT , ϕi_S⁰_,_S =hT,(−1)^|α|D^αϕiˆ _S⁰_,_S =hT,ϕc_αi_S⁰_,_S =hT , ϕˆ _αi_S⁰_,_S, and the other part is proven the same way.

The Fourier transformation on L¹(Rⁿ)

Theorem 1.2.15. The Fourier transformation is linear continuous from L¹(Rⁿ) into L^∞(Rⁿ) and for u∈L¹(Rⁿ), we have

ˆ u(ξ) =

Z

e^−2iπx·ξu(x)dx, kˆuk_L^∞₍_Rⁿ₎ ≤ kuk_L¹₍_Rⁿ₎. (1.2.21) Proof. Formula (1.1.1) can be used to define directly the Fourier transform of a function inL¹(Rⁿ) and this gives aL^∞(Rⁿ) function which coincides with the Fourier transform: for a test functionϕ∈S(Rⁿ), andu∈L¹(Rⁿ), we have by the definition (1.2.17) above and Fubini theorem

hˆu, ϕi_S⁰_,_S = Z

u(x) ˆϕ(x)dx= Z Z

u(x)ϕ(ξ)e^−2iπx·ξdxdξ = Z

u(ξ)ϕ(ξ)dξe with u(ξ) =e R

e^−2iπx·ξu(x)dx which is thus the Fourier transform of u.

The Fourier transformation on L²(Rⁿ) Theorem 1.2.16 (Plancherel formula).

The Fourier transformation can be extended to a unitary operator of L²(Rⁿ), i.e.

there exists a unique bounded linear operator F : L²(Rⁿ)−→ L²(Rⁿ), such that for u∈S(Rⁿ), F u= ˆu and we have F^∗F =F F^∗ = Id_L²₍_Rⁿ₎. Moreover

F^∗ =CF =F C, F²C = Id_L²₍_Rⁿ₎, (1.2.22) where C is the involutive isomorphism ofL²(Rⁿ)defined by (Cu)(x) =u(−x). This gives the Plancherel formula: for u, v ∈L²(Rⁿ),

Z

Rⁿ

ˆ

u(ξ)ˆv(ξ)dξ= Z

u(x)v(x)dx. (1.2.23)

Proof. For test functions ϕ, ψ∈S(Rⁿ), using Fubini theorem and (1.2.6), we get⁵ ( ˆψ,ϕ)ˆ _L²₍_Rⁿ₎ =

Z

ψ(ξ) ˆˆ ϕ(ξ)dξ= Z Z

ψ(ξ)eˆ ^2iπx·ξϕ(x)dxdξ = (ψ, ϕ)_L²₍_Rⁿ₎.

4Ifϕ∈S, we havebϕ(ξ) =ˇ R

e^−2iπx·ξϕ(−x)dx=R

e^2iπx·ξϕ(x)dx= ˆϕ(−ξ) = ˇϕ(ξ).ˆ

5We have to pay attention to the fact that the scalar product (u, v)_L2 in the complex Hilbert spaceL²(Rⁿ) is linear with respect touand antilinear with respect tov: forλ, µ∈C,(λu, µv)_L2= λ¯µ(u, v)_L2.

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Next, the density of S in L² shows that there is a unique continuous extension F of the Fourier transform to L² and that extension is an isometric operator (i.e.

satisfying for all u ∈ L²(Rⁿ),kF uk_L² = kuk_L², i.e. F^∗F = Id_L²). We note that the operator C defined by Cu= ˇu is an involutive isomorphism of L²(Rⁿ) and that for u∈S(Rⁿ),

CF²u=u=F CF u=F²Cu.

By the density ofS(Rⁿ) in L²(Rⁿ), the bounded operators CF², Id_L²₍_Rⁿ₎, F CF, F²C, are all equal. On the other hand for u, ϕ∈S(Rⁿ), we have

(F^∗u, ϕ)_L² = (u, F ϕ)_L² = Z

u(x) ˆϕ(x)dx

= Z Z

u(x) ¯ϕ(ξ)e^2iπx·ξdxdξ = (CF u, ϕ)_L², so thatF^∗u=CF ufor allu∈S and by continuityF^∗ =CF as bounded operators onL²(Rⁿ), thus F F^∗ =F CF = Id. The proof is complete.

Some standard examples of Fourier transform

Let us consider the Heaviside function defined onRbyH(x) = 1 forx >0,H(x) = 0 forx≤0 ; as a bounded measurable function, it is a tempered distribution, so that we can compute its Fourier transform. With the notation of this section, we have, with δ₀ the Dirac mass at 0, ˇH(x) = H(−x),

Hb+Hbˇ = ˆ1 =δ₀, Hb −Hbˇ =sign,d 1 iπ = 1

2iπ2δb₀(ξ) =D\sign(ξ) = ξsignξ.d We note that R 7→ ln|x| belongs to S⁰(R) and⁶ we define the so-called principal value of 1/x onR by

pv 1 x

= d

dx(ln|x|), (1.2.24) so that, hpv1

x, φi=− Z

φ⁰(x) ln|x|dx=− lim

→0₊

Z

|x|≥

φ⁰(x) ln|x|dx

= lim

→0+

Z

|x|≥

φ(x)1

xdx+ φ()−φ(−) ln

| {z }

→0

= lim

→0₊

Z

|x|≥

φ(x)1

xdx. (1.2.25)

This entails ξ signξd −_iπ¹pv(1/ξ)

= 0 and from Remark 1.2.17 below, we get signξd − 1

iπpv(1/ξ) = cδ₀, with c= 0 since the lhs is odd⁷.

6Forφ∈S(R), we have hln|x|, φ(x)i_S⁰(R),S(R)=R

Rφ(x) ln|x|dx.

7A distributionT onRⁿ is said to be odd (resp. even) when ˇT =−T (resp. T).

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Remark 1.2.17. Let T ∈ S⁰(R) such that xT = 0. Then we have T = cδ₀. Let φ ∈S(R) and let χ₀ ∈C_c^∞(Rⁿ) such that χ₀(0) = 1. We have

φ(x) =χ₀(x)φ(x) + (1−χ₀(x))φ(x).

Applying Taylor’s formula with integral remainder, we define the smooth function ψ by

ψ(x) = (1−χ₀(x))

x φ(x)

and, applying Leibniz’ formula, we see also that ψ belongs to S(R). As a result hT, φi_S⁰₍_R_),_S₍_R₎ =hT, χ₀φi=hT, χ₀ φ−φ(0)

i+φ(0)hT, χ₀i=φ(0)hT, χ₀i, since the function x 7→ χ₀(x) φ(x)−φ(0)

/x belongs to C_c^∞(R). As a result T = hT, χ₀iδ₀.

We obtain

sign(ξ) =d 1 iπpv1

ξ, (1.2.26)

pv(\1

πx) =−isignξ, (1.2.27)

Hˆ = δ₀ 2 + 1

2iπpv(1

ξ) = 1 (x−i0)

1

2iπ. (1.2.28)

Let us consider now for 0 < α < n the L¹_loc(Rⁿ) function u_α(x) = |x|^α−n (|x| is the Euclidean norm of x); since uα is also bounded for |x| ≥ 1, it is a tempered distribution. Let us calculate its Fourier transform v_α. Since u_α is homogeneous of degree α−n, we get (Exercise 1.5.9) that v_α is a homogeneous distribution of degree −α. On the other hand, if S ∈ O(Rⁿ) (the orthogonal group), we have in the distribution sense⁸ since uα is a radial function, i.e. such that

v_α(Sξ) =v_α(ξ). (1.2.29)

The distribution|ξ|^αv_α(ξ) is homogeneous of degree 0 onRⁿ\{0}and is also “radial”, i.e. satisfies (1.2.29). Moreover onRⁿ\{0}, the distributionvαis aC¹function which coincides with⁹

Z

e^−2iπx·ξχ0(x)|x|^α−ndx+|ξ|^−2N Z

e^−2iπx·ξ|Dx|^2N χ1(x)|x|^α−n dx,

where χ₀ ∈ C_c^∞(Rⁿ) is 1 near 0 and χ₁ = 1−χ₀, N ∈N, α+ 1< 2N. As a result

|ξ|^αv_α(ξ) = c_α on Rⁿ\{0} and the distribution onRⁿ (note that α < n) T =v_α(ξ)−c_α|ξ|^−α

8ForM ∈Gl(n,R),T ∈S⁰(Rⁿ), we definehT(M x), φ(x)i=hT(y), φ(M⁻¹y)i|detM|⁻¹.

9 We haveuc_α=χ[₀u_α+χ[₁u_α and forφsupported inRⁿ\{0}we get,

hχ[1uα, φi=hχ[1uα|ξ|^2N, φ(ξ)|ξ|^−2Ni=h|Dx\|^2Nχ1uα, φ(ξ)|ξ|^−2Ni.

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is supported in {0} and homogeneous (on Rⁿ) with degree −α. From Exercises 1.5.7(1), 1.5.5, 1.5.8, the condition 0 < α < n gives v_α = c_α|ξ|^−α. To find c_α, we compute

Z

Rⁿ

|x|^α−ne^−πx²dx=c_α Z

Rⁿ

|ξ|^−αe^−πξ²dξ which yields

2⁻¹Γ(α

2)π⁻^α² = Z +∞

0

r^α−1e^−πr²dr =c_α Z +∞

0

r^n−α−1e^−πr²dr

=c_α2⁻¹Γ(n−α

2 )π⁻⁽^n−α² ⁾. We have proven the following lemma.

Lemma 1.2.18. Letn∈N^∗ andα∈(0, n). The functionu_α(x) =|x|^α−nisL¹_loc(Rⁿ) and also a temperate distribution on Rⁿ. Its Fourier transform vα is also L¹_loc(Rⁿ) and given by

vα(ξ) =|ξ|^−απⁿ²^−α Γ(^α₂) Γ(^n−α₂ ).

Fourier transform of Gaussian functions

Proposition 1.2.19. Let A be a symmetric nonsingular n×n matrix with complex entries such that ReA≥0. We define the Gaussian function v_A on Rⁿ by v_A(x) = e^−πhAx,xi. The Fourier transform of v_A is

vc_A(ξ) = (detA)^−1/2e^−πhA⁻¹^ξ,ξi, (1.2.30) where (detA)^−1/2 is defined according to Formula (1.6.8). In particular, when A =

−iB with a symmetric real nonsingular matrix B, we get

Fourier(e^iπhBx,xi)(ξ) = vd−iB(ξ) =|detB|^−1/2eⁱ^π⁴^sign^Be^−iπhB⁻¹^ξ,ξi. (1.2.31) Proof. We use the notations of Section 1.6 (in the subsection Logarithm of a nonsingular symmetric matrix). Let us define Υ^∗₊as the set of symmetricn×n complex matrices with a positive definite real part (naturally these matrices are nonsingular since Ax= 0 for x∈Cⁿ implies 0 = RehAx,xi¯ =h(ReA)x,xi, so that Υ¯ ^∗₊⊂Υ₊).

Let us assume first that A ∈ Υ^∗₊; then the function v_A is in the Schwartz class (and so is its Fourier transform). The set Υ^∗₊ is an open convex subset of C^n(n+1)/2 and the function Υ^∗₊ 3 A 7→ cv_A(ξ) is holomorphic and given on Υ^∗₊∩R^n(n+1)/2 by (1.2.30). On the other hand the function

Υ^∗₊3A7→e⁻¹²^{trace Log}^Ae^−πhA⁻¹^ξ,ξi,

is also holomorphic and coincides with previous one on R^n(n+1)/2. By analytic con- tinuation this proves (1.2.30) forA∈Υ^∗₊.

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If A ∈ Υ₊ and ϕ ∈ S(Rⁿ), we have hcv_A, ϕi_S⁰_,_S = R

v_A(x) ˆϕ(x)dx so that Υ₊ 3 A 7→ hcv_A, ϕi is continuous and thus (note that the mapping A 7→ A⁻¹ is an homeomorphism of Υ₊), using the previous result on Υ^∗₊,

hvcA, ϕi= lim

→0+

hv[A+I, ϕi= lim

→0+

Z

e⁻¹²trace Log(A+I)

e^−πh(A+I)⁻¹^ξ,ξiϕ(ξ)dξ, and by continuity of Log on Υ₊ and dominated convergence,

hcv_A, ϕi= Z

e⁻¹²^{trace Log}^Ae^−πhA⁻¹^ξ,ξiϕ(ξ)dξ, which is the sought result.

Multipliers of S⁰(Rⁿ)

Definition 1.2.20. The space OM(Rⁿ) of multipliers of S(Rⁿ) is the subspace of the functions f ∈C^∞(Rⁿ) such that,

∀α ∈Nⁿ,∃C_α >0,∃N_α∈N, ∀x∈Rⁿ, |(∂_x^αf)(x)| ≤C_α(1 +|x|)^N^α. (1.2.32) It is easy to check that, for f ∈ OM(Rⁿ), the operator u 7→ f u is continuous from S(Rⁿ) into itself, and by transposition from S⁰(Rⁿ) into itself: we define for T ∈S⁰(Rⁿ), f ∈OM(Rⁿ),

hf T, ϕi_S⁰_,_S =hT, f ϕi_S⁰_,_S,

and ifpis a semi-norm ofS, the continuity onS of the multiplication byf implies that there exists a semi-norm q on S such that for all ϕ ∈ S, p(f ϕ) ≤ q(ϕ). A typical example of a function inOM(Rⁿ) ise^iP^(x)whereP is a real-valued polynomial:

in fact the derivatives of e^iP^(x) are of type Q(x)e^iP^(x) where Q is a polynomial so that (1.2.32) holds.

Definition 1.2.21. Let T, S be tempered distributions on Rⁿ such that ˆT belongs to OM(Rⁿ). We define the convolution T ∗S by

T[∗S = ˆTS.ˆ (1.2.33)

Note that this definition makes sense since ˆT is a multiplier so that ˆTSˆis indeed a tempered distribution whose inverse Fourier transform is meaningful. We have

hT ∗S, φi_S⁰₍_Rⁿ_),_S₍_Rⁿ₎ =hT[∗S,φiˆˇ _S⁰₍_Rⁿ_),_S₍_Rⁿ₎ =hS,ˆ Tˆφiˆˇ _S⁰₍_Rⁿ_),_S₍_Rⁿ₎.

Proposition 1.2.22. Let T be a distribution on Rⁿ such that T is compactly supported. Then Tˆ is a multiplier which can be extended to an entire function on Cⁿ such that if suppT ⊂B(0, R¯ ₀),

∃C₀, N₀ ≥0,∀ζ ∈Cⁿ, |Tˆ(ζ)| ≤C₀(1 +|ζ|)^N⁰e^2πR⁰^|^Im^ζ|. (1.2.34) In particular, for S ∈S⁰(Rⁿ), we may define according to (1.2.33) the convolution T ∗S.

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Proof. Let us first check the caseR₀ = 0: then the distributionT is supported at{0}

and from Exercise 1.5.5 is a linear combination of derivatives of the Dirac mass at 0. Formulas (1.2.18), (1.2.20) imply that ˆT is a polynomial, so that the conclusions of Proposition1.2.22 hold in that case.

Let us assume that R₀ > 0 and let us consider a function χ is equal to 1 in neighborhood of suppT (this impliesχT =T) and

hT , φib _S⁰,S =hχT , φic _S⁰,S =hT, χφiˆ _S⁰,S. (1.2.35) On the other hand, defining for ζ ∈Cⁿ (with x·ζ =P

x_jζ_j for x∈Rⁿ),

F(ζ) =hT(x), χ(x)e^−2iπx·ζi_S⁰_,_S, (1.2.36) we see that F is an entire function (i.e. holomorphic on Cⁿ): calculating

F(ζ+h)−F(ζ) =hT(x), χ(x)e^−2iπx·ζ(e^−2iπx·h −1)i

=hT(x), χ(x)e^−2iπx·ζ(−2iπx·h)i +hT(x), χ(x)e^−2iπx·ζ

Z 1 0

(1−θ)e^{−2iθπx·h}dθ(−2iπx·h)²i, and applying to the last term the continuity properties of the linear form T, we obtain that the complex differential ofF is

X

1≤j≤n

hT(x), χ(x)e^−2iπx·ζ(−2iπx_j)idζ_j. Moreover the derivatives of (1.2.36) are

F^(k)(ζ) = hT(x), χ(x)e^−2iπx·ζ(−2iπx)^ki_S⁰_,_S. (1.2.37) To evaluate the semi-norms of x7→χ(x)e^−2iπx·ζ(−2iπx)^k in the Schwartz space, we have to deal with a finite sum of products of type

x^γ(∂^αχ)(x)e^−2iπx·ζ(−2iπζ)^β

≤(1 +|ζ|)^|β| sup

x∈Rⁿ

|x^γ(∂^αχ)(x)e^2π|x||^Im^ζ||.

We may now choose a function χ₀ equal to 1 on B(0,1), supported in B(0,^R_R⁰⁺²

0+) such thatk∂^βχ₀k_L^∞ ≤c(β)^−|β| with = _1+|ζ|^R⁰ . We find with

χ(x) =χ₀(x/(R₀+)) (which is 1 on a neighborhood of B(0, R₀)), sup

x∈Rⁿ

|x^γ(∂^αχ)(x)e^2π|x||^Im^ζ|| ≤(R₀+ 2)^|γ| sup

y∈Rⁿ

|(∂^αχ₀)(y)e^2π(R⁰^+2)|^Im^ζ||

≤(R0+ 2)^|γ|e^2π(R⁰^+2)|^Im^ζ|c(α)^−|α|

= (R₀+ 2 R0

1 +|ζ|)^|γ|e^2π(R⁰⁺²

R0 1+|ζ|)|Imζ|

c(α)(1 +|ζ|

R₀ )^|α|

≤(3R0)^|γ|e^2πR⁰^|^Im^ζ|e^4πR⁰c(α)R^−|α|₀ (1 +|ζ|)^|α|

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yielding

|F^(k)(ζ)| ≤e^2πR⁰^|^Im^ζ|C_k(1 +|ζ|)^N^k,

which implies that Rⁿ 3ξ7→F(ξ) is indeed a multiplier. We have also hT, χφiˆ _S⁰_,_S =hT(x), χ(x)

Z

Rⁿ

φ(ξ)e^−2iπxξdξi_S⁰_,_S.

Since the function F is entire we have for φ ∈ C_c^∞(Rⁿ), using (1.2.37) and Fubini Theorem on `¹(N)×L¹(Rⁿ),

Z

Rⁿ

F(ξ)φ(ξ)dξ =X

k≥0

hT(x), χ(x)(−2iπx)^ki Z

suppφ

ξ^k

k!φ(ξ)dξ. (1.2.38) On the other hand, since ˆφ is also entire (from the discussion on F or directly from the integral formula for the Fourier transform of φ ∈C_c^∞(Rⁿ)), we have

hT, χφiˆ =hT(x), χ(x)X

k≥0

( ˆφ)^(k)(0)x^k/k!i

=hT(x), χ(x) lim

N→+∞

X

0≤k≤N

( ˆφ)^(k)(0)x^k/k!

| {z }

convergence inC^∞_c (Rⁿ)

i

= lim

N→+∞

X

0≤k≤N

hT(x), χ(x)x^k/k!i Z

Rⁿ

φ(ξ)(−2iπξ)^kdξ.

Thanks to (1.2.38), that quantity is equal to R

RⁿF(ξ)φ(ξ)dξ. As a result, the tempered distributions ˆT and F coincide on C_c^∞(Rⁿ), which is dense in S(Rⁿ) (see Exercise 1.5.3) and so ˆT =F, concluding the proof.

1.3 The Poisson summation formula

Wave packets

We define for x∈Rⁿ, (y, η)∈Rⁿ×Rⁿ

ϕ_y,η(x) = 2^n/4e^−π(x−y)²e^{2iπ(x−y)·η} = 2^n/4e^{−π(x−y−iη)}²e^−πη², (1.3.1) where for ζ = (ζ₁, . . . , ζ_n)∈Cⁿ, ζ² = X

1≤j≤n

ζ_j². (1.3.2) We note that the functionϕ_y,ηis inS(Rⁿ) and withL² norm 1. In fact,ϕ_y,ηappears as a phase translation of a normalized Gaussian. The following lemma introduces the wave packets transformas a Gabor wavelet.

Lemma 1.3.1. Let u be a function in the Schwartz class S(Rⁿ). We define (W u)(y, η) = (u, ϕ_y,η)_L²₍_Rⁿ₎= 2^n/4

Z

u(x)e^−π(x−y)²e−2iπ(x−y)·η

dx (1.3.3)

= 2^n/4 Z

u(x)e^{−π(y−iη−x)}²dxe^−πη². (1.3.4)

Elements of Graduate Analysis