Fourier Transform of tempered distributions

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−1ξ,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)2}dxe^−πξ2 =e^−πξ2, where the second equality is obtained by taking the ξ-derivative of R

e^{−π(x+iξ)2}dx : we have indeed

d dξ

Z

e^{−π(x+iξ)2}dx

= Z

e^{−π(x+iξ)2}(−2iπ)(x+iξ)dx

=i Z d

dx e^{−π(x+iξ)2}

dx= 0.

For a >0, we obtain R

Re^−2iπxξe^−πax2dx= a^−1/2e^{−πa−1ξ2}, 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·(tP ξ)}e^−πhDy,yidy (with η=^tP ξ) = Y

1≤j≤n

Z

R

e^−2iπyjηje^−πλjy2jdy_j = Y

1≤j≤n

λ^−1/2_j e^{−πλ−1j η2j}

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

= (detA)^−1/2e^{−πhA−1ξ,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−n}dy

= 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.

1.2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 9

Proposition 1.2.5. Using the notation D_xj = 1

2iπ

∂

∂xj

, D^α_x =

n

Y

j=1

D_x^αj

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 transfor-mation 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),

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 topo-logical 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)

1.2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 11

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_ku(P φ)≤C_up_ku+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 neighbor-hood 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.

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+1p0 |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_kn,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.

1.2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 13

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 as-sumption 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 transforma-tion 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.

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.

1.2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 15

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).

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.

1.2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 17

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|^−απ^n2−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−1ξ,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|^−1/2e^iπ4signBe^{−iπhB−1ξ,ξi}. (1.2.31) Proof. We use the notations of Section 1.6 (in the subsection Logarithm of a non-singular 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^{−12trace LogA}e^{−πhA−1ξ,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^∗₊.

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)−1ξ,ξi}ϕ(ξ)dξ, and by continuity of Log on Υ₊ and dominated convergence,

hcv_A, ϕi= Z

e^{−12trace LogA}e^{−πhA−1ξ,ξi}ϕ(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|)^Nα. (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 sup-ported. 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|Imζ|. (1.2.34) In particular, for S ∈S⁰(Rⁿ), we may define according to (1.2.33) the convolution T ∗S.

1.2. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS 19

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||Imζ||.

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||Imζ|| ≤(R₀+ 2)^|γ| sup

y∈Rⁿ

|(∂^αχ₀)(y)e^{2π(R0+2)|Imζ|}|

≤(R0+ 2)^|γ|e^{2π(R0+2)|Imζ|}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|Imζ|e^4πR0c(α)R^−|α|₀ (1 +|ζ|)^|α|

yielding

|F^(k)(ζ)| ≤e^2πR0|Imζ|C_k(1 +|ζ|)^Nk,

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 tem-pered 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.

Dans le document Elements of Graduate Analysis (Page 7-20)