Lecture Notes on Real Analysis Université Pierre et Marie Curie (Paris 6) Nicolas Lerner

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1 Lecture Notes on Real Analysis Université Pierre et Marie Curie (Paris 6) Nicolas Lerner March 16, 211

2 98 CHAPTER 3. INTRODUCTION TO THE THEORY OF DISTRIBUTIONS Proof. We start with n 3, noting that the function x 2 n is L 1 loc and homogeneous with degree 2 n, so that Δ x 2 n is homogeneous with degree n (see the remark (2)). Moreover, the function x 2 n = f(r 2 ), r 2 = x 2, f(t) = t 1 n 2 + is smooth outside and we can compute there Δ(f(r 2 )) = j j (f (r 2 )2x j ) = j f (r 2 )4x 2 j + 2nf (r 2 ) = 4r 2 f (r 2 ) + 2nf (r 2 ), so that with t = r 2, Δ(f(r 2 )) = 4t(1 n 2 )( n 2 )t n n(1 n 2 )t n 2 = t n n 2 (1 )( 2n + 2n) =. 2 As a result, Δ x 2 n is homogeneous with degree n and supported in {}. From the theorem 3.3.4, we obtain that Δ x 2 n = cδ }{{} + c j,α δ (α). homogeneous 1 j m α =j degree n }{{} homogeneous degree n j The lemma implies that for 1 j m, = α =j c j,αδ (α) and Δ x 2 n = cδ. It remains to determine the constant c. We calculate, using the previous formulas for the computation of Δ(f(r 2 )), here with f(t) = e πt, ( c = Δ x 2 n, e π x 2 = x 2 n e π x 2 4 x 2 π 2 2nπ ) dx = S n 1 + = S n 1 ( 1 2π [e πr2 = S n 1 ( n + 2), r 2 n+n 1 e πr2 (4π 2 r 2 2nπ)dr (4π 2 r 2 2nπ)] π + e πr2 8π 2 rdr ) giving (3.6.6). For the convenience of the reader, we calculate explicitely S n 1. We have indeed + 1 = dx = S n 1 r n 1 e πr2 dr R n e π x 2 }{{} = S n 1 π (1 n)/2 r=t 1/2 π 1/2 + t n 1 2 e t 1 2 t 1/2 dtπ 1/2 = S n 1 π n/2 2 1 Γ(n/2). Turning now our attention to the Cauchy-Riemann equation, we see that 1/z is also L 1 loc (R2 ), homogeneous of degree 1, and satisfies (z 1 ) = on the complement of {}, so that the same reasoning as above shows that (π 1 z 1 ) = cδ. To check the value of c, we write c = (π 1 z 1 ), e πz z = R 2 e πz z π 1 z 1 πzdxdy = 1, which gives (3.6.7). We are left with the Laplace equation in two dimensions and we note that with z = 1 2 ( x i y ), z = 1 2 ( x + i y ), we have in two dimensions Δ = 4 z z = 4 z z. (3.6.8)

3 3.6. SOME FUNDAMENTAL SOLUTIONS 99 Solving( the equation ) 4 E that z ln(z z) = z 1 = 1 z πz 1 leads us to try E = ln z and we check directly7 2π Δ( 1 2π ln z ) = π ( ) ln(z z) = π 1 ( ) z 1 = δ. z z z Hypoellipticity Definition Let P be a linear operator of type (3.6.1). We shall say that P is hypoelliptic when for all open subsets Ω of R n and all u D (Ω), we have singsupp u = singsupp P u. (3.6.9) It is obvious that singsupp P u singsupp u, so the hypoellipticity means that singsupp u singsupp P u, which is a very interesting piece of information since we can then determine the singularities of our (unknown) solution u, which are located at the same place as the singularities of the source f, which is known when we try to solve the equation P u = f. Theorem Let P be a linear operator of type (3.6.1) such that P has a fundamental solution E satisfying singsupp E = {}. (3.6.1) Then P is hypoelliptic. In particular the Laplace and the Cauchy-Riemann operators are hypoelliptic. N.B. The condition (3.6.1) appears as an iff condition for the hypoellipticity of the operator P since it is also a consequence of the hypoellipticity property. Proof. Assume that (3.6.1) holds, let Ω be an open subset of R n and u D (Ω). We consider f = P u D (Ω), x / singsupp f, χ Cc (Ω), χ = 1 near x. We have from the proposition that χu = χu P E = (P χu) E = ([P, χ]u) E + Cc (Rn ) {}}{ (χf) E }{{} C (R n ) and thus, using the the proposition for singular supports, we get singsupp(χu) singsupp([p, χ]u) + singsupp E = singsupp([p, χ]u) supp(u χ), and since χ is identically 1 near x, we get that x / supp(u χ), implying x / singsupp(χu), proving that x / singsupp u and the result. 7 ( Noting that ln(x 2 + y 2 ) and its first derivatives are L 1 loc (R2 ), we have for ϕ Cc (R 2 ), ) z ln z 2, ϕ = 1 ( x ϕ+i y ϕ) ln(x 2 +y 2 )dxdy = ϕ(x, y)(xr 2 iyr 2 )dxdy = (x+iy) 1 ϕ(x, y)dxdy. 2 R 2

4 1 CHAPTER 3. INTRODUCTION TO THE THEORY OF DISTRIBUTIONS 3.7 Appendix The Gamma function The gamma function Γ is a meromorphic function on C given for Re z > by the formula Γ(z) = + e t t z 1 dt. (3.7.1) For n N, we have Γ(n + 1) = n!; another interesting value is Γ(1/2) = π. The functional equation Γ(z + 1) = zγ(z) (3.7.2) is easy to prove for Re z > and can be used to extend the Γ function into a meromorphic function with simple poles at N and Res(Γ, k) = ( 1)k. For instance, for k! 1 < Re z with z we define Γ(z) = Γ(z + 1), where we can use (3.7.1) to define Γ(z + 1). z More generally for k N, 1 k < Re z k, z k, we can define Γ(z) = Γ(z + k + 1) z(z + 1)... (z + k). There are manifold references on the Gamma function. One of the most comprehensive is certainly the chapter VII of the Bourbaki volume Fonctions de variable réelle [2] LF spaces The Schwartz kernel theorem Coordinate transformations and pullbacks

5 14 CHAPTER 4. INTRODUCTION TO FOURIER ANALYSIS The Fourier transformation on S (R n ) Definition Let T be a tempered distribution ; the Fourier transform ˆT of T is the tempered distibution defined by the formula ˆT, ϕ S,S = T, ˆϕ S,S. (4.1.1) The linear form ˆT is obviously a tempered distribution since the Fourier transformation is continuous on S. Thanks to the lemma 3.1.7, if T S, the present definition of ˆT and (4.1.1) coincide. Note that for T, ϕ S, we have ˆT, ϕ = T (x)e 2iπx ξ ϕ(ξ)dxdξ = T, ˆϕ. This definition gives that δ = 1, (4.1.11) since δ, ϕ = δ, ϕ = ϕ() = ϕ(x)dx = 1, ϕ. Theorem The Fourier transformation is an isomorphism of S (R n ). Let T be a tempered distribution. Then we have 3 T = ˇˆT. (4.1.12) With obvious notations, we have the following extensions of (4.1.7), D α x T (ξ) = ξ α ˆT (ξ), (D α ξ ˆT )(ξ) = ( 1) α xα T (x)(ξ). (4.1.13) Proof. Using the notation ( ˇϕ)(x) = ϕ( x) for ϕ S, we define Š for S S by (see the remark 3.4.4), Š, ϕ S,S = S, ˇϕ S,S and we obtain for T S ˇˆT, ϕ S,S = ˆT, ˇϕ S,S = ˆT, ˆˇϕ S,S = T, ˆˇϕ S,S = T, ϕ S,S, where the last equality is due to the fact that ϕ ˇϕ commutes 4 with the Fourier transform and (4.1.4) means ˇˆϕ = ϕ, a formula also proven true on S by the previous line of equality. The formula (4.1.7) is true as well for T S since, with ϕ S and ϕ α (ξ) = ξ α ϕ(ξ), we have D α T, ϕ S,S = T, ( 1) α D α ˆϕ S,S = T, ϕ α S,S = ˆT, ϕ α S,S, and the other part is proven the same way. The following lemma will be useful. Lemma Let T S (R n ) be a homogeneous distribution of degree m. Then its Fourier transform is a homogeneous distribution of degree m n Proof. We check (ξ D ξ ) ˆT = ξ xt = (D x xt ) = n 2iπ ˆT 1 2iπ (x x T ) = so that the Euler equation (3.4.6) ξ ξ ˆT = (n + m) ˆT is satisfied. (n + m) 2iπ 3 According to the remark 3.4.4, Ť is the distribution defined by Ť, ϕ = T, ˇϕ and if T S, Ť is also a tempered distribution since ϕ ˇϕ is an involutive isomorphism of S. 4 If ϕ S, we have ˇϕ(ξ) = e 2iπx ξ ϕ( x)dx = e 2iπx ξ ϕ(x)dx = ˆϕ( ξ) = ˇˆϕ(ξ). ˆT,

6 4.1. FOURIER TRANSFORM OF TEMPERED DISTRIBUTIONS The Fourier transformation on L 1 (R n ) Theorem The Fourier transformation is linear continuous from L 1 (R n ) into L (R n ) and for u L 1 (R n ), we have û(ξ) = e 2iπx ξ u(x)dx, û L (R n ) u L 1 (R n ). (4.1.14) Proof. The formula (4.1.1) can be used to define directly the Fourier transform of a function in L 1 (R n ) and this gives an L (R n ) function which coincides with the Fourier transform: for a test function ϕ S (R n ), and u L 1 (R n ), we have by the definition (4.1.1) above and the Fubini theorem û, ϕ S,S = u(x) ˆϕ(x)dx = u(x)ϕ(ξ)e 2iπx ξ dxdξ = ũ(ξ)ϕ(ξ)dξ with ũ(ξ) = e 2iπx ξ u(x)dx which is thus the Fourier transform of u The Fourier transformation on L 2 (R n ) We refer the reader to the section 5.3 in Chapter Some standard examples of Fourier transform Let us consider the Heaviside function defined on R by H(x) = 1 for x >, H(x) = for x ; it is obviously a tempered distribution, so that we can compute its Fourier transform. With the notation of this section, we have, with δ the Dirac mass at, Ȟ(x) = H( x), Ĥ + Ȟ = ˆ1 = δ, Ĥ Ȟ = ŝign, 1 iπ = 1 2iπ 2 δ (ξ) = D sign(ξ) = ξŝignξ so that ξ ( ŝignξ 1 iπ pv(1/ξ)) = and from the theorem 3.2.8, we get ŝignξ 1 iπ pv(1/ξ) = cδ, with c = since the lhs is odd. We obtain ŝign(ξ) = 1 iπ pv 1 ξ, (4.1.15) pv( 1 πx ) = i sign ξ, (4.1.16) Ĥ = δ iπ pv(1 ξ ) = 1 (x i) 1 2iπ. (4.1.17) Let us consider now for < α < n the L 1 loc (Rn ) function u α (x) = x α n ( x is the Euclidean norm of x); since u α 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 from the lemma that v α is a homogeneous distribution

7 16 CHAPTER 4. INTRODUCTION TO FOURIER ANALYSIS of degree α. On the other hand, if S O(R n ) (the orthogonal group), we have in the distribution sense (see the definition 3.4.3), since u α is a radial function, v α (Sξ) = v α (ξ). (4.1.18) The distribution ξ α v α (ξ) is homogeneous of degree on R n \{} and is also radial, i.e. satisfies (4.1.18). Moreover on R n \{}, the distribution v α is a C 1 function which coincides with e 2iπx ξ χ (x) x α n dx + ξ 2N e 2iπx ξ D x 2N( χ 1 (x) x α n) dx, where χ C c (R n ) is 1 near and χ 1 = 1 χ, N N, α + 1 < 2N. As a result ξ α v α (ξ) = c α on R n \{} and the distribution on R n (note that α < n) T = v α (ξ) c α ξ α is supported in {} and homogeneous (on R n ) with degree α. From the theorem and the lemma 3.4.8, the condition < α < n gives v α = c α ξ α. To find c α, we compute x α n e πx2 dx = u α, e πx2 = c α ξ α e πξ2 dξ which yields 2 1 Γ( α + + α 2 )π 2 = r α 1 e πr2 dr = c α We have proven the following lemma. r n α 1 e πr2 dr = c α 2 1 Γ( n α )π n α 2. 2 Lemma Let n N and α ], n[. The function u α (x) = x α n is L 1 loc (Rn ) and also a temperate distribution on R n. Its Fourier transform v α is also L 1 loc (Rn ) and given by v α (ξ) = ξ α π n 2 α Γ( α) 2 Γ( n α) The Poisson summation formula Wave packets We define for x R n, (y, η) R n R n ϕ y,η (x) = 2 n/4 e π(x y)2 e 2iπ(x y) η = 2 n/4 e π(x y iη)2 e πη2, (4.2.1) where for ζ = (ζ 1,..., ζ n ) C n, ζ 2 = ζj 2. (4.2.2) 1 j n We note that the function ϕ y,η is in S(R n ) and with L 2 norm 1. In fact, ϕ y,η appears as a phase translation of a normalized Gaussian. The following lemma introduces the wave packets transform as a Gabor wavelet.

8 4.2. THE POISSON SUMMATION FORMULA 17 Lemma Let u be a function in the Schwartz class S(R n ). We define (W u)(y, η) = (u, ϕ y,η ) L 2 (R n ) = 2 n/4 u(x)e π(x y)2 e 2iπ(x y) η dx (4.2.3) = 2 n/4 u(x)e π(y iη x)2 dxe πη2. (4.2.4) For u L 2 (R n ), the function T u defined by (T u)(y + iη) = e πη2 W u(y, η) = 2 n/4 u(x)e π(y+iη x)2 dx (4.2.5) is an entire function. The mapping u W u is continuous from S(R n ) to S(R 2n ) and isometric from L 2 (R n ) to L 2 (R 2n ). Moreover, we have the reconstruction formula u(x) = W u(y, η)ϕ y,η (x)dydη. (4.2.6) R n R n Proof. For u in S(R n ), we have W u(y, η) = e 2iπyη Ω1(η, y) where Ω 1 is the Fourier transform with respect to the first variable of the S(R 2n ) function Ω(x, y) = u(x)e π(x y)2 2 n/4. Thus the function W u belongs to S(R 2n ). It makes sense to compute 2 n/2 (W u, W u) L 2 (R 2n ) = lim u(x 1 )u(x 2 )e π[(x 1 y) 2 +(x 2 y) 2 +2i(x 1 x 2 )η+ɛ 2 η 2] dydηdx 1 dx 2. (4.2.7) ɛ + Now the last integral on R 4n converges absolutely and we can use the Fubini theorem. Integrating with respect to η involves the Fourier transform of a Gaussian function and we get ɛ n e πɛ 2 (x 1 x 2 ) 2. Since 2(x 1 y) 2 + 2(x 2 y) 2 = (x 1 + x 2 2y) 2 + (x 1 x 2 ) 2, integrating with respect to y yields a factor 2 n/2. We are left with (W u, W u) L 2 (R 2n ) = lim u(x 1 ) u(x 2 )e π(x 1 x 2 ) 2 /2 ɛ n e πɛ 2 (x 1 x 2 ) 2 dx 1 dx 2. (4.2.8) ɛ + Changing the variables, the integral is lim u(s + ɛt/2) u(s ɛt/2)e πɛ2 t 2 /2 e πt2 dtds = u 2 L ɛ 2 (R n ) + by Lebesgue s dominated convergence theorem: the triangle inequality and the estimate u(x) C(1 + x ) n 1 imply, with v = u/c, v(s + ɛt/2) v(s ɛt/2) (1 + s + ɛt/2 ) n 1 (1 + s + ɛt/2 ) n 1 (1 + s + ɛt/2 + s ɛt/2 ) n 1 (1 + 2 s ) n 1.

9 18 CHAPTER 4. INTRODUCTION TO FOURIER ANALYSIS Eventually, this proves that i.e. W u 2 L 2 (R 2n ) = u 2 L 2 (R n ) (4.2.9) W : L 2 (R n ) L 2 (R 2n ) with W W = id L 2 (R n ). (4.2.1) Noticing first that W u(y, η)ϕ y,η dydη belongs to L 2 (R n ) (with a norm smaller than W u L 1 (R 2n )) and applying Fubini s theorem, we get from the polarization of (4.2.9) for u, v S(R n ), (u, v) L 2 (R n ) = (W u, W v) L 2 (R 2n ) = W u(y, η)(ϕ y,η, v) L 2 (R n )dydη = ( W u(y, η)ϕ y,η dydη, v) L 2 (R n ), yielding the result of the lemma u = W u(y, η)ϕ y,η dydη Poisson s formula The following lemma is in fact the Poisson summation formula for Gaussian functions in one dimension. Lemma For all complex numbers z, the following series are absolutely converging and e πm2 e 2iπmz. (4.2.11) m Z e π(z+m)2 = m Z Proof. We set ω(z) = m Z e π(z+m)2. The function ω is entire and 1-periodic since for all m Z, z e π(z+m)2 is entire and for R > sup e π(z+m)2 sup e πz2 e πm2 e 2π m R l 1 (Z). z R z R Consequently, for z R, we obtain, expanding ω in Fourier series 5, ω(z) = k Z 1 e 2iπkz ω(x)e 2iπkx dx. 5 Note that we use this expansion only for a C 1-periodic function. The proof is simple and requires only to compute Re 1 k N e2iπkx sin π(2n+1)x = sin πx. Then one has to show that for a smooth 1-periodic function ω such that ω() =, 1 sin λx lim ω(x)dx =, λ + sin πx which is obvious since for a smooth ν (here we take ν(x) = ω(x)/ sin πx), 1 ν(x)sin λxdx = O(λ 1 ) by integration by parts.

10 4.3. FOURIER TRANSFORMATION AND CONVOLUTION 19 We also check, using Fubini s theorem on L 1 (, 1) l 1 (Z) 1 ω(x)e 2iπkx dx = m Z 1 m+1 e π(x+m)2 e 2iπkx dx = e πt2 e 2iπkt dt m Z m = e πt2 e 2iπkt = e πk2. R So the lemma is proven for real z and since both sides are entire functions, we conclude by analytic continuation. It is now straightforward to get the n-th dimensional version of the previous lemma: for all z C n, using the notation (4.2.2), we have = e πm2 e 2iπm z. (4.2.12) m Z n m Z n e π(z+m)2 Theorem (The Poisson summation formula). Let n be a positive integer and u be a function in S(R n ). Then we have u(k) = û(k), (4.2.13) k Z n k Z n where û stands for the Fourier transform of u. In other words the tempered distribution D = k Z n δ k is such that D = D. Proof. We write, according to (4.2.6) and to Fubini s theorem k Z n u(k) = W u(y, η)ϕ y,η (k)dydη k Z n = W u(y, η) ϕ y,η (k)dydη. k Z n Now, (4.2.12), (4.2.1) give k Z n ϕ y,η(k) = k Z n ϕ y,η(k), so that (4.2.6) and Fubini s theorem imply the result. 4.3 Fourier transformation and convolution Fourier transformation on E (R n ) Theorem Let u E (R n ). Then û is an entire function on C n. Proof. We have for ϕ D(R n ), according to the definition (3.4.14), û, ϕ = u, ˆϕ = u(x), e 2iπx ξ ϕ(ξ)dξ = u(x) ϕ(ξ), e 2iπx ξ E (R 2n ),E (R 2n ) = ϕ(ξ), u(x), e 2iπx ξ, }{{} ũ(ξ)

11 11 CHAPTER 4. INTRODUCTION TO FOURIER ANALYSIS an identity which implies û = ũ and moreover the function ũ is indeed entire, since with ζ C n, and ũ(ζ) = u(x), e 2iπx ζ the function ũ is C (C n ) from the corollary 3.4.2, and we can check that ũ = (a direct computation of ũ(ζ +h) u(ζ) provides elementarily the holomorphy of ũ). Definition The space O M (R n ) of multipliers of S (R n ) is the subspace of the functions f E (R n ) such that, α N n, C α >, N α N, x R n, ( α x f)(x) C α (1 + x ) Nα. (4.3.1) It is easy to check that, for f O M (R n ), the operator u fu is continuous from S (R n ) into itself, and by transposition from S (R n ) into itself: we have for T S (R n ), f O M (R n ), ft, ϕ S,S = T, fϕ S,S, and if p is a semi-norm of S, the continuity on S of the multiplication by f implies that there exists a semi-norm q on S such that for all ϕ S, p(fϕ) q(ϕ). A typical example of a function in O M (R n ) is e ip (x) where P 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 (4.3.1) holds. Lemma Let u E (R n ). Then û belongs to O M (R n ). Proof. We have already seen that û(ξ) = u(x), e 2iπx ξ is a smooth function so that (D α ξ u)(ξ) = u(x), e 2iπx ξ x α ( 1) α which implies (Dξ αu)(ξ) C sup β N x K sought result. β x (e 2iπx ξ x α ) C 1 (1 + ξ ) N, proving the Convolution and Fourier transformation Theorem Let u S (R n ), v E (R n ). Then the convolution u v belongs to S (R n ) and û v = ûˆv. (4.3.2) N.B. We note that both sides of the equality (4.3.2) make sense since the lhs is the Fourier transform of u v which belongs to S (this has to be proven) and ˆv belongs to O M (R n ) so that the product of û S with ˆv makes sense. Proof. Let us prove first that u v belongs to S. We have for ϕ D(R n ) and χ D(R n ) equal to 1 near the support of v, u v, ϕ D (R n ),D(R n ) = u(x) v(y), ϕ(x + y)χ(y) D (R 2n ),D(R 2n ). Now if ϕ S (R n ) the function (x, y) ϕ(x+y)χ(y) = Φ(x, y) belongs to S (R 2n ): it is a smooth function and x α y β γ x ρ yφ is a linear combination of terms of type (x + y) ω ( ν ϕ)(x + y)y λ ( µ χ)(y)

12 4.4. SOME FUNDAMENTAL SOLUTIONS 111 which are bounded as product of bounded terms. Moreover, if Φ S (R 2n ), the function ψ(x) = v(y), Φ(x, y) is smooth (see the corollary 3.4.2(2)) and belongs to S (R n ) since x α ( β x ψ)(x) = v(y), x α β x Φ(x, y) and for some compact subset K of R n, x α ( x β ψ)(x) = v(y), x α x β Φ(x, y) C sup x α x β y γ Φ(x, y) = p(φ), γ N y K where p is a semi-norm on S (R 2n ). As a result, we can extend u v to a continuous linear form on S (R n ) so that u v S (R n ). Let w S such 6 that ŵ = ûˆv. For ϕ S (R n ), we have On the other hand, we have ˆv(ξ) ˇˆϕ(ξ) = v(x), e 2iπx ξ so that w, ϕ S,S = ûˆv, ˇˆϕ S,S = û, ˆv ˇˆϕ S,S. ϕ(y)e 2iπy ξ dy = v(x) ϕ(y), e 2iπ(y x) ξ = v(x), ϕ(y), e 2iπ(y x) ξ = v(x), ˇϕ(y), e 2iπ(y+x) ξ = (v ˇϕ)(ξ), w, ϕ = û, (v ˇϕ) = ǔ, v ˇϕ = u( x), v(x y), ϕ( y) which gives w = u v and (4.3.2) The Riemann-Lebesgue lemma = u(x), v(y x), ϕ(y) = (u v), ϕ, Lemma Let u L 1 (R n ). Then from (4.1.14) û(ξ) = e 2iπx ξ u(x)dx; moreover û belongs to C () (Rn ), where C () (Rn ) stands for the space of continuous functions on R n tending to at infinity. In particular û is uniformly continuous. Proof. This follows from the Riemann-Lebesgue lemma (see e.g. the lemma in [9]); moreover, û(ξ + h) û(ξ) = u(x) e 2iπx h 1 dx = σ u (h), and the Lebesgue dominated convergence theorem implies that lim h σ u (h) =, implying as well the uniform continuity. 4.4 Some fundamental solutions The heat equation The heat operator is the following constant coefficient differential operator on R t R n x where the Laplace operator Δ x on R n is defined by (3.6.3). 6 Take w = ˇ ûˆv. t Δ x, (4.4.1)

13 112 CHAPTER 4. INTRODUCTION TO FOURIER ANALYSIS Theorem We define on R t R n x the L 1 loc function E(t, x) = (4πt) n/2 H(t)e x 2 4t. (4.4.2) The function E is C on the complement of {(, )} in R R n. The function E is a fundamental solution of the heat equation, i.e. t E Δ x E = δ (t) δ (x). Proof. To prove that E L 1 loc (Rn+1 ), we calculate for T, T + T t n/2 r n 1 e r2 4t dtdr }{{} = r=2t 1/2 ρ + t n/2 2 n 1 t (n 1)/2 ρ n 1 e ρ2 2t 1/2 dtdρ = 2 n T + ρ n 1 e ρ2 dρ < +. Moreover, the function E is obviously analytic on the open subset of R 1+n {(t, x) R R n, t }. Let us prove that E is C on R (R n \{}). With ρ defined in (3.1.1), the function ρ 1 defined by ρ 1 (t) = H(t)t n/2 ρ (t) is also C on R and E(t, x) = H( x 2 4t )( x 2 ) n/2e ( x 2 4t x n π n/2 = x n π n/2 4t ) ρ 1, 4t x 2 which is indeed smooth on R t (R n x\{}). We want to solve the equation t u Δ x u = δ (t)δ (x). If u belongs to S (R n+1 ), we can consider its Fourier transform v with respect to x (well-defined by transposition as the Fourier transform in (4.1.1)), and we end-up with the simple ODE with parameters on v, t v + 4π 2 ξ 2 v = δ (t). (4.4.3) It remains to determine a fundamental solution of that ODE: we have d dt + λ = e tλ d dt etλ, ( d dt + λ) (e tλ H(t)) = ( e tλ d dt etλ) (e tλ H(t)) = δ (t), (4.4.4) so that we can take v = H(t)e 4π2 t ξ 2, which belongs to S (R t R n ξ ). Taking the inverse Fourier transform with respect to ξ of both sides of (4.4.3) gives 7 with u S (R t R n ξ ) t u Δ x u = δ (t) δ (x). (4.4.5) To compute u, we check with ϕ D(R), ψ D(R n ), + u, ϕ ˇψ = v x, ϕ ψ = v, ϕ ˆψ = ϕ(t) R n 4π2t ξ 2 ˆψ(ξ)e dtdξ. We can use the Fubini theorem in that absolutely converging integral and use (4.1.2) to get + ( ) u, ϕ ˇψ = ϕ(t) (4πt) n/2 e π x 2 4πt ψ(x)dx dt = E, ϕ ˇψ, R n where the last equality is due to the Fubini theorem and the local integrability of E. We have thus E = u and E satisfies (4.4.5). The proof is complete. 7 The Fourier transformation obviously respects the tensor products.

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