(3) A bilinear map B : S(R n ) S(R m ) B is continuous (for the product topology) if and only if there exist C, N and M such that

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1 The material here can be found in Hörmander Volume 1, Chapter VII but he ha already done almot all of ditribution theory by thi point(!) Johi and Friedlander Chapter 8. Recall that S( ) i a complete metric pace. We know that convergence with repect to thi metric repreent convergence with repect to each of the underlying norm (1) φ N = up (1 + x ) N Dx α φ ; x, α N note that thee norm increae with N. Conider the ame ort of thing for continuity of map Lemma 1. (1) A linear map A : S( ) B to a normed pace i continuou iff and only if there exit C and N uch that (2) Aφ B C φ N φ S( ). (2) A linear map P : S( ) S(R m ) i continou if and only if for each N there exit C = C(N ) and N = N(N ) uch that (3) P φ N C φ N φ S( ). (3) A bilinear map B : S( ) S(R m ) B i continuou (for the product topology) if and only if there exit C, N and M uch that (4) B(φ, ψ) B C φ N ψ M. (4) A bilinear map G : S( ) S(R m ) S(R k ) i continuou if and only if for each N there exit C, N and M uch that (5) G(φ, ψ) N C φ N ψ M Proof. Let me quickly go through the proof for B, the other are imilar. Firt check that (4) implie continuity we are in a metric pace etting o equential continuity i the ame thing. If (φ n, ψ n ) (φ, ψ) in the product metric pace then φ n φ and ψ n ψ (and converely). Moreover, convergence in S( ) implie (i equivalent to) convergence in each norm N ince if ɛ > 0 i mall, ɛ < 1, chooing δ = 2 N 1 ɛ, then d(φ n, φ) < δ = φ n φ N 1 2 ɛ(1 + φ n φ N ) = φ n φ N ɛ. In the other direction, ue the topological definition. Continuity at 0 (which i equivalent to continuity everywhere for a bilinear form) implie that there exit δ > 0 uch that d(ψ, 0) < δ, d(φ, 0) < δ = B(φ, ψ) B < 1. Since 2 k+1 < δ for ome k we can arrange the inequalitie on the left by demanding (6) ψ k < 1 4 δ = N 2 N ψ N 1 + ψ N < 1 2 δ δ plitting the um in two at k. By caling uing the bilinearity it follow that (7) B(φ, ψ) B C φ + k ψ k, C = 8δ 2. 1

2 2 So continuity for Fréchet pace (or countably normed one for that matter) i jut like continuity on normed pace with extra qualifier! (Didn t do thi) Even though we do not need to ue it at the moment I want to recall becaue it i ignificant later one thing completelne doe for u: Propoition 1. A bilinear map B : S( ) S(R m ) C (or more generally a above) i continuou if and only if it i eparately continuou the map B(, ψ) : S( ) C and B(φ, ) : S(R m ) C are continou for each fixed φ and ψ. Proof. I will not do it in cla. The point i Baire Theorem (what ued to be called Baire Catergory argument) jut like the uniform boundedne principle. For each N look at the et (8) D(N) = {φ S( ); B(φ, ψ) N φ N ψ N ψ S(R m )}. Continuity in the firt variable how thi i cloed and continuity in the econd how that the D(N) cover S( ). Baire Theorem then how that one at leat of the D(N) ha non-empty interior and tranlating and caling around an interior point give the continuity etimate. Now, we know what tempered ditribution are, they are the continuou linear map u : S( ) C, uch that for ome N, (9) C = up u(φ) <. φ N =1 We denote the linear pace the dual pace a S ( ). They are called tempered or temperate ditribution (meaning in ome ene they have polynomial bound but beware of thi). The firt point i that there i an injection (10) I : L 2 ( ) S ( ) which i actually the tranpoe of the incluion S( ) L 2 ( ). The latter i continuou a follow from ( ) ( ) 1 (11) φ L 2 up (1 + x ) 1 2 (n+1) φ(x) (1 + x ) n 1 2 dx ince thi integral i finite. Another way of putting thi i that the L 2 norm i continuou on S( ) φ L 2 C φ k, k > 1 2 n. Thu if u L 2 ( ) we define (12) I(u) S ( ), I(u)(φ) = u(x)φ(x). Not only i thi well-defined but the map I i an injection. In fact we will quite oon drop the I from the notation altogether and regard thi map a an identification. Lemma 2. The map I in (11), (12) i an injection.

3 Proof. I will give a direct proof a bit later on uing convolution. However, you probably know that in one dimenion the eigenfunction of the harmonic ocillator give an orthonormal bai of L 2 (R). They are all the product of polynomial and a gauian, o are in S(R) which i therefore dene. Moreover, their product in n variable give an orthonormal bai of L 2 ( ) proving the denity in higher dimenion. You hould go through a imilar argument to ee that L 1 ( ) S ( ) i injective uing the ame formula (12) maybe leaving denity of S( ) in L 1 ( ) until later! Now we can pecialize to (13) I : S( ) S ( ), I(ψ)(φ) = ψ(x)φ(x) and thi i really the heart of the matter. Lemma 3. If ψ S( ) then for all φ S( ) 3 (14) I(x α ψ)(φ) = I(ψ)(x α φ), I( β x ψ)(φ) = I(ψ)(( 1) α β x φ), I(µψ)(φ) = I(ψ)(µφ), µ S( ). Proof. Thi i jut manipulation under the integral, in the econd cae involving integration by part α time. In fact it uffice to take α = e j and iterate and in that cae (15) ( j ψ)φ = lim ( j ψ)φ R [ R,R] n ( ) = lim ψ( j φ) + ψφ xj=r ψφ xj= R = ψ( j φ) R [ R,R] n [ R,R] n 1 ince the integrand, and hence the n 1 fold integral, vanihe rapidly at infinity. The econd of thee identitie i what i called the weak formulation of differentiation. We certainly know the map (13) i injective (becaue we can evaluate at φ = ψ) and then β ψ S ( ) i the unique point in the image of S( ) which atifie thi identity the right hand ide determine β ψ a a ditibution. Thi i the fundamental point of ditribution theory. We can ue the identitie in (14) a definition. (16) Definition 1. If u S ( ) then there are uniquely defined element β u, x α u and µu defined by the identitie (x α u)(φ) = u(x α φ), ( β x u)(φ) = u(( 1) α β x φ), (µu)(φ) = u(µφ), µ S( ) for all φ S( ).

4 4 (17) (18) Of coure we need to note that thee are indeed ditribution, however thi follow from the fact that the map S( ) φ x α φ, S( ) φ β φ, S( ) S( ) (µ, φ) µφ S( ) are all continuou and thi follow readily epecially if you do thi week homework. The important point here i that thi definition i conitent with the pointwie notion: x α I(ψ) = I(x α ψ) β x I(ψ) = I( β ψ), µi(ψ) = I(µψ) where on the left we ue the ditributional notion and on the right the claical one. Thi i why we can afely drop the I a little bit later. So, we have now defined the differentiation of an arbitrary tempered ditribution. For intance the derivative of an element of L 2 are well-defined, they cannot in general be function but they are ditribution. We need to improve on the embedding I to how that if a function ha claical derivative in an appropriate ene then thee are equal to it ditributional derivative o far thi ha only been hown for element of S( ). On Thurday I will prove the Fourier inverion formula and then come back to propertie of ditribution. We define the Fourier tranform of a function u L 1 ( ) a (19) F(u)(ξ) = û(ξ) = e ix ξ u(x)dx. The boundedne and continuity of the ocillating exponential mean that thi exit a a Lebegue integral for each ξ and define a bounded function which i continuou and vanihe at infinity (20) F : L 1 (R) {û C 0 ( ); up u(x) 0 a R }. x R The continuity of û follow from continuity-in-the-mean of L 1 function that lim t 0 f(x + t) f(x) dx = 0 and the vanihing at infinity from a denity argument below. For the moment we are intereted rather in howing that (21) F : S( ) S( ) i continuou, then that it i in fact an iomorphim with a continuou invere given by the Fourier inverion formula (22) u = Gû, G(v)(x) = (2π) n e ix ξ v(ξ)dξ. Firt then, continuity of (21). We know that S( ) L 1 ( ) ince (1 + x ) n 1 L 1 (R ) and (23) u(x) u n+1 (1 + x ) n 1.

5 5 If we formally differentiate through the integral we find that ξj û(ξ) = ξj e ix ξ u(x) = F(ix j u) (24) and ince x j u S( ) the function on the right exit and i at leat bounded. So, how to jutify the exchange of limit involved in differentiating under the integral? We can appeal to tandard theorem (either for the Riemann integral over big rectangle or directly for the Lebegue integral) or we can jut do it. Namely, look at the difference quotient û(ξ + e j ) u(x) e ix (ξ+e j) e ix ξ e ix j 1 = u(x) = e ix ξ u(x). Taylor formula with (Legendre?) remainder or an appropriate application of the Fundamental Theorem of Calculu give e ixj 1 ix j = 1 r 0 0 d 2 dt 2 eitxj dtdr and the integrand of the RHS i 2 x 2 j o the integral i globally (in eitxj x) bounded by 2 x 2 and hence (25) eixj 1 ix j x 2. Since x 2 u(x) L 1 we can pa to the limit and jutify (26) ξj û(ξ) = F(ix j u) u S( ) meaning that the Fouier tranform ha partial derivative, they are given by thi formula and hence are globally bounded. Now we can iterate and conclude that derivative of all order exit, are continuou and are all bounded:- (27) u S( ) = û C ( ), α ξ û = F(i α x α u). Next we need to do the ame thing for ξ β û but that i quite a lot eaier ince it jut involve integration by part. So now we can ee that F : S( ) S( ) i a continuou linear map.