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Real Variables # 9: Hilbert Spaces Brandon Behring Real Variables # 9: Hilbert Spaces Exercise 3 jjf + gjj2 + jjf − gjj2 = 2(jjfjj2 + jjgjj2) Proof: We have jjf + gjj2 = (f + g; f + g) = (f; f) + (g; f) + (f; g) + (g; g) = jjfjj2 + (f; g) + (f; g) + jjgjj2 = jjfjj2 + jjgjj2 + 2<(f; g) and switching g with −g we have jjf − gjj2 = jjfjj2 + jjgjj2 − 2<(f; g) thus jjf + gjj2 + jjf − gjj2 = 2(jjfjj2 + jjgjj2): Exercise 4a l2 is complete. Proof: Let us assume that C is complete. Take a Cauchy sequence (k) 1 (k) (k) 2 xk = (xi )i=1 = (x1 ; x2 ; ···) 2 l ; then for any > 0 there exists N such that for all m; n > N 1 X (m) (n) 2 2 jjxm − xnjjl2 = jxi − xi j < : i=0 (m) (n) 2 so for all i 2 N we must have jxi − xi j < . Choosing a specific i, we (k) have that the sequence of the i-th entries xi is a Cauchy sequence in C and thus must converge to some xi 2 C by the completeness of C. Doing this for all entries we construct a new sequence x = (x1; x2; ··· :::) 1 We must now verify x 2 l2. We have 2 jjxm − xjjl2 = lim jjxm − xnjjl2 < : n!1 2 2 thus xm − x 2 l . Since l is a vector space (Minkowski) we have x = 2 (x − xm) + xm 2 l . Exercise 4b l2 is separable. Proof: Just as we assumed C is complete, we assume the complex ratio- nals are dense in C. We must show that any element of l2 is arbitrarily close to an element of the form α = (α1; α2; ··· ; αn; 0; 0;:::) where αi are of the form αk = ak + ibk with ak; bk 2 Q. 2 Take an arbitrary element x = (x1; x2; x3;:::) 2 l . Since this converges in the norm, the tail end of must converge. That is, for sufficiently large N we must have for all > 0 that 1 X jx j2 < i 2 k=N+1 Additionally, since the Q is dense in R, we must have that for any xi there is an αi arbitrary close to it. Therefore, we can find > 0 such that N X jx − α j2 < : i i 2 k=1 Finally, 2 2 jjx − αjjl2 = jj(x1 − α1; x2 − α2; ··· ; xn − αn; xn+1; ···)jj N 1 X 2 X 2 = jxi − αij + jxij k=1 k=N+1 = + = . 2 2 Exercise 5 Establish the following relations L2(Rd) and L1(Rd): (a) Neither the inclusion L2(Rd) ⊂ L1(Rd) nor the inclusion L1(Rd) ⊂ L2(Rd) is valid. Proof: Let us take the case d = 1. The general dimensions work exactly the same but obscure what is actually happening. For big p, 2 Lp behaves badly for large values of f. While for small p, Lp behaves badly for small values of f. Let us consider ( p1 for jxj ≤ 1 f(x) = jxj 0 for jxj > 1: Then Z jjfjj1 = jfjdx Z +1 1 = p dx −1 jxj +1p = 2 · j−1 jxjdx = 4 < 1 so f 2 L1(R); however, Z 2 2 jjfjj2 = jfj dx Z +1 1 = dx −1 jxj 1 = 2 · j0 log jxjdx = 1 so f2 = L2(R). Next, take 0 for jxj ≤ 1 g(x) = jxj−3=4 for jxj > 1: Then Z jjgjj1 = jgjdx Z 1 = 2 jxj−3=4dx 1 1 1=4 = 6 · j1 jxj dx = 1 so f2 = L1(R); however, Z 2 2 jjgjj2 = jgj dx 3 Z 1 = 2 jxj−3=2dx 1 1 −1=2 = −4 · j1 jxj dx = 4 < 1 so f 2 L2(R). (b) Let f 2 L2 be supported on a set of finite measure E. Then 1=2 jjfjj1 ≤ m(E) jjfjj2: Proof: Applying the Cauchy-Schwarz inequality we have Z jjfj1 = jfj E Z = jfjχEdx R = (jfj; χE) ≤ jjfjj2jjχEjj2 Z 1=2 2 = jjfjj2 jχEj dx R 1=2 = m(E) jjfjj2 (c) If f is bounded and f 2 L1 then f 2 L2 with 1=2 1=2 jjfjj2 ≤ M jjfjj1 : Proof: Using jf(x)j < M we have Z 2 2 jjfj2 = jfj dx ZR ≤ Mjfjdx R = Mjjfjj1 Thus 1=2 1=2 jjfjj2 ≤ M jjfjj1 : 4 Exercise 6 The following are dense subspaces of L2(Rd). (a) The simple functions Proof: It is sufficient to consider f ≥ 0. The simple approxima- tion theorem (Chapter 1, Theorem 4.1) gives us a sequence of non- 1 negative simple functions fφkgk=1 that converges to f. Because we have 2 2 2 2 3 2 jφn − f j ≤ 2 (jφnj + jfj ) ≤ 2 jfj 2 and by definition of L2 that jfj is integrable, the Dominated Con- vergence theorem tells us that 2 jjφn − fjjL (R) ! 0: It should be clear how to apply this to Lp(R). (b) The step functions. Proof: A simple function is a linear combination of characteristic Pn functions of sets on a finite measure, φ(x) = i=1 aiχEi . A step function is nothing more than a finite combination of characteristic function of sets over cubes. However, any measurable set of finite measure can be approximated as close as we would like by a union of cubes.This line of reasoning is worked out in Chapter 1 Theorem 4.3 which gives us an almost disjoint family of rectangles fRjg with M 2 m(E∆[j=1 Rj) < . We now have that a simple function χE can be P approximated by = j xRj and differ at most on a set of measure 2 and thus M 1=2 jjχE − jjL2 = m(E∆ [j=1 Rj) = . (c) The continuous functions of compact support. Proof: Any step function can be approximated in LP by a continu- ous function. This follows by the same argument in the proof of the case in L1 ( Theorem 2.2.4). 1 2 d Exercise 7 Let fφkgk=1 is an orthonormal basis for L (R ). Prove that the collection fφk;jg1≤;k;j<1 with φk;j(x; y) = φk(x)φj(y) is an orthonormal basis of L2(R2 × Rd). Proof: Assuming (φi; φj) = δij, we have using Fubini's theorem Z Z (φij; φmn) = φij(x; y)φmn(x; y)dxdy ZR ZR = φi(x)φj(y)φm(x)φn(y)dxdy ZR R Z = φi(x)φm(x)dx φj(y)φn(y)dxdy R R = (φi; φm)(φj; φn) = δimδjn 5 so fφk;jg1≤;k;j<1 is an orthonormal series. By Theorem 2.3 we have 4 equivalent ways to verify an orthornormal set is in fact a basis We will choose option (iv), which our context amounts to fφk;jg1≤;k;j<1 is a basis 2 2 d 2 2 d for L (R × R ) if F (x; y) 2 L (R × R ) and (F; φk;j) = 0 for all k; j, then F = 0. Let us assume we have (F; k;j) = 0 for all k; j Z Z (F; φk;j) = F (x; y)φk;k(x; y)dxdy R R Z Z = F (x; y)φk(y)dy φk(x)dx R R Z = Fj(x)φk(x) R = (Fj; φk) = 0 where Z Fj(x) = (F (x; y); φj) = F (x; y)φj(y)dy: R 1 2 d Because fφkgk=1 is an orthonormal basis for L (R ), we have that Fj(x) 2 2 d L (R ) and (Fj; φk) = 0 for all k implies Fj(x) = 0 for all j. However, we 2 2 d now have that F (x; y) 2 L (R × R ) and (F (x; y); φj) = 0. Since φj is a basis, this implies F (x; y) = 0. Exercise 8 Let η(t) be a fixed strictly positive continuous function on [a; b]. 2 Define Hη = L ([a; b]; η) to be the space of all measurable functions f on [a; b] such that Z b jf(t)j2η(t)dt < η: a Define the inner product on Hη by Z b (f; g)η = f(t)g(t)η(t)dt: a (a1) Show that Hη is a Hilbert space. Let us verify the conditions of a Hilbert space. (i) Hη is a vector space over C. Take f; g 2 Hη, then Z 2 jjaf + bgjjHη = jaf + bgj η(t)dt Z Z ≤ 4jaj jfj2η(t)dt + 4jbj jgj2η(t)dt = 4jajjjfjjHη + 4jbjjjgjjHη < 1: 6 so af + bg 2 Hη. (ii) Hη is equipped with an inner product (f; g)η, so that f 7! (f; g) is a linear on Hη for every fixed g 2 Hη, Z b (cf1 + df2; g)η = [cf1(t) + df2(t)] g(t)η(t)dt a Z b Z b = c f1(t)g(t)η(t)dt + d f2(t)g(t)η(t)dt a a = c(f1; g)η + d(f2; g)η and that (f; g)η = (g; f)η because η(t) is real Z b (g; f)η(f; g)η = g(t)f(t)η(t)dt a Z b = f(t)g(t)η(t)dt = (f; g)η: a Finally, Z b (f; f) = jfj2η(t)dt ≥ 0 a because η(t) (and jfj) is positive. (iii) We have Z b 2 2 jjfjjη = (f; f) = jfj η(t)dt = 0 a if and only if jfj2η(t) = 0. Since η(t) > 0, we must have f = 0. (iv) This a redundant condition as it follows from (i) and (ii). (v) We will examine this in part (b) as we examine the limitations on function η. (vi) For separability follow the same strategy as for L2.
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