Functional Analysis I Solutions to Exercises James C. Robinson Contents 1 Examples I page 1 2 Examples II 5 3 Examples III 9 4 Examples IV 15 iii 1 Examples I 1. Suppose that v= n X αj ej and v= j=1 m X βk fk . k=1 with αj , βk ∈ K and ej , fk ∈ E. Relabel βk and fk so that fj = ej for j = 1, . . . , n0 , and fj ∈ / {e1 , . . . , en } for j > n0 , i.e. 0 v= n X m X βk ek + βk fk , k=n0 +1 k=1 with the understanding that the first sum is zero if n0 = 0, and the second zero if n0 = m. It follows that n X 0 αj e j − j=1 n X βk ek − m X βk fk = 0, k=n0 +1 k=1 or 0 n m n X X X αj ej + βk fk = 0. (αj − βj )ej − k=n0 +1 j=1 k=n0 +1 Since E is linearly independent, it follows that αj = βj = 0 for j > n0 , from which n0 = n, and that αj = βj for j = 1, . . . , n. So the expansion is unique. 2. If x, y ∈ `lim (C) then x + y ∈ `lim (C) and αx ∈ `lim (C), since lim (xn + yn ) = lim xn + lim yn n→∞ n→∞ n→∞ 1 and lim αxn = α lim xn . n→∞ n→∞ 2 1 Examples I 3. Write kxkp`p = X |xj |p = X j |xj |p−α |xj |α . j Hölder’s inequality with exponents a and b yields 1/a 1/b X X kxkp`p ≤ |xj |(p−α)a |xj |αb j = j kxkα`αb . kxkp−α `(p−α)a We want to choose α, and conjugate exponents a, b (a−1 + b−1 = 1) so that (p − α)a = q and αb = r, so that the factors on the right are powers of kxk`q and kxk`r . Thus b = r/α; since a, b are conjugate it follows that a = r/(r − α), and then the condition that (p − α)a = q gives α = (p − q)r/(r − q), a = (r − q)/(r − p), b = (r − q)/(p − q). The result follows as stated on substituting for α. 4.(i) If there exist a, b with 0 < a ≤ b such that akxk1 ≤ kxk2 ≤ bkxk1 then 1 1 kxk2 ≤ kxk1 ≤ kxk2 , b a while (ii) if akxk1 ≤ kxk2 ≤ bkxk1 and αkxk2 ≤ kxk3 ≤ βkxk2 then aαkxk1 ≤ kxk3 ≤ bβkxk1 . 5. Since (by the previous question) being equivalent is an equivalence relation (!), we only need to show that k · k2 and k · k∞ are equivalent to k · k1 . Examples I 3 We have n X |xj |2 ≤ j=1 2 |xj | ≤ n n X j=1 which gives kxk2 ≤ kxk1 ≤ √ |xj |2 , j=1 n kxk2 , and max |xj | ≤ j=1,...,n n X n X j=1 |xj | ≤ n max |xj |, j=1,...,n i.e. kxk∞ ≤ kxk1 ≤ nkxk∞ . 6. First we show that if f is continuous and K is closed then f −1 (K) = {x ∈ X : f (x) ∈ K} is closed. (In fact this is in the proof of Corollary 3.12 in the notes.) Take xn ∈ f −1 (K), and suppose that xn → x; since f is continuous, f (xn ) → f (x). But f (xn ) ∈ K, and K is closed, so f (x) ∈ K. It follows that x ∈ f −1 (K), and so f −1 (K) is closed. Now suppose that whenever K is closed, f −1 (K) is closed. Take xn ∈ X with xn → x, and suppose that f (xn ) 6→ f (x). Then for some subsequence xnj , we must have |f (xnj ) − f (x)| > . So f (xnj ) is contained in the closed set Z = Y \ B(f (x), ), where B(f (x), ) = {y ∈ Y : ky − f (x)kY < } is open. It follows that f −1 (Z) is closed. Since f (xnj ) ∈ Z, xnj ∈ f −1 (Z); since xnj → x and f −1 (Z) is closed, it follows that x ∈ f −1 (Z), so that |f (x) − f (x)| > , clearly a contradiction. 7. Let Y be closed subset of the compact set K. If yn ∈ Y then yn ∈ K, so there is a subsequence ynj such that ynj → y ∈ K. Since Y is closed, y ∈ Y , and so Y is compact. 8. If {un } is a Cauchy sequence in U then, since U ⊂ V and the norm on U is simply the restriction of the map u 7→ kukV to U , it follows that {un } is also a Cauchy sequence in V . Since V is complete, un → u for some u ∈ V . But since U is closed, we know that if un → u then u ∈ U . So U is complete. 4 1 Examples I 9. As it stands the question is ambiguous, since one needs to specify a norm on c0 (K). To be interesting the question requires the `∞ norm; any element of `p with p < ∞ must be an element of c0 (K). To show that c0 (K) (with the `∞ norm) is complete, we will first show that `∞ (K) is complete (this case was omitted from the proof of Proposition 4.5 in the notes, put the proof is simpler), and then that c0 (K) is closed, which will show that c0 (K) is complete by the previous question. So first suppose that xk = (xk1 , xk2 , · · · ) is a Cauchy sequence in `∞ (K). Then for every > 0 there exists an N such that kxn − xm kp`∞ = sup |xnj − xm j |< for all n, m ≥ N . (1.1) j In particular {xkj }∞ k=1 is a Cauchy sequence in K for every fixed j. Since K is complete (recall K = R or C) it follows that for each k ∈ N xkj → ak for some ak ∈ R. Set a = (a1 , a2 , · · · ). We want to show that a ∈ `∞ and also that kxk − ak`∞ → 0 as k → ∞. First, since {xk } is Cauchy we have from (1.1) that kxn − xm k`∞ < for all n, m ≥ N , and so sup |xnj − xm j | ≤ . j Letting m → ∞ we obtain sup |xnj − aj | ≤ , j and so xk − a ∈ `∞ . But since `∞ is a vector space and xk ∈ `∞ , this implies that a ∈ `∞ and kxk − ak`∞ ≤ . / c0 (K). Now take xn ∈ c0 (K) such that xn → x in `∞ . Suppose that x ∈ Then there exists a δ > 0 and a sequence nj → ∞ such that |xnj | > δ for every j. Now choose N large enough that kxn − xk`∞ < δ/2 for all n ≥ N. In particular it follows that |xnnj | > δ/2 for all n ≥ N (and every j); but then xn ∈ / c0 (K), a contradiction. 2 Examples II 1. If x = (x1 , x2 , . . .) ∈ `2 (R), then ∞ X |xj |2 < ∞. j=1 So kx − (x1 , x2 , . . . , xn , 0, 0, . . .)k2`2 = ∞ X |xj |2 → 0 j=n+1 as n → ∞. So `f (R) is dense in `2 (R). 2. If x ∈ X ∩ Y then there exist X and Y such that {y ∈ B : ky − xk < X } ⊂ X and {y ∈ B : ky − xk < Y } ⊂ Y. Taking = min(X , Y ) it follows that {y ∈ B : ky − xk < } ⊂ X ∩ Y, and so X ∩ Y is open. Now, given z ∈ B and > 0, since X is dense there exists an x ∈ X such that kx − zk < /2. Since X is open, there is a δ < /2 such that {x0 ∈ B : kx0 − xk < δ} ⊂ X. Since Y is dense, there is a y ∈ Y such that ky − xk < δ/2. By the above, it follows that we also have y ∈ X. So we have found y ∈ X ∩ Y such that ky − zk ≤ ky − xk + kx − zk < δ/2 + /2 < , and so X ∩ Y is dense. A refinement of this argument allows one to prove the powerful Baire Category Theorem: a countable intersection of open and dense sets is dense. 5 6 2 Examples II 3. We have to assume that (Y, k · kY ) is a Banach space. Then if xn ∈ X and xn → x (with x ∈ V ) we know that {xn } is Cauchy in V . So, since kF (xn ) − F (xm )kY ≤ Lkxn − xm k it follows that {F (xn )} is a Cauchy sequence in Y . Since Y is complete, we know that limn→∞ F (xn ) exists and is an element of Y . If xn → v, and yn → v, then kF (xn ) − F (yn )k ≤ Lkxn − yn k. Taking limits as n → ∞ on both sides implies that lim F (xn ) = lim F (yn ). n→∞ n→∞ That F so defined is continuous is clear: for any v1 , v2 ∈ V , one can find sequences in X that converge to v1 , v2 , and then from the definition kF (v1 ) − F (v2 )k ≤ Lkv1 − v2 k. 4. Let V = R, X = Q, and expand each the interval Iq = (q − √ q ∈ Q to √ d(q), q + d(q)), where d(q) = |q − 2|/2. Then 2 ∈ / Iq for any q, so S I does not cover R. q q∈Q (j) 5. Fix n ∈ N, and cover Aj with a collection of intervals Ik P (j) −(n+j) . Then ∪ A is covered by j j k |Ik | < 2 [ (j) Ik , such that j,k (j) P∞ and j,k |Ik | < j=n+1 2−j = 2−n . If Pj occurs almost everywhere then it fails on a set Aj of measure zero. Since the union of the Aj still has measure zero, every Pj occurs simultaneously almost everywhere. 6. Use φn ↑ f to mean that φn is an increasing sequence that converges almost everywhere to f . (i) is essentially Lemma 5.5. (ii) If f, g ∈ Linc (R) then there are sequences {φn }, {ψn } ∈ Lstep (R) such that φn ↑ f and ψn ↑ g. Then {φn + ψn } is an increasing sequence of step functions with φn + ψn ↑ f + g. So f + g ∈ Linc (R), and since the integral is additive on Lstep (R), Z Z Z Z Z Z f + g = lim φn + ψn = lim ( φn + ψn ) = f + g. P n→∞ n→∞ step (iii) If {φn } ∈ Lstep (R) with φn ↑ f Rthen λφn ∈ R L (R) and λφn ↑ f . The result follows from the fact that λφn = λ φn . Examples II 7 (iv) Follows from the same properties for step functions, along similar lines to the above. 7. For x ∈ [n, n + 1) we have Z x Z = 0 n x Z f (r) dr + f (r) dr = 1 n n−1 X j=1 (−1)n (−1)j + (x − n) , j n and so x Z lim x→∞ 0 f (x) dx = ∞ X (−1)j j! j=1 < ∞. But the same calculation with |f (x)| gives Z x Z ≥ 0 n |f (r)|, dr = 1 n−1 X j=1 1 , j which diverges as n → ∞. [Functions can be ‘integrable’ in some sense even when they are not ‘Lebesgue integrable’.] R 8. Consider fn = n|f |. Then fn is a monotonic sequence, and fn = R n |f | = 0 for every n. So there exists a g ∈ L1 such that fn → g almost everywhere. The limit as n → ∞ of fn is zero where f = 0, and +∞ where fn 6= 0. Since g is defined almost everywhere, it follows that f = 0 almost everywhere. 9. If f ∈ Linc (R) then there exists a sequence {φn } ∈ Lstep (R) such that φn ↑ f (almost everywhere), and we define Z Z f = lim φn . n→∞ Since f ≥ φn , Z Z |f − φn | = Z f− φn , which tends to zero as n → ∞ by the definition of Now take a step function φ(x) = n X R f. cj χIj (x) j=1 with the endpoints of Ij being aj and bj . Consider the ‘continuous version’ 8 2 Examples II of χI (x), for an interval I with endpoints a, b, 0 x<a−δ (x − a + δ)/δ a − δ ≤ x ≤ a XI (x; δ) = 1 a<x<b (b + δ − x)/δ b ≤ x ≤ b + δ 0 x > b + δ. Now if δ < /(n maxj |cj |) and φ (x) = n X cj XIj (x; δ), j=1 then Z |φ (x) − φ(x)| dx < . So if g ∈ L1 (R), we have g = f1 − f2 ∈ Linc (R). Approximate f1 and f2 to within /4 (wrt the L1 norm) by step functions h1 and h2 , and then approximate h1 and h2 by continuous functions c1 and c2 to within /4 (again wrt the L1 norm). So c1 − c2 ∈ C 0 (R) approximates g to within , i.e. C 0 (R) is dense in L1 (R). P 10. With gn = ( nk=1 |fk |)2 we have !2 !2 2 Z Z X n ∞ n n X X X kfk kL2 ≤ kfk kL2 gn = ( |fk |)2 = |fk | ≤ k=1 k=1 L2 k=1 k=1 using the definition of the L2 norm and the triangle inequality. It follows P that gn → g almost everywhere for some g ∈ L1 . It follows that ∞ k=1 fk 2 is absolutely convergent almost everywhere to some f ∈ L . 11. Note that we have !2 2 2 n ∞ ∞ X X X f− fk = fk ≤ |uk | ≤ |f |2 . k=1 k=n+1 k=n+1 So we can use the DCR to deduce that Z n n X X 2 lim |f − fk | = lim |f − uk |2 = 0. n→∞ k=1 n→∞ k=1 3 Examples III 1. Expanding the right-hand side gives (x + y, x + y) − (x − y, x − y) + i(x + iy, x + iy) − i(x − iy, x − iy) = kxk2 + (x, y) + (y, x) + kyk2 − kxk2 + (x, y) + (y, x) − kyk2 + i(kxk2 + i(y, x) − i(x, y) + kyk2 − kxk2 + i(y, x) − i(x, y) − kyk2 ) = 4 real(x, y) + 4 imag(x, y) = 4(x, y) 2. Write Z 2 t |f (r)| dt t Z 2 −1/2 1/2 r |f (r)| dr Z t Z t −1 2 ≤ r dr r|f (r)| dr = r s s s s 1/2 ≤ K(log t − log s) using the Cauchy-Schwarz inequality. [K 2 in examples sheet is a misprint.] 3. Suppose that E is orthonormal and that n X αj e j = 0 j=1 for some αj ∈ K and ej ∈ E. Then taking the inner product of both sides with ek (with k = 1, . . . , n) gives n X ( αj ej , ek ) = 0 ⇒ αk = 0, j=1 so αk = 0 for each k, i.e. E is linearly independent. 9 10 3 Examples III 4.(i) Suppose that d = 0. Then there exists a sequence yn ∈ Y such that lim kx − yn k = 0, n→∞ i.e. yn → x. Since Y is a closed linear subspace, it follows that x ∈ Y – but this contradicts the fact that x ∈ / Y. (ii) By definition there exists a sequence yn ∈ Y such that d = limn→∞ kx− yn k. So for n sufficiently large, clearly kx − yn k < 2d. Let z be one such yn . (iii) We have kx̂ − yk = x−z 1 1 1 −y = x − (z + kx − zky) > d= , kx − zk kx − zk 2d 2 since z + kx − zky ∈ Y and d = inf{kx − yk : y ∈ Y }. 5. Take x1 ∈ B with kx1 k = 1. Let Y1 be the linear subspace {αx1 : α ∈ K}. Using the result of question 7 there exists an x2 ∈ B with kx2 k = 1 such that kx2 − x1 k > 12 . Now let Y2 = Span(x1 , x2 ), and use question 7 to find x3 ∈ B with kx3 k = 1 with kx3 − yk > 21 for every y ∈ Span(x1 , x2 ) – in particular 1 1 kx3 − x1 k > and kx3 − x2 k > . 2 2 Continuing in this way we obtain a sequence {xn } ∈ B such that kxn k = 1 and 1 kxn − xm k > 2 for all n 6= m. It follows that the unit ball is not compact, since no subsequence of the {xn } can be a Cauchy sequence. 6. The unit ball is closed and bounded. If B is finite-dimensional then this implies that the unit ball is compact. We have just shown, conversely, that if B is infinite-dimensional then the unit ball is not compact. 7. Take x, y ∈ A and suppose that λx + (1 − λ)y ∈ A for all λ ∈ [0, 1] such that 2k λ ∈ N. Then given µ with 2k+1 µ ∈ N, 2k µ ∈ / N we have µ = l2−(k+1) with l odd, so that µ= (l − 1) −(k+1) (l + 1) −(k+1) 2 + 2 . 2 2 With λ1 = (l − 1)2−(k+1) and λ2 = (l + 1)2−(k+1) we have 2k λ1 , 2k λ2 ∈ N, and so 1 µx + (1 − µ)y = [(λ1 x + (1 − λ1 )y) + (λ2 x + (1 − λ2 )y)] , 2 Examples III 11 with λj x + (1 − λj )y ∈ A, i.e. µx + (1 − µ)y ∈ A. It follows by induction that λx + (1 − λ)y ∈ A for any λ ∈ [0, 1] such that 2k λ ∈ N for some k ∈ N. Since any λ ∈ [0, 1] can be approximated by a sequence λj such that 2j λj ∈ N, it follows since A is closed that λx + (1 − λy) ∈ A for all λ ∈ [0, 1], i.e. A is convex. 8. We have x ∈ (X ⊥ )⊥ if for every y ∈ X ⊥ , (x, y) = 0. So clearly X ⊆ (X ⊥ )⊥ . If we do not have equality here then there exists a z ∈ / X with (z, y) = 0 for every y ∈ X ⊥ . But we know that z = x + ξ with x ∈ X and ξ ∈ X ⊥ ; but then (x + ξ, ξ) = 0, which implies that kxik2 = 0, i.e. x ∈ X. So (X ⊥ )⊥ = X as claimed. 9. We have !r r r !r 5 5 3 3 e04 = x3 − x3 , (3x2 − 1) (3x2 − 1) − x3 , x x 8 8 2 2 1 1 √ − x3 , √ 2 2 Z 1 Z Z 5 3x 1 4 1 1 3 = x3 − (3x2 − 1) (3t5 − t3 ) dt − t dt − t dt 8 2 −1 2 −1 −1 3x . = x3 − 5 Then ke04 k2 7 1 3t 2 t 6t5 3t3 8 3 = t − dt = − + = 5 7 25 25 −1 7 × 25 −1 Z 1 and so r e4 = 7 (5x3 − 3x). 8 10. Approximation of sin x by a third degree polynomial f3 (x): note that (sin x, e1 ) and (sin x, e3 ) will be zero since sin x is odd and e1 and e3 are even, so Z Z 3x 1 7(5x3 − 3x) 1 3 f3 (x) = t sin t dt + (5t − 3t) sin t dt. 2 −1 8 −1 Now, Z 1 t sin t dt = −1 [−t cos t]1−1 Z 1 + −1 cos t dt = [sin t]1−1 = 2 sin 1 12 3 Examples III and Z 1 1 t sin t dt = −t3 cos t −1 + 3 −1 1 Z 1 = 3t2 sin t −1 − 6 3t2 cos t dt −1 1 Z t sin t dt −1 = −6 sin 1, giving 1 Z (5t3 − 3t) sin t dt = −36 sin 1; −1 it follows that 195x − 315x3 7(27x − 45x3 ) = sin 1 . f3 (x) = sin 1 3x + 2 2 11. First way: given orthonormal polynomials ẽj (x) on [−1, 1] put √ ej (x) = 2 ẽj (2x − 1), since then Z 1 Z ej (x)ek (x) dx = 2 0 1 Z 1 ẽj (2x − 1)ẽk (2x − 1) dx = ẽj (y)ẽk (y) dy. −1 0 Doing this gives e1 (x) = 1, e2 (x) = √ 3 (2x − 1), √ e3 (x) = √ 5 [3(4x2 − 4x + 1) − 1] = 5 (6x2 − 6x + 1), 2 and √ 7 5(8x3 − 12x2 + 6x − 1) − 3(2x − 1) √2 = 7 [20x3 − 30x2 + 12x − 1]. e4 (x) = Or the painful way: e1 (x) = 1, then Z 0 e2 (x) = x − 0 1 1 t dt = x − , 2 and then ke02 k2 = Z 0 1 " (x − 12 )3 (x − 12 )2 dx = 3 #1 = 0 1 12 Examples III 13 √ 3 (2x − 1). Now set Z 1 Z 0 2 2 e3 (x) = x − 3(2x − 1) t (2t − 1) dt − which gives e2 (x) = 1 t2 dt 0 0 4 t 1 1 t3 = x − 3(2x − 1) − − 2 3 0 3 2x − 1 1 1 = x2 − − = x2 − x + , 2 3 6 2 and then ke03 k2 1 Z 1 (t − t + )2 dt = 6 2 = 0 Z 0 1 t4 − 2t3 + 4t2 t 1 − + dt 3 3 36 1 t = − + − + 5 2 9 6 36 0 1 , = 180 √ and so e3 (x) = 5(6x2 − 6x + 1). Finally, we have Z 1 0 3 2 3 2 e4 (x) = x − 5(6x − 6x + 1) t (6t − 6t + 1) dt 0 Z 1 Z 1 3 −3(2x − 1) t (2t − 1) dt − t3 dt t5 t4 4t3 t2 0 0 9 1 6x2 − 6x + 1 − (2x − 1) − = x3 − 4 20 4 20x3 − 30x2 + 12x − 1 = 20 with Z 1 1 = (20t3 − 30t2 + 12t − 1)2 dt 400 0 Z 1 1 = 400t6 + 900t4 + 144t2 + 1 − 1200t5 + 480t4 − 40t3 − 720t3 400 0 2 +60t − 24t dt 1 400 900 144 1200 480 40 720 60 24 = + + +1− + − − + − 400 7 5 3 6 5 4 4 3 2 1 = , 400 × 7 √ and so e4 (x) = 7 (20x3 − 30x2 + 12x − 1) as above. ke04 k2 14 3 Examples III 1/2 1/2 12. Given x, y ∈ `2w , set x̂j = wj xj and ŷj = wj yj . Then (x, y)`2w = (x̂, ŷ)`2 , so the fact that (·, ·)`2w is an inner product on `2w follows from the fact that (·, ·)`2 is an inner product on `2 . Similarly the completeness of `2w follows from the completeness of `2 – if {xn } is a Cauchy sequence in `2w then {x̂n } is a Cauchy sequence in `2 , so there exists a y ∈ `2 such that 1/2 x̂n → y. Setting zj = yj /wj gives a z ∈ `2w such that xn → z in `2w . 4 Examples IV 1. If kxkX ≤ 1 then kAxkY = kxkX A x kxkX ≤ A x , kxkX and so since k(x/kxkX )kX = 1 we have sup kAxkY ≤ sup kAzkY ≤ sup kAzkY . kxkX ≤1 kzkX =1 kzkX ≤1 Rearranging the above equality we have x kAxkY = A kxkX kxkX , Y and so, again since k(x/kxkX )kX = 1, sup x6=0 kAxkY = sup kAxkY . kxkX kxk=1 2. Given x ∈ `2 , we have kT xk2`2 = X |xj |2 j j2 ≤ X |xj |2 , j so kT k ≤ 1. Now, for any fixed n one can take xn = (1, · · · , 1, 0, · · · ), an element of `2 whose first n entries are 1 and whose others are zero. Then T xn = (1, 1/2, 1/3, . . . , 1/n, 0, · · · ) which is in `2 . {T xn } forms a sequence in `2 that converges to (1, 1/2, 1/3, 1/4, . . .), P which is an element of `2 since n (1/n2 ) is finite. However, the ‘preimage’ 15 16 4 Examples IV of this would be the sequence consisting all of 1s, which is not an element of `2 . So the range of T is not closed. 3. Clearly fφ is linear. Given u ∈ C 0 ([a, b]) we have Z b φ(t)u(t) dt |fφ (u)| = a Z b ≤ max |u(t)| |φ(t)| dt t∈[a,b] a b Z |φ(t)| dt, = kuk∞ (4.1) a Rb and so kfφ k ≤ a |φ(t)| dt. Now since φ is continuous we can find a sequence of continuous functions un that approximate the sign of φ (+1 if φ > 0, −1 if φ < 0) increasingly closely in the L2 norm: if φ(s) 6= 0 then there is an interval around s on which φ has the same sign. Since this interval has non-zero length, and the union of these intervals lies within [a, b], there are at most a countable number, {Ij }∞ j=1 . Let un be the continuous function defined on Ij = (lj , rj ) as (t − lj )/δj,n lj ≤ t ≤ lj + δj,n un (t) = sign(φ(t)) × 1 lj + δj,n < t < rj − δj,n (rj − t)/δj,n rj − δj,n ≤ t ≤ rj where δj,n = min(2−(n+j) , (rj − lj )/2), and zero when φ(t) = 0. Then Z b |un (t) − sign(φ(t))|2 dt ≤ a ∞ X 2−(n+j) = 2−(n−1) , j=1 i.e. un → sign(φ) in L2 (a, b). Note that kun k∞ = 1. It follows that Z b φ(t)un (t) dt |fφ (un )| = a Z = ≥ b Z |φ(t)| dt − φ(t)(un (t) − sign(φ(t))) dt a Z b ≥ a Z a a Z b φ(t)(un (t) − sign(φ(t))) dt |φ(t)| dt − ≥ b b |φ(t)| dt − kφkL2 kun − sign(φ)kL2 . a Since φ ∈ C 0 ([a, b]) we know that kφkL2 < ∞, and we have just shown Examples IV 17 that kun − sign(φ)kL2 → 0 as n → ∞. It follows since kun k∞ = 1 that we cannot improve the bound in (4.1) and therefore Z b |φ(t)| dt. kfφ k = a 4. Take x, y ∈ L2 (0, 1), then Z 1 Z t (T x, y) = K(t, s)x(s) ds y(t) dt 0 Z 0 1Z t K(t, s)x(s)y(t) ds dt = 0 Z 0 1Z 1 = K(t, s)x(s)y(t) dt ds Z 1 Z 1 K(t, s)y(t) dt x(s) ds = 0 0 t t = (x, T ∗ y) (draw a picture to see how the limits of integration change) where Z 1 T ∗ (y)(s) = K(t, s)y(t) dt. t 5. If x ∈ (range(T ))⊥ then (x, y) = 0 for all y ∈ range(T ), i.e. (x, T z) = 0 for all z ∈ H. Since (T ∗ )∗ = T , this is the same as (T ∗ x, z) = 0 for all z ∈ H. This implies that T ∗ x = 0, i.e. that x ∈ Ker(T ∗ ). This argument can be reversed, which gives the required equality. Now – with apologies for the mistake in the question: we will show that λ̄ is an eigenvalue of T ∗ – suppose that range(T − λI) 6= H. Then there exists some non-zero u ∈ (range(T − λI))⊥ . By the equality we have just proved, this gives a non-zero u in Ker((T − λI)∗ ), i.e. a non-zero u such that (T − λI)∗ u = 0. Since (T − λI)∗ = T ∗ − λ̄I, this gives a non-zero u with T ∗ u = λ̄u. So λ̄ is an eigenvalue of T ∗ . 18 4 Examples IV 6.(i) If T ∈ B(H, H) then T ∗ ∈ B(H, H) and kT ∗ kop = kT kop . So if {xn } is a bounded sequence in H, with kxn k ≤ M , say, it follows that kT ∗ xn k ≤ kT ∗ kop kxn k ≤ M kT ∗ kop . So {T ∗ xn } is also a bounded sequence in H. Since T is compact, it follows that {T (T ∗ xn )} = {(T T ∗ )xn } has a convergent subsequence, which shows that T T ∗ is compact. (ii) Following the hint, we have kT ∗ xk2 = (T ∗ x, T ∗ x) = (T T ∗ x, x) ≤ kT T ∗ xkkxk. Now, if {T T ∗ xn } is Cauchy then given any > 0, there exists a N such that for all n, m ≥ N , kT T ∗ xn − T T ∗ xm k = kT T ∗ (xn − xm )k ≤ . It follows that for all n, m ≥ N , kT ∗ xn − T ∗ xm k2 = kT ∗ (xn − xm )k2 ≤ kT T ∗ (xn − xm )kkxn − xm k. Since {xn } is a bounded sequence, with kxn k ≤ M , say, we have kT ∗ xn − T ∗ xm k2 ≤ M for all n, m ≥ N. It follows that {T ∗ xn } is Cauchy. (iii) So suppose that {xn } is a bounded sequence in H. Part (i) shows that T T ∗ is compact, so {T T ∗ xn } has a subsequence {T T ∗ xnj } that is Cauchy. But part (ii) shows that this implies that {T ∗ xnj } is Cauchy too. So T ∗ is compact. 7.(i) If for every z ∈ H we have (xn , z) → (x, z) (xn , z) → (y, z) and then clearly (x, z) = (y, z) for every z ∈ H. But then (x − y, z) = 0 for every z ∈ H; in particular we can take z = x − y, which shows that kx − yk2 = 0, i.e. that x = y. (ii) Take y ∈ H, and suppose that (en , y) does not tend to zero as n → ∞. Then for some > 0, thee exists a sequence nj → ∞ such that |(enj , y)| > . But then X X kyk2 = |(y, ej )|2 ≥ |(y, enj )|2 = ∞, n j contradicting the fact that y ∈ H with {ej } an orthonormal basis. Examples IV 19 (iii) For some fixed z ∈ H, consider the map f : H → K given by u 7→ (T u, z). This map is clearly linear, u + λv 7→ (T (u + λv), z) = (T u, z) + λ(T v, z), and it is bounded, |(T u, z)| ≤ kT ukkzk ≤ kT kop kzk kuk. So f ∈ H ∗ . It follows that there exists a y ∈ H such that (T u, z) = f (u) = (u, y) for every u ∈ H. So if un * u, in particularly for this choice of y we have (un , y) → (u, y) ⇒ (T un , z) → (T u, z). Since this holds whatever our choice of z, it follows that T un * T u. (iv) **This part requires you to know that any weakly convergence sequence is bounded, which is not at all obvious (but true). Suppose that T xn 6→ T x. Then there exists a δ > 0 and a subsequence T xnj such that kT xnj − T xk > δ for all j = 1, 2, 3, . . . . (4.2) Since T is compact and {xnj } is bounded there exists a further subsequence such that T xnj → w for some w ∈ H. But since convergence implies weak convergence, we must have T xnj * w. But we already know that T xnj * T x, so it follows from the uniqueness of weak limits that w = T x, and so T xnj → T x, contradicting (4.2). 8. (i) For an x with K(x, ·) ∈ L2 (a, b), use the fact that {φj } is a basis for L2 to write ∞ X K(x, y) = ki (x)φi (y). i=1 Since Z ki (x) = b K(x, y)φi (y) dy, a 20 4 Examples IV we have Z b |ki (x)|2 dx = Z b Z 2 b K(x, y)φi (y) dy dx a a Z b Z b Z b 2 2 |K(x, y)| dy |φi (y)| dy dx ≤ a a Z = a a b |φi (y)|2 dy × a Z bZ a b |K(x, y)|2 dy dx, a L2 (a, b). and so ki ∈ P (ii) Since ki ∈ L2 (a, b), we can write ki (x) = j κij φj , where κij = Rb a ki (x)φj (x) dx, i.e. X K(x, y) = κij φi (y)φj (x), i,j and so {φi (y)φj (x)} is a basis for L2 ((a, b) × (a, b)). 9. Clearly n X T nx = λnj (x, ej )ej . j=1 If |λj | < 1 for all j = 1, . . . , n then kT k < 1, and so (I − T )−1 = I + T + T 2 + · · · , i.e. −1 (I − T ) = I+ = I+ n X j=1 n X λj (x, ej )ej + n X λ2j (x, ej )ej + · · · j=1 (λj + λ2j + · · · )(x, ej )ej j=1 = I+ n X j=1 λj (x, ej )ej . 1 − λj 10.(i) Since x = f + αT x, the solution is given by x = (I − αT )−1 f . So if kαT kop < 1 we can write ∞ X x = (I + αT + α T + α T + · · · )f = (αT )j f. 2 2 3 3 j=0 So this expansion is valid for |α| < 1/kT kop . (ii) Suppose that Z b n−1 (T x)(t) = Kn−1 (t, s)x(s) ds. a Examples IV 21 Then n Z b K(t, s)(T n−1 x)(s) ds a Z b Z b K(t, s) Kn−1 (s, r)x(r) dr ds = (T x)(t) = a a Z bZ b = K(t, s)Kn−1 (s, r)x(r) dr ds Z b Z b K(t, s)Kn−1 (s, r) ds x(r) dr, = a a a a i.e. b Z n Kn (t, r)x(r) dr. (T x)(t) = a with b Z K(t, s)Kn−1 (r, s) ds Kn (t, r) = a as claimed. 11. We have ∞ ∞ X X (T u, v) = λj (u, ej )ej , v = λj (u, ej )(ej , v) j=1 = u, ∞ X j=1 λj (ej , v)ej = u, j=1 ∞ X λj (v, ej )ej j=1 = (u, T v) and so T is self-adjoint. Now consider Tn u defined by Tn u = n X λj (u, ej )ej . j=1 Since the range of T is contained in the span of {ej }nj=1 it is finitedimensional, and so Tn is compact. Since kTn u − T uk2 = 2 ∞ X λj (u, ej )ej j=n+1 = ∞ X j=n+1 |λj |2 |(u, ej )|2 , 22 4 Examples IV we have from the fact that λn → 0 that given any > 0 there exists an N such that for all n ≥ N we have |λj | < , and hence 2 ∞ X 2 kTn u − T uk ≤ 2 2 |(u, ej )| ≤ j=n+1 ∞ X |(u, ej )|2 = 2 kuk2 , j=1 i.e. kTn − T kop ≤ , and so Tn → T . It follows from Theorem 13.8 that T is compact. 12. We have Z b Tu = K(x, y)u(y) dy a Z bX ∞ = λj ej (x)ej (y)u(y) dy a j=1 ∞ X = λj ej (ej , u), j=1 and so T ek = λk ek . To show that these are the only eigenvalues and eigenvectors, if u ∈ P 2 L (a, b) with u = w + ∞ k=1 (u, ek )ek and T u = λu then, since w ⊥ ej for all j, Tu = Tw + = = ∞ Z X ∞ X (u, ek )T ek k=1 b λj ej (x)ej (y)(u, ek )ek (y) dy j,k=1 a ∞ X λj (u, ej )ej (x) j=1 and λu = ∞ X λ(u, ej )ej (x). j=1 Taking the inner product with each ek yields λ(u, ek ) = λk (u, ek ), so either (u, ek ) = 0 or λ = λk . Examples IV 23 (i) Let (a, b) = (−π, π) and consider K(t, s) = cos(t−s). Then K is clearly symmetric, so the corresponding T is self-adjoint. We have cos(t − s) = cos t cos s − sin t sin s; recall that cos t and sin t are orthogonal in L2 (−π, π). We have Z π K(t, s) sin s ds T (sin t) = −π Z π cos t cos s sin s − sin t sin2 s ds = −π = −2π sin t and Z π K(t, s) cos s ds T (cos t) = Z−π π = cos t cos2 s − sin t sin s cos s ds −π = 2π cos t. (ii) Let (a, b) = (−1, 1) and let K(t, s) = 1 − 3(t − s)2 + 9t2 s2 . Then in fact ! r ! ! r ! r 8 3 3 5 2 5 2 K(t, s) = 4 t s + (3t − 1) (3s − 1) , 2 2 5 8 8 q q and so, since { 32 t, 58 (3t2 −1)} are orthonormal (they are some of the Legendre polynomials from Chapter 6) the integral operator associated with K(t, s) has r T (t) = 4t and 8 T (3t2 − 1) = (3t2 − 1). 5 13. This equation is the equation in Theorem 14.2 with p = 1 and q = 0. So it suffices to show that (i) u1 (x) = x satisfies the equation −d2 u/dx2 = 0 with u(0) = 0 (which is clear), (ii) that u2 (x) = (1 − x) satisfies the same equation with u(1) = 0 (which is also clear), and (iii) that Wp (u1 , u2 ) = 1: Wp (u1 , u2 ) = [u01 u2 − u02 u1 ] = 1(1 − x) − (−1)x = 1. Or one could follow the proof of Theorem 4.12 with this particular choice of G to show that u(x) as defined does indeed satisfy −d2 u/dx2 = f . 24 4 Examples IV 14.(i) If x 6= 0 then kxk > 0, from which it follows that kT xk > 0, and so x ∈ / Ker(T ). So Ker(T ) = {0} and T −1 exists for all y ∈ range(T ). Then we have kT (T −1 x)k2 ≥ αkT −1 xk2 , i.e. kT −1 xk2 ≤ α−1 kxk2 , so T −1 is bounded. (ii) Let λ = α + iβ. Then k(T − λI)xk2 = (T x − αx − iβx, T x − αx − iβx) = k(T x − αx)k2 − (T x − αx, iβx) − (iβx, T x − αx) + |β|2 kxk2 = kT x − αxk2 + i(T x − αx, βx) − i(βx, T x − αx) + |β|2 kxk2 . Now, since T is self-adjoint and α is real, it follows that (T x − αx, βx) = (βx, T x − αx), and so k(T − λI)xk2 = kT x − αxk2 + β 2 kxk2 ≥ β 2 kxk2 , i.e. T − λI is bounded below as claimed. (iii) If λ ∈ C and T − λI has a densely defined inverse then (T − λI)−1 ∈ B(H, H), i.e. λ ∈ R(T ). So σc (T ) ⊆ R. (iv) If the range of T − λI is not dense in H then by Question 3 we know that λ̄ is an eigenvalue of T ∗ . If T is self-adjoint, we must have λ̄ an eigenvalue of T . But all eigenvalues of T are real, so in fact λ is an eigenvalue of T . So λ ∈ σp (T ) rather than σr (T ), which implies that σr (T ) is empty. 15.(i) If y ∈ `q then |l(x)| = ∞ X xj yj ≤ kxkp kykq j=1 using Hölder’s inequality. So klkp∗ ≤ kykq as claimed. (ii) Any x = (x1 , x2 , x3 , . . .) ∈ `p can be written as x= ∞ X xj ej , j=1 for the elements ej defined in the question. Since l is bounded it is Examples IV 25 continuous, so n X l(x) = l lim n→∞ = lim l n X n→∞ = = ∞ X j=1 ∞ X xj ej j=1 n X n→∞ xj ej j=1 = lim xj l(ej ) j=1 xj l(ej ) x j yj , j=1 if we define yj = l(ej ). [Should you come across something similar in an exam you should at least mention continuity of l as the justification for taking it inside the infinite sum.] (iii) We have kx(n) kpp = n X |yj |q yj p j=1 = n X |yj |p q − 1 = n X j=1 j=1 and so, as claimed, 1/p n X kx(n) kp = |yj |q . j=1 Also l(x (n) )= n X |yj |q j=1 yj yj = n X |yj |q . j=1 Since |l(x(n) | ≤ klkp∗ kx(n) kp , we have n X j=1 |yj |q ≤ klkp∗ n X j=1 1/p |yj |q , |yj |q , 26 4 Examples IV from which it follows, since 1 − 1/p = 1/q, that 1/q n X |yj |q ≤ klkp∗ . j=1 Since this holds for every n, we must have y ∈ `q with kykq ≤ klkp∗ .