Functional Analysis Exam [MA3001]

Transkript

1 Functional Analysis Exam [MA3001] Technische Universität München Prof Dr Michael Wolf , 10:30-12: , MI HS 1, Friedrich L Bauer Hörsaal (5602EG001) Score Signature (marker) Please fill in with BLOCK CAPITALS (except for the signature) Family name: First name: Matriculation number: Major: Row: Signature: Information: Place: Check whether you have all 6 problems The duration of the exam is 90 minutes You may use one handwritten (on both sides) DIN-A4-sheet (210x297mm) paper Other tools are not allowed You can use the page of each problem as well as the opposite page for the solution If you need more space, please ask for additional paper, and indicate clearly that your solution continues elsewhere Grade /15 /9 /9 /9 /15 /6 /63 Left the room from to Notes: 1

2 (1) Let X be the space of all polynomials of one variable and complex coefficients, and define p c n k=0 (k + 1) a k for every p X, p(x) = n k=0 a kx k a) Show that c is a norm on X (Homogeneity, positivity, triangle inequality: 1 point each) b) Prove that (X, c ) is not complete (3 points) c) Compute the norm of the linear functional T : X C, T (p) p (1), (ie, the derivative of p at 1), when X is equipped with c (4 points) d) Compute the norm of the above linear functional when X is equipped with C(), defined as p C() sup x p(x), p X (2 points) e) Prove or disprove: c and C() are equivalent norms on X (3 points) Total: 15 points (German translation in the footnote 1 ) Solution: a) Positivity: It is obvious from the definition that p c 0 for all p X If p c = 0 for some polynomial p(x) = a 0 + a 1 x + + a n x n then a i = 0 for all i = 1,, n, and so p = 0 Homogeneity: Let p X be given by p(x) = n k=0 a kx k Then λp is given by (λp)(x) = n k=0 λa kx k, and hence λp c = n k=0 (k + 1) λa k = λ n k=0 (k + 1) a k = λ p c Triangle inequality: Let p, q X be given by p(x) = n k=0 a kx k, q(x) = n k=0 b kx k Then p + q is given by (p + q)(x) = n k=0 (a k + b k )x k, and hence p + q c = n k=0 (k + 1) a k + b k n k=0 (k + 1)( a k + b k ) = p c + q c b) Obviously, the polynomials {1} {x n : n N} form a countably infinite algebraic basis in X Since we know that every infinite-dimensional Banach space is more than countable dimensional, (X, c ) is not a Banach space c) Let p(x) = n k=0 a kx k Then T (p) = p n (1) = ka k k=1 n k a k k=1 n (k + 1) a k = p c Since this holds for all p X, we have T 1 On the other hand, for every n N, let p n (x) x n, so that T (p n ) = p n(1) = n Since p n c = n + 1, we have T (p n ) T sup n N p n c n = sup n N n + 1 = 1 1 Sei X der Raum aller Polynome in einer Variablen mit komplexen Koeffizienten und definiere p c n k=0 (k + 1) a k für jedes p X, p(x) = n k=0 a kx k a) Zeigen Sie, dass c eine Norm auf X ist b) Beweisen Sie, dass (X, c ) nicht vollständig ist c) Berechnen Sie die Norm des linearen Funktionals T : X C, T (p) p (1), (sprich, die Ableitung von p an der Stelle 1), wenn X mit der Norm c ausgestattet ist d) Berechnen Sie die Norm des oben gegeben Funktionals, falls X mit der Norm C() ausgestattet ist, die durch p C() sup x p(x), p X definiert ist e) Beweisen oder widerlegen Sie: c und C() sind equivalente Normen auf X k=1 2

3 Thus, T = 1 d) For every n N, let p n (x) x n Then T (p n ) = n, while p n C() = 1 Thus, T (p n ) n T sup = sup n N p n C() n N 1 = +, and hence T = + e) Since T is bounded wrt c and unbounded wrt C(), it follows immediately that the two norms cannot be equivalent 3

4 (2) Let X C([0, 1]) be the space of real-valued continuous functions on [0, 1], equipped with the norm f 2 ( f(t) 2 dt) 1/2, f X Let F (f) (f(t)) 2 dt, f X Determine the (Fréchet-)derivative (DF )(f) of F at any point f X, and compute the norm of (DF )(f) (9 points) (German translation in the footnote 2 ) Solution: Let f, h X Then F (f + h) F (f) = (f(t) + h(t)) 2 dt (f(t)) 2 dt = 2h(t)f(t) + (h(t)) 2 dt The part linear in h is 2h(t)f(t) dt, so we guess (DF )(f)(h) = 2h(t)f(t) dt We have (DF )(f)(h) 2 f 2 h 2 by the Cauchy-Schwarz inequality, and thus (DF )(f) is bounded with (DF )(f) 2 f 2 When f = 0, this gives (DF )(f) = 0 When f 0, choose h f/ f 2, so that h 2 = 1, and (DF )(f)(h) = 2 (f(t)) 2 dt = 2 f 2 2 f 2 f = 2 f 2 2 So we get that for all f X, (DF )(f) = 2 f 2 Finally, F (f + h) F (f) (DF )(f) = and thus (h(t)) 2 dt = h 2 2, F (f + h) F (f) (DF )(f) h 2 2 lim = lim = lim 0< h 0 h 2 0< h 0 h h 2 = 0, 2 0< h 2 0 proving that indeed h 2h(t)f(t) dt is the derivative of F at f 2 Es sei X C([0, 1]) der Raum der reellwertigen stetigen Funktionen auf [0, 1], ausgestattet mit der Norm f 2 ( f(t) 2 dt) 1/2, f X Sei F (f) (f(t)) 2 dt, f X Bestimmen Sie die (Fréchet-)Ableitung (DF )(f) von F an jedem Punkt f X und berechnen sie die Norm (DF )(f) 4

5 (3) Let c 00 { x K N : #{n : x(n) 0} < + } be the space of sequences that take zero value at all but finitely many indeces Recall that the p-norms are defined on c 00 as x p ( n N x(n) p) 1/p for all 1 p < + Define the linear functional f(x) + 1 n α x(n), x c 00, where α is a fixed real number For what values of α is this linear functional bounded, if c 00 is equipped with 3? (You can use without proof that c 00 is dense in l p (N, K)) (9 points) (German translation in the footnote 3 ) Solution: The Hölder conjugate q of p = 3 is given by 1 = 1 1 = 2, ie, q = 3 Define q y K N by y(n) 1 Then n α { + + y(n) q 2α 1 = +, if = 1, 3 n 2α 2α 3 < +, if > 1 3 Thus, if α > 3/2, f y (x) + y(n)x(n) defines a bounded linear functional on l 3(N, K) Since (c 00, 3 ) is a normed subspace of (l 3, 3 ), and f is the restriction of f y onto c 00, f is bounded ( NM ) 1/3 If α 3/2 then for every M > 0 there exists an N M such that y(n) 3 > M As we have seen in the exercise class, there exists an x M c 00 such that x M (n) = 0 for n > N M, x M 3 = 1 and ( NM f(x M ) f sup M N x M Hence, f is not bounded y(n) 3 ) 1/3 = NM x M (n)y(n) = f(x) Thus, sup M = + M N 3 Es sei c 00 { x K N : #{n : x(n) 0} < + } der Raum der Folgen die den Wert 0 an allen bis auf endlich vielen Indizes annehmen Die p-normen sind auf c 00 definiert als x p ( n N x(n) p) 1/p für alle 1 p < + Definiere das lineare Funktional f(x) + 1 n α x(n), x c 00, wobei α eine feste reelle Zahl ist Für welche Werte von α ist dieses lineare Funktional beschränkt, falls c 00 mit 3 ausgestattet ist? (Sie können ohne Beweis verwenden, dass c 00 dicht in l p (N, K) liegt) 5

6 (4) Let (X, ) be a Banach space a) Let M X a bounded set Determine the interior of M in the weak topology (5 points) b) Let S be a subspace in X that is closed in the norm topology, and let (x n ) n N be a weakly convergent sequence in S Show that the weak limit of (x n ) n N is in S (4 points) (Total: 9 points) (German translation in the footnote 4 ) Solution: a) Assume first that X is infinite-dimensional We have seen in the lecture that in this case every non-empty weakly open set is unbounded Hence, the largest weakly open set contained in a bounded set is the empty set By the definition of the interior, this means that the interior of M in the weak topology is empty When X is finite-dimensional, the weak topology and the norm topology coincide, and hence the weak interior is the same as the interior in the norm topology (+2 points extra for the discussion of the finite-dimensional case) b) Since S is convex, its weak closure coincides with its norm closure, as we have seen in the lecture Since S is norm closed, this means that S is also weakly closed, and hence it contains the (weak) limit of all weakly convergent sequence in S 4 Sei (X, ) ein Banachraum a) Es sei M X eine beschränkte Menge Bestimmen sie das Innere von M in der schwachen Topologie b) Es sei S ein Unterraum von X der in der Normtopologie abgeschlossen ist und (x n ) n N sei eine schwach konvergente Folge in S Zeigen sie das der schwache Grenzwert von (x n ) n N in S liegt 6

7 (5) Let 1 p < +, and let l p l p (N, C) be equipped with its usual norm x p ( + x(n) p) 1/p Define the left shift operator S : lp l p by (Sx)(n) x(n + 1), n N, x l p a) Compute the norm of S (4 points) b) Determine the point spectrum of S (3 points) c) For every λ on the complex unit circle, and every N N, let x λ,n (λ, λ 2,, λ N, 0, ), ie, x λ,n = N λn e n, where e n (m) δ n,m Compute (λi S)x λ,n and its norm, and use this to determine the continuous spectrum of S (7 points) d) Determine the residual spectrum of S (1 point) (Total: 15 points) (German translation in the footnote 5 ) Solution: a) For every x l p, ( + ) 1/p ( + ) 1/p ( + ) 1/p Sx p = (Sx)(n) p = x(n) p x(n) p = x p, n=2 and thus S 1 On the other hand, for every n 2, e n p = 1 and Se n p = e n 1 p = 1, showing that S 1 Hence, S = 1 b) Assume that Sx = λx for some λ C and x C N Then x(2) = λx(1), x(3) = λx(2), yields that x = c(1, λ, λ 2, ) = c + λn e n For this x we have + x(n) p = c p + λ p, that is finite if and only if λ < 1 Thus, σ p (S) = B 1 (0) = {λ C : λ < 1} c) We have (λi S)x λ,n = (0,, 0, λ N+1, 0 ) = λ N+1 e N, and thus ran(λi S) span{e N : N N} = c 00 Since c 00 is dense in l p, we see that ran(λi S) is dense in l p Moreover, λ / σ p (S), and hence either λ is in the continuous spectrum, or it is in the resolvent However, we have (λi S)x λ,n p = 1, while x λ,n p = N 1/p Thus, if λi S is invertible then x λ,n p = (λi S) 1 (λi S)x λ,n p, implying that x (λi S) 1 λ,n p sup = sup N 1/p = +, N N (λi S)x λ,n p N N a contradiction Hence, λi S is not invertible, and by the above, λ σ c (S) This shows that {λ : λ = 1} σ c (S) On the other hand, {λ C : λ < 1} = σ p (S), while S = 1 yields σ(s) {λ : λ 1} Putting these together, we get {λ : λ = 1} = σ c (S) 5 Sei 1 p < + und l p l p (N, C) sei mit seiner gewöhnlichen Norm x p ( + x(n) p ) 1/p ausgestattet Definiere den Linksshiftoperator S : l p l p durch (Sx)(n) x(n + 1), n N, x l p a) Berechnen Sie die Norm von S b) Bestimmen sie das Punktspektrum von S c) Für jedes λ auf dem komplexen Einheitskreis und jedes N N, sei x λ,n (λ, λ 2,, λ N, 0, ), sprich x λ,n = N λn e n, wobei e n (m) δ n,m Berechnen Sie (λi S)x λ,n und dessen Norm Benutzen Sie dies um das kontinuierliche Spektrum von S zu bestimmen d) Bestimmen Sie das Residualspektrum von S 7

8 d) By the above, σ r (S) = σ(s) \ (σ p (S) σ c (S)) {λ : λ 1} \ (σ p (S) σ c (S)) = 8

9 (6) Let H be a Hilbert space, and let V B(H) be a contraction, ie, V 1 Prove that ker(i V ) = ker(i V ) (Hint: Prove the inclusion in one direction) (German translation in the footnote 6 ) (6 points) Solution: Assume that x ker(i V ), ie, V x = x Then 0 (I V )x 2 = x 2 V x, x x, V x + V x 2 = x 2 x, V x V x, x + V x 2 = x 2 x, x x, x + V x 2 = V x 2 x 2 0, where the last inequality is due to the fact that V = V 1 Hence, (I V )x 2 = 0, ie, x ker(i V ) This proves ker(i V ) ker(i V ) Since this is true for every contraction V, applying it to V yields ker(i V ) ker(i V ) = ker(i V ), finishing the proof 6 Sei H ein Hilbertraum und sei V B(H) eine Kontraktion, das heißt V 1 Beweisen Sie, dass ker(i V ) = ker(i V ) (Hinweis: Beweisen Sie die Inklusion in eine Richtung) 9