Functional Analysis F3/F4/NVP (005 Homework assignment 3 All students should solve the following problems: 1. Section 4.8: Problem 8.. Section 4.9: Problem 4. 3. Let T : l l be the operator defined by T x = ( ξ 1 + ξ, ξ, ξ 3 + ξ 4, ξ 4, ξ 5 + ξ 6, ξ 6, ξ 7 + ξ 8,..., x = (ξ j l. Determine the four sets ρ(t, σ p (T, σ c (T, σ r (T. 4. Prove the following (partial converse to the spectral theorem for compact, self-adjoint operators: If H is a Hilbert space, {e n } is a total orthonormal sequence in H, and {λ n } is a sequence of real numbers such that lim n λ n = 0, then there exists a unique bounded linear operator T : H H such that T (e n = λ n e n for all n. This operator is compact and self-adjoint. Students taking Functional Analysis as a 6 point course should also solve the following problems: 5. Section 8.: Problem 8. 6. Let H be the Hilbert space L (, and let T : H H be the multiplication operator T x(t = sin(t x(t. Verify that T is a bounded self-adjoint operator, and find its spectral family. Solutions should be handed in by Thursday, March 3, 16.00. (Either give the solutions to me directly or put them in my mailbox, third floor, House 3, Polacksbacken. 1
Functional Analysis F3/F4/NVP Comments to homework assignment 3 (to be updated! 1.Many students claim that if (x n is a weak Cauchy sequence in X then there must exist some x in X such that (x n converges weakly to x. This is not true (even if X is a Banach space. Example: Let X = c 0 (see p. 70, exercise 1 and let a 1, a, a 3,... be any bounded sequence of complex numbers. Then the following is a weak Cauchy sequence in X = c 0 : x 1 = (a 1, 0, 0, 0,..., x = (a 1, a, 0, 0, 0,..., x 3 = (a 1, a, a 3, 0, 0, 0,..., x 4 = (a 1, a, a 3, a 4, 0, 0, 0,... (Proof: one uses the fact given in exercise 8 on p. 16. In explicit terms this exercise tells us: For every f (c 0 there is some (b n l 1 such that f((ξ n = b nξ n. It follows that if x 1, x, x 3,... is the sequence in c 0 given above then lim j f(x j = b na n. Hence x 1, x, x 3,... is indeed weak Cauchy. However, for most choiches of numbers a 1, a, a 3,..., the above sequence (x n does not converge weakly to an element x c 0.
3 Functional Analysis F3/F4/NVP Solutions to homework assignment 3 1. Assume that (x n is a weak Cauchy sequence in X. Take an arbitrary f X. Then (f(x n is a Cauchy sequence in K, and hence it is a bounded sequence. Hence, for every f X there exists some constant c f such that f(x n c f for all n. Now we imitate the argument from p. 58 in the book (part (c: For every x X we define g x X as usual by g x (f = f(x for all f X (cf. p. 39 and Lemma 4.6-1. We now have for all n, g xn (f = f(x n c f. This tells us that the sequence (g xn in X = B(X, K is pointwise bounded. We also know that X is a Banach space by Theorem.10-4. Hence the Uniform Boundedness Theorem (Theorem 4.7-3 can be applied, and this gives that there exists a constant c such that g xn c for all n. But x n = g xn by Lemma 4.6-1; hence we have proved that x n c for all n, i.e. the sequence (x n is bounded.. Let X be an arbitrary normed space, and let (f n be a sequence in X. Assume that (f n converges weakly to f X. We wish to prove that (f n is weak* convergent to f. Let x be an arbitrary vector in X. Let g x X be the image of x under the canonical mapping of X into X (cf. p. 39 40, i.e. g x (h = h(x, h X. Since g x X and (f n converges weakly to f X, we have lim n g x (f n = g x (f. This means lim n f n (x = f(x. Since this holds for all x X, (f n is weak* convergent to f. Now assume that X is reflexive. We wish to prove that weak* convergence in X implies weak convergence in X. Let (f n be an arbitrary sequence in X which is weak* convergent to some f X. Let g be an arbitrary element in X. Then since X is reflexive, there is an element x X such that g x = g. Since (f n is weak* convergent to f we have lim n f n (x = f(x. This can also be written: lim n g x (f n = g x (f. But g x = g, hence we have lim n g(f n = g(f. But g X was arbitrary; hence (f n is weakly convergent to f.
4 3. T is a bounded linear operator l l, since, for all (ξ n l, T ((ξ n = ξ 1 + ξ + ξ + ξ 3 + ξ 4 + ξ 4 +... (1 + 1 ( ξ 1 + ξ + ξ + (1 + 1 ( ξ 3 + ξ 4 + ξ 4 +... 3 ξ n = 3 (ξ n. (This also shows T 3. In the computation above we used Schwarz inequality on C, which says the following: ( a 1 b 1 + a b ( a 1 + a ( b 1 + b, a 1, a, b 1, b C. (We used this with a 1 = a = 1 and b 1 = ξ 1, b = ξ, etc. To get started, we investigate when T λ = T λi is injective. Let us write x = (ξ n l and y = (T λix = (η n l. Then ( y = (η n = ((1 λξ 1 + ξ, (1 λξ, (1 λξ 3 + ξ 4, (1 λξ 4,... Now if λ = 1 then T x = 0 for every vector x = (ξ n l with ξ = ξ 4 = ξ 6 =... = 0, and there exist many such vectors which are 0, e.g. x = (1, 0, 0, 0,... This shows that if λ = 1 then T λi is not injective; hence 1 σ p (T. From now on we assume λ 1. Then for any x = (ξ n l, formula (** implies: ξ 1 = 1 η 1 1 η ( ξ = 1 η ( ξ 3 = 1 η 3 1 η ( 4 ξ 4 = 1 η 4... Now if some x = (ξ n l would give T x = T x = (η n then each ξ n is given by the same formula as ξ n in (***, i.e. ξ n = ξ n for all n and hence x = x. This proves that T λi is injective (under our present assumption λ 1. From the above computations we also obtain explicit formulas for (T λi 1 and D ((T λi 1 : D ( { } (T λi 1 = (η n l if ξ n is computed using (*** then ξ n < and (T λi ((η 1 n = (ξ n, (η n D ( (T λi 1.
To decide whether λ belongs to σ c (T or σ r (T or ρ(t we must determine whether D ((T λi 1 is dense in l and whether (T λi 1 is bounded. In the present problem it turns out to be easiest to consider the first question first. This means trying to bound (ξ n in terms of (η n in (***. For every (η n l we have, if (ξ n is computed using (***: ξ n = 1 η 1 1 η ( + 1 η + 1 η 3 1 4 η ( 4 + 1 5 η 4 ( 1 1 + ( η ( 1 + η + 1 η ( 1 + 1 + ( η ( 3 + η 4 + 1 η 4 + Here we used again the inequality (* above. It follows from the above that if we let C λ be the constant C λ = 1 1 + ( + 1, then ξ n C λ η n. In particular, for every (η n l we have ξ n <, i.e. (ξ n l. In view of our earlier formular for D ((T λi 1, this proves that D ( (T λi 1 = l, i.e. D ((T λi 1 is dense in l in the trivial way that it is equal to l. The above computation also proves that (T λi 1 ((η n = ξ n C λ (η n for all (η n l, i.e. the operator (T λi 1 is bounded, with norm C λ. Together, this implies that λ ρ(t. This is true for all λ 1. Hence we have proved: σ p (T = {1}; σ c (T = ; σ r (T = ; ρ(t = C \ {1}. 4. Let {e n } be a total sequence in H, and {λ n } is a sequence of real numbers such that lim n λ n = 0. By the main Theorem 3.5- (in the +...
6 version I proved in class, we know that H = { α n e n α n K, α n < } Now if T is a bounded linear operator T : H H with T (e n = λ n e n for all n, we have for all sequences (α n in K with α n < : ( ( N N T α n e n = T α n e n = lim T α n e n N = lim N lim N N α n λ n e n = α n λ n e n. The last sum converges since ( α nλ n sup n λ n α n < (here sup n λ n < since lim n λ n = 0. This proves uniqueness, i.e. that there is only one possible bounded linear operator T : H H with T (e n = λ n e n for all n. On the other hand, the formula T ( α ne n = α nλ n e n truly defines a bounded linear operator T : H H. (Proof: Linearity is checked by a straightforward computation. Furthermore, we have T ( α ne n = α nλ n e n = α nλ n (sup n λ n α n = (sup n λ n α ne n. Hence T is bounded with norm T sup n λ n. We now prove that T is self-adjoint. Let x, y be two arbitrary vectors in H. Then we have x = α ne n and y = β ne n for some α n, β n K with α n <, β n <. Now, by a generalization of the Parseval relation which I have mentioned in class (compare p.175, exercise 4, T x, y = λ n α n e n, β n e n = λ n α n β n and x, T y = α n e n, λ n β n e n = α n λ n β n But all numbers λ n are assumed to be real; hence T x, y = x, T y. Since x, y were arbitrary this proves that T is self-adjoint. Finally, we now prove that T is compact, by imitating the argument in Example 8.1-6, p.409 in the book. For each m Z + we define the
operator T m : H H by T m α n e n = m λ n α n e n. This operator is clearly linear, and it is bounded since T m ( m m α n e n = λ n α n sup λ n α n 1 n m ( sup λ n α n e n, 1 n m for all vectors x = α ne n H. (This shows that T m sup1 n m λ n = sup 1 n m λ n. Furthermore the range R(T m Span{e 1, e,..., e m } is finite dimensional, hence by Theorem 8.1-4(a, T m is compact. For each m and each x = (T T m α n e n = hence T T m α ne n H we have ( λ n α n n=m+1 sup λ n n m+1 ( sup λ n α n e n, n m+1 sup n m+1 λ n = sup λ n. n m+1 n=m+1 This number tends to 0 as m, since lim n λ n = 0. Hence T T m 0 as m. Hence by Theorem 8.1-5, the operator T is compact. 7 α n