1. Let P be the space of all polynomials (of one real variable and with real coefficients) with the norm

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1 Uppsala Universitet Matematiska Institutionen Andreas Strömbergsson Prov i matematik Funktionalanalys Kurs: F3B, F4Sy, NVP Skrivtid: 9 14 Tillåtna hjälpmedel: Manuella skrivdon, Kreyszigs bok Introductory Functional Analysis with Applications och Strömbergssons häfte Spectral theorem for compact, self-adjoint operators. 1. Let P be the space of all polynomials (of one real variable and with real coefficients) with the norm { } p = sup p(t) : 1 t 1. Define f : P R by f(p) = p(). Prove that f is an unbounded linear functional. Also define T : P P by T (p) = p (the derivative of p) and prove that T is an unbounded linear operator.. Let Y = {x = (ξ j ) l ξ 1 = 0, ξ + ξ 3 = 0, ξ 1 + ξ 3 + ξ 4 = 0} and let P : l l be the orthogonal projection onto Y. Determine P. 3. Let Y = { (ξ j ) l 1 at most finitely many ξ j 0 }. Show that Y is not complete. Determine the closure of Y in l 1. (5p) 4. Define T : l l 1 by T ((ξ 1, ξ, ξ 3, )) = ( 1 ξ 1, (ξ 1 + ξ ), 3 (ξ 1 + ξ + ξ 3 ), ). Prove that T is a bounded linear operator T : l l 1 and compute the norm T.

2 5. Let H be a Hilbert space and let f 1, f H, f 1 0 and f 0. Assume that the following condition holds: (A) x H : f 1 (x) = f 1 x = f (x) = 0. Prove that then the following is also true: (B) x H : f (x) = f x = f 1 (x) = 0. (Hint: Apply Riesz s Theorem both to f 1 and f and then try to interpret the condition (A).) 6. Let X be a normed space and let x 1, x, x 3, X be a sequence of points in X. Assume that for each f X the sequence of numbers f(x 1 ), f(x ), f(x 3 ), is bounded. Prove that the sequence x 1, x, x 3, is bounded. (5p) 7. Let T : l l be the operator given by (η j ) = T ((ξ k )), where η j = j k ξ k. Prove that T is compact and self-adjoint. Find all the eigenvalues and eigenvectors of T. GOOD LUCK! Note: If you are taking the 6p-course, please contact me to decide a date for further examination.

3 3 Solutions 1. If a, b R and p, q P then f(ap + bq) = [ap + bq]() = ap() + bq() = af(p) + bf(q), and T (ap + bq) = (ap + bq) = a(p ) + b(q ) = at (p) + bt (q). This proves that f and T are linear. Given an integer n 1 we consider the polynomial p(t) = t n. Note that p = 1. Furthermore we have f(p) = p() = n. Hence f is unbounded, since there is no constant C such that n C 1 holds for all n 1. We also note T (p)(t) = nt n 1 and thus T (p) = n. Hence T is unbounded, since there is no constant C such that n C 1 holds for all n 1.. (Very similar to exam :.) Note that for x = (ξ j ) l we have ξ 1 = x, e 1, ξ +ξ 3 = x, e +e 3 and ξ 1 +ξ 3 +ξ 4 = x, e 1 +e 3 +e 4. Hence x = (ξ j ) belongs to Y if and only if x is orthogonal to e 1, e + e 3 and e 1 + e 3 + e 4. That is: This can be rewritten as: Y = {e 1, e + e 3, e 1 + e 3 + e 4 }. Y = ( Span{e 1, e + e 3, e 1 + e 3 + e 4 } ) = ( Span{e1, e + e 3, e 3 + e 4 } ). But Span{e 1, e + e 3, e 3 + e 4 } is a closed subspace of l since it is finite dimensional (Theorem.4-3), and hence by Lemma 3.3-6, Y = ( Span{e 1, e + e 3, e 3 + e 4 } ) = Span{e1, e + e 3, e 3 + e 4 }. Using Gram-Schmidt we find an orthogonal basis f 1 = e 1, f = e + e 3, f 3 = e + e 3 + e 4 in Y. The orthogonal projection P onto Y is then given by P ((ξ j )) = (ξ j), f 1 f 1, f 1 f 1 + (ξ j), f f, f f + (ξ j), f 3 f 3, f 3 f 3 = 1 ( ) 3ξ 1, ξ + ξ 3 ξ 4, ξ + ξ 3 + ξ 4, ξ + ξ 3 + ξ 4, 0, 0, 0, Let x n = ( 1,, 3,, n, 0, 0, 0, ). Then x 1, x,... Y. We also define x = ( 1,, 3, ) l 1.

4 4 (The fact that x l 1 follows from k = 1 <.) We now claim that x n x as n ; this follows from x x n = (0, 0,, 0, n 1, n, ) = k = n 0 as n. k=n+1 Note that x / Y, by the definition of Y. Hence by Theorem 1.4-6(b), Y is not closed (as a subspace of l 1 ). Hence by Theorem 1.4-7, Y is not complete. We claim that the closure of Y is all of l 1, i.e. Y = l 1. To prove this, let us fix an arbitrary vector x = (ξ 1, ξ, ) l 1. Then form the sequence x 1, x,... where x n = (ξ 1, ξ,, ξ n, 0, 0, 0, ). Then x 1, x, Y and x n x as n, since x x n = ξ k 0 as n. k=n+1 (Detailed proof of the last statement: Let S n = n ξ k. Then S = lim n S n = ξ k exists as a real number, since x = (ξ n ) l 1. Hence (S n ) is a Cauchy sequence, and thus for any ε > 0 there is N such that S m S n ε whenever m n N, that is, m k=n+1 ξ k ε whenever m n N. For fixed n we may let m in the last statement to obtain k=n+1 ξ k ε whenever n N. But such ε > 0 was arbitrary; hence we have proved lim n k=n+1 ξ k = 0.) Hence by Theorem 1.4-6(a), x Y. This is true for every x l 1. Hence Y = l T is obviously a linear operator. We have, for all (ξ k ) l, if (η k ) = T ((ξ k )), j j (η j ) l 1 = η j = ξ k j ξ k = ξ k = j ξ k 1 k (ξ k ) l 1 k = (ξ k ) l Hence T is a bounded, and T. On the other hand, taking (ξ j ) = (1, 1, 1, ) we clearly obtain equality in each step in the above computation, hence T ((1, 1, 1, )) =, and this shows that T =. Answer: T =. j=k j

5 5. By Riesz s Theorem (Theorem in the book) there are vectors z 1, z H such that f 1 (x) = x, z 1 and f (x) = x, z for all x H. Since f 1, f 0 we must have z 1, z 0. We also see in that theorem that f 1 = z 1 and f = z. Now the condition (A) can be written: 5 (A) x H : x, z 1 = z 1 x = x, z = 0. However we know from Schwarz inequality (Lemma 3.-1(a)) that x, z 1 = z 1 x holds if and only if x and z 1 are linearly independent, i.e. if and only if x = cz 1 for some c C (here we used the fact z 1 0). Hence condition (A) can be rewritten as: (A) x H : [ x = cz1 for some c C ] = x, z = 0. This can also be written as: (A) c C : cz 1, z = 0. Clearly this holds if and only if z 1, z = 0. Similarly one proves that condition (B) is equivalent to z, z 1 = 0. Hence condition (A) indeed implies condition (B). 6. Define g n X by g n (f) = f(x n ) for all f X. It follows from Lemma that each g n is indeed a bounded linear functional on X, and g n = x n. Now the assumption in the problem says that for each f X the sequence g 1 (f), g (f), g 3 (f), is bounded. Now X is a Banach space by Theorem.10-4, and g 1, g, g 3, is a sequence of bounded linear operators X C. Hence by the Uniform Boundedness Theorem (Theorem 4.7-3), there is a constant c such that g n c for all n. By what we have noted above, this means x n c for all n. In other words, the sequence x 1, x, x 3, is bounded. 7. The quickest solution is probably to compute the spectral decomposition of T on a scratch paper and then use this result to prove everything at once. But we first have to prove that T is a bounded

6 6 linear operator: T is clearly linear. If (η j ) = T ((ξ k )) then (η j ) = η j = j k ξ k = j k ξ k ( ) = j k ξ k = 1 k ξ k 3 ( 1 ) k ξ k (ξ k). Hence T is bounded and T 1. Let f 3 1 = 3 ( 1,, 3, ) l. Note that f 1 = 1, since f 1 = 3 j = /4 = 1. By the Gram-Schmidt process (recall that l is separable) we can now extend f 1 to a total orthonormal set f 1, f, f 3,... in l. We compute (η j ) = T (f 1 ): η j = j k ξ k = 3 3 j k = 3 j. Hence T (f 1 ) = 1 3 f 1. Note also that for n, writing f n = (ξ j ), the fact f n, f 1 = 0 implies k ξ k = 0 Hence T (f n ) = 0 for all n. Now we have exactly the situation from Homework assignment 3, problem 4: T is a bounded linear operator on l, {f 1 } is a total orthonormal sequence in l, and T (f n ) = λ n f n with λ 1 = 1and λ 3 = λ 3 = = 0 (thus lim n λ n = 0). Hence by that problem, T is compact and self-adjoint. [Note that in the present case the fact that T is compact can also be seen directly using Theorem 8.1-4(a), for by the above formulas we have T (H) = Span{f 1 }, i.e. the range of T is finite dimensional.] We also see that the eigenvalues of T are 1 (with eigenspace = 3 Span{f 1 }) and 0 (with eigenspace = Span{f, f 3, } = {f 1 } ).

7 [Proof that there are no other eigenvectors and eigenvalues: Assume that x = n=1 a nf n l is an eigenvector of T with eigenvalue λ. Then n=1 a n < and λx = T x = 1 a 3 1f 1 + 0a f + 0a 3 f 3 + = 1 a 3 1f 1, hence λa 1 = 1a 3 1 and for each n, λa n = 0. We also know that x 0, i.e. there is at least one n for which a n 0. It follows that either λ = 1 3 and x = a 1 f 1 or else λ = 0 and x = a f +a 3 f 3 + Span{f, f 3, }, as claimed.] 7

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