Functional Analysis Exercise Class

Transcription

1 Functional Analysis Exercise Class Week: January 18 Deadline to hand in the homework: your exercise class on week January 5 9. Exercises with solutions (1) a) Show that for every unitary operators U, V, the operators UV and U 1 are also unitaries. Conclude that the unitary operators on a Hilbert space form a group with respect to the operator multiplication and inverse. b) Show that every orthogonal projection is a partial isometry. c) Is it true that the product of two partial isometries/orthogonal projections is a partial isometry/orthogonal projection? a) Let U, V be unitaries, so that UU = U U = V V = V V = I. Then (UV ) (UV ) = V U UV = V V = I, (UV )(UV ) = UV V U = UU = I, so UV is a unitary. By definition, U 1 = U, and hence, (U 1 ) U 1 = (U ) U = UU = I, U 1 (U 1 ) = U U = I, so U 1 is a unitary. The conclusion about the group structure is obvious. b) Let P be a orthogonal projection; then (ker P ) = ran P, and we have P x = x for every x ran P. In particular, P is an isometry on (ker P ), and hence, by definition, it is a partial isometry. Alternatively, one can use the characterization that an operator A is a partial isometry if and only of A A is a orthogonal projection. Obviously, if P is a orthogonal projection then so is P P = P. c) No, in general the product of two orthogonal projections is not an orthogonal projection; by the above, this implies that the product of two partial isometries is not a partial isometry. As a concrete example, consider P := [ ], Q := 1 [ ], the projections onto the subspaces spanned by (1, 0) T and 1 (1, 1) T, respectively, in K. Then P Q = 1 [ ] 1 1, 0 0 which is easily seen not to be an orthogonal projection (for instance, it is not selfadjoint). 1

2 () Let H be a complex Hilbert-space. a) Show that if T B (H) is self-adjoint, and T 1, then U ± := T ± i(i T ) 1/ are unitaries. b) Prove that every T B (H) is the linear combination of at most 4 unitaries. (Hint: Use the following fact from the lecture: (I T ) 1/ commutes with every operator that commutes with I T.) a) As T 1, we have that T T 1, and hence, by the lecture, T I, i.e., I T 0. Hence, (I T ) 1/ is well-defined and commutes with every operator that commutes with I T (see the lecture). Since T commutes with I T, we see that (I T ) 1/ commutes with T. Thus, U ±U ± = (T ± i(i T ) 1/ )(T i(i T ) 1/ ) = T ± i(i T ) 1/ T T i(i T ) 1/ + (I T ) = T + (I T ) = I, and similarly, U ± U ± = I, i.e., U is a unitary. b) By a Lemma from the lecture we can decompose any T B (H) as T = A + ib with Hermitian operators A, B B (H). Now we apply (a) to the operators à = A and A B =. (If A = 0 then we define à := 0, and similarly for B.) This shows that B B U ± := à ± i(i à ) 1/ and V ± := B ± i(i B ) 1/ are unitaries. A direct computation shows that à = 1(U + + U ) and B = 1(V + + V ). Thus, we obtain the desired decomposition as T = A + ib = A (U + + U ) + i B (V + + V ). (3) A linear operator T on a Hilbert space H is called symmetric if x, T y = T x, y for all x, y H. Show that T is continuous and hence Hermitian. Using the closed graph theorem, it suffices to prove that for a sequence (x n ) n in H with x n 0 and T x n z, we have z = 0. But z, z = z, lim T x n = lim z, T x n = lim T z, x n = T z, lim x n = T z, 0 = 0. (4) (Multiplication operators) Let (, F, µ) be a σ-finite measure space, let H := L (, F, µ) be the Hilbert space of square-integrable functions with its standard inner product, i.e., f, g := f(x)ḡ(x), dµ(x).

3 Let φ L (, F, µ), and define the multiplication operator M φ f := φf, f L (, F, µ). Show that M φ is bounded, and the following hold: a) φ M φ is an algebra morphism such that M 1 = I b) (M φ ) = M φ c) M φ is normal. d) M φ1 = M φ φ 1 = φ µ-almost everywhere. e) M φ is self-adjoint φ(x) R for almost every x. f) M φ is unitary φ(x) = 1 for almost every x. g) M φ is positive φ(x) 0 for almost every x. h) M φ is an orthogonal projection φ is the indicator function of a measurable set. Remark: If you are not familiar with measure theory, then instead of the above abstract setting, take H to be the completion of C R ([0, 1]), take φ to be continuous, and skip (4)h) First, we show that φ L (, F, µ) guarantees that M φ is bounded. Indeed, M φ f = φ(x)f(x) dµ(x) φ f(x) dµ(x) = φ f, for any f L (, F, µ), showing that M φ φ. a) Let φ 1, φ L (, F, µ) and λ K. Then for every f L (, F, µ), M λφ1 +φ f (λm φ1 + M φ )f = (λφ 1 + φ )f (λφ 1 f + φ f) = 0, and hence M λφ1 +φ = λm φ1 + M φ. The proof of M φ1 φ = M φ1 M φ and M 1 = I goes the same way. b) For every f, g L (, F, µ), we have f, M φ g = M φ f, g = φ(x)f(x)g(x) dµ(x) = showing that (M φ ) = M φ. c) By the above, M φm φ = M φ M φ = M φ = M φ M φ = M φ M φ, and hence M φ is normal. f(x)φ(x)g(x) dµ(x) = f, M φ g, 3

4 d) By the linearity of φ M φ, it is enough to show that M φ = 0 = φ = 0 µ-almost everywhere. Let g(x) := φ(x)/ φ(x) when φ(x) 0, and g(x) := 0 otherwise. Then g is measurable. Since the measure space is σ-finite, there exist measurable sets n, n N, of finite measure such that n n =. For every n N, let f n := g1 n L (, F, µ), where 1 n is the indicator function of n. Assume that M φ = 0; then 0 = M φ f n = φ 1 n, n N, and hence φ(x) = 0 for almost every x n. Using that n n =, the assertion follows. e) We have M φ = (M φ ) = M φ φ = φ µ-almost everywhere. f) We have I = (M φ ) M φ = M φ φ = 1 µ-almost everywhere, and similary for I = M φ (M φ ). g) We have M φ f, f = φ(x) f(x) dµ(x), and hence if φ 0 µ-almost everywhere then M φ 0. Assume now that M φ 0. According to the lecture, this implies that M φ is self-adjoint, and hence, by the above, φ is real-valued µ-almost everywhere. Assume that := {x : φ(x) < 0} has non-zero measure. Then there exists a measurable subset A of finite measure (we are using the σ-finiteness here), and thus 1 A L (, F, µ). Moreover, M φ f, f = φ(x) f(x) dµ(x) = φ(x) dµ(x) < 0, contradicting the assumption that M φ 0. h) By definition, M φ is an orthogonal projection if and only if Mφ = M φ = Mφ, and by the above, this is equivalent to φ being real-valued, and φ(x) = φ(x) for µ-almost every x, which in turn is equivalent to φ taking only 0 or 1 value µ-almost everywhere, which is equivalent to φ being an indicator function of some set A. Since φ is measurable, A = {x : φ(x) = 1} is measurable, too. (5) (Lax-Milgram) Let B : H H K be a sesquilinear form that is bounded, i.e., B := sup{ B(x, y) : x = 1, y = 1} < +. Show the following: a) There exists a bounded linear operator B B(H) such that B(x, y) = x, By, x, y H, and B = B. b) If there exists a c > 0 such that B(x, x) c x, x H, then B is invertible, and B 1 c 1. Last year s lecture notes: page 61 at M5/Allgemeines/MA3001_014W/Hilbert4.pdf A 4

5 Homework with solutions (1) Show that the product of two orthogonal projections is an orthogonal projection if and only if they commute. Let P, Q be orthogonal projections. Then (P Q) = Q P = QP, and hence (P Q) = P Q QP = P Q, i.e., if P Q is an orthogonal projection then P and Q commute. To prove the opposite direction, assume that P and Q commute. Then, by the above, (P Q) = P Q, and (P Q) = P QP Q = P Q = P Q, and thus P Q is indeed an orthogonal projection. () Let A B(H). Show that the following are equivalent: (i) A is a self-adjoint unitary. (ii) There exists an orthogonal projection P such that A = P I. (iii) There exist orthogonal closed subspaces H 1, H such that H = H 1 H, and for every x = x 1 + x, x i H i, Ax = x 1 x, i.e., A is a reflection. (ii)= (i): Let A = P I. Then A is obviously self-adjoint, and A A = A = (P I)(P I) = 4P P P + I = I, and thus A is a unitary. (i)= (ii): Let A be a self-adjoint unitary, and define P := 1 (A + I). Then P is obvously self-adjoint, and P = 1 4 ( A + A + I ) = 1 4 (I + A + I) = 1 (A + I) = P, where we used that A = A A = I due to the self-adjointness and unitarity of A. Thus, P is an orthogonal projection. (ii)= (iii): Let H 1 := ran P, H := ker P. Then H 1 H = H, so that every x H can be uniquely decomposed as x = x 1 + x, x i H i. Moreover, we get Ax = (P I)(x 1 + x ) = P (x 1 + x ) (x 1 + x ) = x 1 (x 1 + x ) = x 1 x. (iii)= (ii): Let P be the projection onto H 1. Let x = x 1 + x, x i H i. Then (P I)x = P (x 1 + x ) (x 1 + x ) = x (x 1 + x ) = x 1 x = Ax, and hence A = P I. Scores: Total: 8 points. Each implication above is worth points. (3) a) Show that every partial isometry V B(H) on a finite-dimensional Hilbert space H, can be extended to a unitary on H. b) Show that every bounded operator on a finite-dimensional Hilbert space can be written as the linear combination of at most two unitaries. (Hint: Use the polar decomposition.) 5

6 a) Let V be a partial isometry, i.e., V is an isometry on (ker V ). This implies that r := dim ran V = dim(ker V ). Let e 1,..., e r be an orthonormal basis in (ker V ), and e r+1,..., e d be an orthonormal basis in ker V ; then e 1,..., e r, e r+1,..., e d is an orthonormal basis in the whole Hilbert space H. Similarly, let f 1,..., f r be an orthonormal basis in ran V, f r+1,..., f d be an orthonormal basis in (ran V ) ; then f 1,..., f r, f r+1,..., f d is an orthonormal basis in H. Define { 0, 1 i r, V e i := f i, r + 1 i d. Then (V + V )e i = f i, i = 1,..., r, and hence V + V is an isometric extension of V. b) By the polar decomposition theorem, every bounded operator A can be written as A = V A, where V is a partial isometry. By the solution of Exercise (), A can be written as A = A (U 1 + U ), where U 1, U are unitaries. By the previous part, if the Hilbert space is finite-dimensional then V can be extended to a unitary U, and hence A = A (UU 1 + UU ). By Exercise (1), UU 1 and UU are unitaries, and hence the assertion follows. 6