MATH 421/510 Assignment 3 Suggested Solutions February 2018 1. Let H be a Hilbert space. (a) Prove the polarization identity: 1 hx; yi = (kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2): 4 Proof. By direct calculation, kx + yk2 − kx − yk2 = 2hx; yi + 2hy; xi: Similarly, kx + iyk2 − kx − iyk2 = 2hx; iyi − 2hy; ixi = −2ihx; yi + 2ihy; xi: Addition gives kx + yk2−kx − yk2+ikx + iyk2−ikx − iyk2 = 2hx; yi+2hy; xi+2hx; yi−2hy; xi = 4hx; yi: (b) If there is another Hilbert space H0, a linear map from H to H0 is unitary if and only if it is isometric and surjective. Proof. We take the definition from the book that an operator is unitary if and only if it is invertible and hT x; T yi = hx; yi for all x; y 2 H. A unitary operator T is isometric and surjective by definition. On the other hand, assume it is isometric and surjective. We first show that T preserves the inner product: If the scalar field is C, by the polarization identity, 1 hT x; T yi = (kT x + T yk2 − kT x − T yk2 + ikT x + iT yk2 − ikT x − iT yk2) 4 1 = (kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2) = hx; yi; 4 where in the second equation we used the linearity of T and the assumption that T is isometric. 1 If the scalar field is R, then we use the real version of the polarization identity: 1 hx; yi = (kx + yk2 − kx − yk2) 4 The remaining computation is similar to the complex case. It remains to show that T is injective. But T x = 0 () hT x; T xi = 0 () hx; xi = 0 () x = 0: Hence T is injective. 2. If E is a subset of a Hilbert space H, then (E?)? is the smallest closed subspace of H containing E. Proof. For each subset A of H, A? is always a subspace by definition. Also, it is ? closed, since if yn ! y in H where hyn; xi = 0, then hy; xi = 0. With A = E , we have (E?)? is a closed subspace. Moreover, it contains E by definition. It remains to show minimality. let K be any closed subspace containing E, and we would like to show (E?)? ⊆ K. Suppose not. Then there is x 2 (E?)? with x2 = K. Since K is a proper closed subspace, by Question 4 of the last homework, there is l 2 H∗ such that l(x) = 1 and l ≡ 0 on K. By the Riesz-Fr´echet theorem, there is y 2 H with l(x) = hx; yi = 1 6= 0. Since x 2 (E?)?, we have y2 = E?. That means hz; yi= 6 0 for some z 2 E. But this is a contradiction since l(z) = hz; yi and l ≡ 0 on E. Therefore (E?)? ⊆ K. 3. Suppose H is a Hilbert space and T 2 L(H; H). (a) There is a unique T ∗ 2 L(H; H), called the adjoint of T , such that hT x; yi = hx; T ∗yi for all x; y 2 H. Proof. We first prove the existence. Fix y 2 H. Define the mapping ly(x) := hT x; yi, which lies in H∗. By the Riesz-Fr´echet theorem, there is a unique ∗ z 2 H with ly(x) = hx; zi for all x 2 H, with klyk = kzk. We define T y := z by the above relations. The uniqueness of z ensures that the mapping is well defined, and satisfies hT x; yi = hx; T ∗yi by construction. One can check that T ∗ is linear and has operator norm 1. This shows the existence. To establish the uniqueness, it suffices to prove the following assertion: if T is a linear operator on a Hilbert space H and hx; T yi = 0 for all x; y 2 H, then T ≡ 0. Indeed, fix y 2 H, and taking x = T y. Then we have hT y; T yi = 0, whence T y = 0. Hence T ≡ 0, so uniqueness is proved. (b) kT ∗k = kT k, kT ∗T k = kT k2,(aS + bT )∗ = aS∗ + bT ∗,(ST )∗ = T ∗S∗, and T ∗∗ = T . 2 Proof. • We first show T ∗∗ = T . Indeed, hT ∗∗x; yi = hy; T ∗∗xi = hT ∗y; xi = hx; T ∗yi = hT x; yi: Since the above holds for all x; y 2 H, T ∗∗ = T . • We then show kT ∗k = kT k. { We first show kT ∗k ≤ kT k < 1. Let x 2 H. If T ∗x = 0, then we have nothing to prove. Otherwise, by the Cauchy-Schwarz inequality, kT ∗xk2 = hT ∗x; T ∗xi = hx; T T ∗xi ≤ kxkkTT ∗xk ≤ kxkkT kkT ∗xk: Thus we have kT ∗xk ≤ kT kkxk. This shows that kT ∗k ≤ kT k. { Since T ∗∗ = T , we have kT k = kT ∗∗k ≤ kT ∗k by the previous direction. Hence kT ∗k = kT k. • We show kT ∗T k = kT k2. { On the one hand, kT ∗T xk ≤ kT ∗kkT kkxk = kT k2kxk; where we have used kT ∗k = kT k in the last equality. Thus kT ∗T k ≤ kT k2 < 1. { On the other hand, kT xk2 = hT x; T xi = hx; T ∗T xi ≤ kxkkT ∗T xk ≤ kxkkT ∗T kkxk: 1 1 2 Hence kT xk ≤ kT ∗T k 2 kxk, so kT k ≤ kT ∗T k 2 . Hence kT k ≤ kT ∗T k. • The other two equalities are direct. (c) Let R and N denote the range and nullspace; then R(T )? = N (T ∗) and N (T )? = R(T ∗). Proof. •R (T )? = N (T ∗): { R(T )? ⊆ N (T ∗): let x 2 R(T )?. Then hT ∗x; yi = hx; T yi = 0 for all y 2 H. Taking y = T ∗x shows that T ∗x = 0, that is, x 2 N (T ∗). { R(T )? ⊇ N (T ∗): let x 2 N (T ∗). Then T ∗x = 0. For any y 2 H, hx; T yi = hT ∗x; yi = 0, which shows that x 2 R(T )?: • Applying the first part of the question to T ∗, we have R(T ∗)? = N (T ∗∗) = N (T ), so (R(T ∗)?)? = N (T )?. But by Question 2 in this homework, (R(T ∗)?)? is the smallest closed subspace of H containing R(T ∗). Since R(T ∗) is already a subspace of H, the smallest closed subspace of H con- taining R(T ∗) is R(T ∗). Hence N (T )? = R(T ∗). (d) T is unitary if and only if T is invertible and T −1 = T ∗. 3 Proof. As in Question 1 (b), we take the definition that an operator is unitary if and only if it is invertible and hT x; T yi = hx; yi for all x; y 2 H. If T is unitary, then T is invertible by definition. To show that T −1 = T ∗, we note that hT x; yi = hT x; T T −1yi = hx; T −1yi; for all x; y 2 H, since T is unitary. This is to say that T −1 is an adjoint of T . By uniqueness of the adjoint operator, we have T −1 = T ∗. On the other hand, suppose T is invertible and T −1 = T ∗. To show T is unitary, it suffices to show it preserves the inner product. But we easily compute hT x; T yi = hx; T ∗T yi = hx; T −1T yi = hx; yi; for all x; y 2 H. hence T is unitary. 4. Let M be a closed subspace of the Hilbert space H, and for x 2 H let P x be the element of M such that x − P x 2 M? as in Theorem 5.24. (a) P 2 L(H; H), P ∗ = P , P 2 = P , R(P ) = M, and N (P ) = M?. P is called the orthogonal projection onto M. Proof. i. P is linear: Theorem 5.24 states that each x 2 H can be uniquely decomposed into x = y + z, where P x := y 2 M and z 2 M?. Using this and the fact that M, M? are subspaces, we can prove linearity. P is bounded: Theorem 5.24 also states that P x is perpendicular to x−P x. By the Pythagorean Theorem, kP xk2 = kxk2 − kx − P xk2 ≤ kxk2; which shows that kP k ≤ 1. Hence P 2 L(H; H). ii. By uniqueness of the adjoint operator, it suffices to show that for all x; x0 2 H, we have hP x; x0i = hx; P x0i: Decompose x = y + z, x0 = y0 + z0 as in Theorem 5.24. This can be proved using the fact that y; y0 2 M and z; z0 2 M?. iii. Note that for each y 2 M, the orthogonal decomposition in Theorem 5.24 is y = y + 0, so P y = y. (This implies that kP k = 1 unless M = f0g). Applying this to y = P x 2 M, we have P 2x = P (P x) = P x for all x 2 H. iv. We have R(P ) ⊆ M by definition. On the other hand, given any x 2 M, P x = x implies that x 2 R(P ). v. Since P = P ∗, N (P ) = N (P ∗). By Question 3(c), N (P ∗) = R(P )?. But R(P ) = M, so R(P )? = M?. Hence N (P ) = M?. (b) Conversely, suppose that P 2 L(H; H) satisfies P 2 = P ∗ = P . Then R(P ) is closed and P is the orthogonal projection onto R(P ). 4 Proof. By Theorem 5.24, it suffices to show that M := R(P ) is closed, and then show that P x = x for all x 2 M and P x = 0 for all x 2 M?.