Shift Operators Jayadeb Sarkar Indian Statistical Institute, Bangalore. [email protected] June 4, 2014 This set of notes are based on the lectures given at the Instructional School for Lecturers (ISL) held at Indian Statistical Institute, Bangalore. The present draft is not perfect (and far from complete) and certainly suffers from obvious flaws. All of them are the authors sole responsibility, including mathematical mistakes and typos. I am very grateful to Tiju Cherian John, ISI Bangalore, for typing the draft in Tex format. I am also thankful to him for reading the entire manuscript, providing me with comments and corrections. Contents 1 Introduction 2 2 Shift Operator on the Hardy Space H2 3 3 Isometries 7 4 Beurling's Theorem and some Consequences 11 5 Bergman space and more 14 1 1 Introduction In these lectures we will try to explore the paper written by A. Beurling in 1948 titled "On 2-problems concerning linear operators on Hilbert spaces". Let us first explain the various terms in the title. 1. Hilbert Spaces under consideration are the Hardy Space H2(D) or the square summable sequence space l2(N) where ( ) X1 X1 2 n 2 H (D) = f 2 Hol(D): f(z) = anz ; an 2 C; janj < 1 n=0 n=0 and D = fz 2 C : jzj < 1g is the open unit disk in the complex plane C, and ( ) X1 2 N f g1 j j2 1 l ( ) = an n=0 : an < n=0 Now onwards we use the following notations H2(D) := H2 l2(N) := l2 Exercise: l2 is a Hilbert space with inner product defined as X1 hf g1 f g1 i an n=0; bn n=0 = anbn n=0 f g1 2 and en n=0 forms an orthonormal basis for l where en is the sequence with 1 in the nth position and zero elsewhere. We state without proof that H2(D) is a Hilbert Space with the inner product defined as * + X1 X1 X1 n n anz ; bnz = anbn (1.1) n=0 n=0 n=0 For a proof please refer Duren [3] and Halmos [1]. That H2 is a Hilbert space also follows readily from the following simple ex- ercise. Exercise: Show that the mapping X : l2 ! H2 X1 f g1 7−! n an n=0 anz n=0 is well defined and one to one and onto. Prove that H2 is a Hilbert space with the inner product defined as in (1.1) (Note that under this isomorphism n 2 2 en 7−! z and thus f1; z; z ; :::g forms an orthonormal basis of H ). 2 2. The Linear Operator under consideration is the shift, S on l or the shift Mz on H2 (see Definition 2.1). 2 3. 2-Problems mentioned are W • Take f 2 H2=l2. When does Snf = H2 (or l2)? W 2 2 ∗ n • For f 2 H =l . When does Cf := (S ) f is generated by the eigenvec- tors? W Here denotes the span closure and that S∗ is the adjoint of S 2 Shift Operator on the Hardy Space H2 2.1 Definition. Define shift S on l2 by f g1 f g S ( an n=0) = 0; a0; a1; ::: ; f g1 2 2 for all an n=0 l . 2.2 Remarks. 1. S is a linear isometry on l2, that is, k f g1 k kf g1 k S( an n=0) = an n=0 : f g1 f g1 2 2 2. For any an n=0; bn n=0 l h ∗ f g1 f g1 i hf g1 f g1 i S ( an n=0); bn n=0 = an n=0;S( bn n=0) hf g1 f gi = an n=0; 0; b1; b2; ::: = hfa1; a2; :::g; fb0; b1; :::gi ) ∗ f g1 f g1 f g S ( an n=0) = an n=1 = a1; a2;::: : ∗ ∼ ∗ 3. ker S = ffa0; 0; 0; :::g : a0 2 Cg = C. In particular, dim[kerS ] = 1 4. S∗S = I but ∗ SS = I − Pker S∗ = I − PC; 2 ∗ where Pker S∗ is the orthogonal projection of H onto ker S . Therefore, that S is not normal. 2.3 Definition. Let H be a Hilbert space and T 2 L(H). ∗n ∗n 1. T is said to be C·0 if T −! 0 in strong operator topology (that is, kT hk ! 0 for all h 2 H). 2. A closed subspace M ⊆ H is said to be invariant subspace of T 2 L(H) (or, T -invariant) if T (M) ⊆ M. 3. A closed subspace M ⊆ H is said to be co-invariant subspace of T 2 L(H) (or, T ∗-invariant) if T ∗(M) ⊆ M. 4. A closed subspace M is said to be T -reducing if T (M);T ∗(M) ⊆ M. 5. T 2 L(H) is said to be irreducible if T has no reducing subspace except f0g and H. 3 2 2.4 Definition. Define shift Mz on H by 2 Mz(f) = zf; 8f 2 H : Note that for all f 2 H2, zf is a function in H2 defined by (zf)(w) = wf(w); for all w 2 D. 2.5 Remarks. ∗ 1. Mz Mz = IH2 ∗ − 2 2. MzMz = IH2 PC where PC is the orthogonal projection of H onto the subspace of all constant functions, denoted by C. P P 1 n 1 n+1 3. Mz( n=0 anz ) = n=0 anz P P ∗ 1 n 1 n−1 4. Mz ( n=0 anz ) = n=1 anz ∗n −! 5. Mz 0 in strong operator topology. 2 6. Mz on H is irreducible. 2 2 7. H ⊖ MzH = C. 2 2 8. That H ⊖ MzH = C satisfies the following relation: _ M1 _ M1 n 2 2 n 2 z (H ⊖ MzH ) = z C = H : n=0 n=0 ⊆ 2 ∗ j 2 HW: Let M H be a Mz -invariant subspace. Then Mz M L(H) is in C·0. [What is the conclusion if that M is Mz-invariant?] W ≥ f n n+1 g Example: Let n 1 be a fixed integer and Mn =W z ; z ; ::: . Then Mn is invari- ∗ f 2 n−1g ∗ ant under Mz but not under Mz . Also, Qn = 1; z; z ; :::; z is Mz invariant but not Mz invariant. Questions: n 2 (1) Is it true that Mn = z H ? n 2 ? (2) Is it true that Qn = (z H ) ? 2.6 Definition. Magic/ kernel Vectors: For each w 2 D define k : D −! C by P w n n kw(z) = n≥0 w z . 2.7 Remarks. 2 2 1=2 1. kw 2 H ; 8w 2 D and kkwk = (1 − jwj ) . P 2 2 n 2. Let f H with f(z) = n≥0 anz then * + X1 X1 X1 n n n n anz ; w z = anw : n=0 n=0 n=0 Consequently, for all f 2 H2 and w 2 D, f(w) = hf; kwi: 2 Hence, that kw reproduce the value of f 2 H at each w 2 D. 4 2 ∗ − 8 2 D 3. kw ker(Mz w), w because, ∗ ∗ 2 2 ··· Mz (kw) = Mz (1 + wz + w z + ) = w + w2z + w3z2 + ··· = wkw: 2 4. Evaluation Functional: Define evw : H −! C, for all w 2 D by evw(f) = f(w): Note that jevw(f)j = jf(w)j = jhf; kwij ≤ kfkkkwk. What is kevwk? 5. The Szego kernel over D is the function, k : D × D −! C defined by k(λ, w) := (1 − λw)−1; for all w; λ in D. Note that k(λ, w) = hkw; kλi; 8λ, w 2 D and that k is holomorphic in the first variable and anti-holomorphic in the second variable. ◦ ∗ 6. Prove that k(z; w) = evz evw. 2 7. Prove that fkw : w 2 Dg ⊆ H is a total set, that is, _ 2 fkw : w 2 Dg = H : To proceed further we recall the following important notion. Let H be a Hilbert space and T 2 L(H) then the spectrum of T; σ(T ) is defined as σ(T ) = fλ 2 C : T − λI is not invertibleg: We recall that a bounded linear operator T on H is invertible if and only if that T is bounded below (that is, there exists c > 0 such that kT fk > ckfk for all f(=6 0) 2 H) and of dense range (that is ranT = H). The approximate point spectrum of T; σa(T ) is defined as σa(T ) = fλ 2 C : 9fn ⊆ H 3 kfnk = 1; k(T − λ)fnk −! 0g, and the point spectrum, σp(T ) is defined as σp(T ) = fλ 2 C : T f = λf for some f =6 0g. Finally, the compression spectrum, Π(T ) is defined as Π(T ) = fλ 2 C : ran(T − λI) $ Hg. 2.8 Theorem. @σ(T ) ⊆ σa(T ). 5 Proof. See Problem 78 in [1]. 2.9 Theorem. 8 2 D ∗ − f g C · (i) w ; ker(Mz w) = span kw = kw. ∗ − In particular, dim[ker(Mz w¯)] = 1. ∗ D (ii) σp(Mz ) = . ∗ D ∗ (iii) σ(Mz ) = = σa(Mz ). P 2 ∗ − 2 2 n Proof. We know that kw ker(Mz w¯). Let f H be such that f(z) = anz ∗ and Mz f = wf. This implies 2 2 a1 + a2z + a3z + ··· = w(a0 + a1z + a2z + ··· ) 2 n Therefore, a1 = wa0; a2 = wa1 = w a2;:::; and an = w a0; 8 n 2 N and hence, ∗ − f g f = a0kw. Consequently, ker(Mz w) = span kw . This completes the proof of part (i). k ∗k ∗ ⊆ D D ⊆ ∗ Since Mz = 1, we have that σ(Mz ) .