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Powers and direct sums
Accepted Manuscript Powers and direct sums
John B. Conway, Gabriel Prˇajiturˇa, Alejandro Rodríguez-Martínez
PII: DOI: Reference:
S0022-247X(13)01086-X 10.1016/j.jmaa.2013.12.007 YJMAA 18111
To appear in:
Journal of Mathematical Analysis and Applications
Received date: 20 May 2010
Please cite this article in press as: J.B. Conway et al., Powers and direct sums, J. Math. Anal. Appl. (2014), http://dx.doi.org/10.1016/j.jmaa.2013.12.007
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POWERS AND DIRECT SUMS ˇ ˇ JOHN B. CONWAY, GABRIEL PRAJITUR A, ´ ´ AND ALEJANDRO RODRIGUEZ-MARTINEZ
1. Introduction If a general theory of multiplicity ever comes to be, it will likely be the case that for any operator A, the operator A(2) = A ⊕ A has twice the multiplicity of A. As in the case of hermitian operators, it might sometimes be the case that A2 has twice the multiplicity of A and sometimes that it does not. Though this paper will not try to begin a general theory of multiplicity, it does explore the relationship between A ⊕ A and A2 . Specifically it studies operators A such that A ⊕ A and A2 are similar, a question that has an intrinsic interest independent of any attempt at multiplicity theory. 1.1. Definition. If A ∈ B(H), A satisfies Condition S if A2 ≈ A ⊕ A. Say that A satisfies Condition U if A2 ∼ = A ⊕ A. (≈ means similar and ∼ = means unitarily equivalent.) Realize that for normal operators Condition S and Condition U are the same since two normal operators are similar if and only if they are unitarily equivalent ([1], Corollary IX.6.11). We will shortly see several examples of operators satisfying Condition S, but for the time being we present only two. 1.2. Example. If S is the unilateral shift, then S satisfies Condition U. 1.3. Example. If A denotes multiplication by the independent variable on L2 [0, 1], then A does not satisfy Condition S. In fact since x2 is oneto-one on [0, 1], A2 is unitarily equivalent to A; however A ⊕ A has uniform multiplicity 2. On the other hand, A(∞) , the direct sum of A with itself an infinite number of times, does satisfy Condition S (and therefore Condition U). The paper is organized as follows. §2 presents some spectral properties of operators that satisfy Condition S. In particular it is shown that such operators have spectral radius 1 and that the only compact 1
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 2 A,
operator satisfying Condition S is the zero operator. §3 gives a complete characterization of the normal operators satisfying Condition S. In §4 the unilateral weighted shifts satisfying Condition S as well as those satisfying Condition U are characterized. This results in a rather simple criterion for a hyponormal weighted shift to satisfy Condition S and the fact that the isometric weighted shift is the only hyponormal weighted shift that satisfies Condition U. 2. Spectral results The proofs of the next two results are straightforward. Recall that σe (A) denotes the essential spectrum of A; that is, the spectrum of the image of A in the Calkin algebra. 2.1. Proposition. If A satisfies Condition S, then σ(A) = σ(A)2 and σe (A) = σe (A)2 . 2.2. Proposition. (a) If A and B satisfy Condition S (or U ), then so does A ⊕ B. (b) A satisfies Condition S (or U) if and only if A∗ does. n (c) If A satisfies Condition S (or U ), then so does A2 for all n ∈ N. Note that a scalar multiple of an operator satisfying Condition S (or U) satisfies the same condition only if the scalar is either 0 or 1. In light of Proposition 2.1 it becomes important for the problem of characterizing the operators satisfying Condition S to characterize the compact subsets K of C that satisfy K = K 2 . At present we cannot do this, but we can make such a characterization when K is a subset of R. 2.3. Proposition. A compact subset K of R satisfies K = K 2 if and only if one of the following holds: (a) K = {0}; (b) K = {1}; (c) K = {0, 1}; (d) for any r in the open (0, 1) there is a compact subset D of interval n=∞ 2n 1/2n D ∪ ∪ {0, 1} [r2 , r] such that K = n=∞ n=0 n=1 D Proof. It is clear that any set K that has the form in parts (a) through (d) must satisfy K = K 2 , so we look at the converse. Assume K = K 2 and assume that none of the conditions (a) through (c) is true; let 0 < r < 1. We first note that K ∩ [r2 , r] = ∅. In fact if s ∈ K and n s = 0, 1, let n be the smallest natural number such that s2 ≤ r. If n n−1 it were the case that s2 ≤ r2 , then we would have that s2 ≤ r,
POWERS AND DIRECT SUMS
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2n
a contradiction; s > r2 . That is D = K ∩ [r2 , r] = ∅. Let n=∞ n=∞ 2n so n L = n=0 D ∪ n=1 D1/2 ∪ {0, 1}. Clearly L ⊆ K and L = L2 . Using an argument similar to the one used to show that K ∩ [r2 , r] = ∅ we can establish the reverse inclusion and thus the equality of K and L. The next result is straightforward. 2.4. Proposition. Assume E is a non-empty, bounded subset of C that satisfies E = E 2 . (a) E ⊆ cl D. (b) If E ∩ D = ∅, then 0 ∈ cl E and cl E ∩ ∂D = ∅. (c) If int E = ∅ and E = −E, then (int E)2 = int E. (d) (cl E)2 = cl E. The next result illustrates the utility of an extra assumption on sets E such that E = E 2 . The authors wish to thank the referee for this suggestion. 2.5. Lemma. If E ⊆ cl D and E = E 2 = −E, then whenever a ∈ E, E contains a dense set of the circle {z : |z| = |a|}. Proof. When a = |a|eiα ∈ E, let Ta = {θ ∈ [0, 2π] : aeiθ ∈ E}. The lemma will follow by establishing the following. Claim. If θ ∈ Ta , then θ + addition is modulo 2π.
kπ 2n
∈ Ta for 0 ≤ k ≤ 2n+1 , where the
This is established by induction. Assume n = 1. If θ ∈ Ta , then |a|ei(α+θ) ∈ E; thus −|a|2 ei(2α+2θ) ∈ E. Taking square roots it follows that i|a|ei(α+θ) = |a|ei(α+θ+π/2) ∈ E and |a|ei(α+θ+3π/2) ∈ E. That is, θ + π2 , θ + 3π ∈ Ta . This says that θ + kπ ∈ Ta for 0 ≤ k ≤ 4 since the 2 2 case k = 2, 4 are trivial. The proof of the induction step is similar. Note that if E is as in the preceding lemma, then so is cl D\E. 2.6. Proposition. If G is a non-empty open subset of D that satisfies G = G2 = −G and G = D, then G is the union of a countable number of pairwise disjoint open annuli. Proof. Since E = cl D\G satisfies E = E 2 = −E and E is closed, we get that when a ∈ G, then the circle {z : |z| = |a|} ⊆ E. This implies that each component of G is an annulus. If K is a compact subset of cl D that satisfies K = K 2 = −K, then we have a characterization of K since G = D\K satisfies the preceding proposition.
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 4 A,
We now turn to some basic spectral properties of operators satisfying Condition S. 2.7. Proposition. If A satisfies Condition S and A = 0, then the spectral radius of A is 1; consequently, A ≥ 1. Moreover if A is an invertible operator satisfying Condition S, then σ(A) ⊆ ∂D. Proof. First observe that Proposition 2.1 implies that σ(A) ⊆ cl D or r(A) ≤ 1. Assume A = 0 and there is an invertible operator R such that A ⊕ A = RA2 R−1 . If c = ( R R−1 )−1 , then A ≤ 1c A2 , so that A2 ≥ c A . Similarly A2 ⊕A2 = RA4 R−1 and so A4 ≥ c A2 . n n n+1 n+1 n By induction we have A2 ⊕A2 = RA2 R−1 and so A 2 ≥ c A2 n+1 for all n ≥ 1. This implies A2 ≥ cn+1 A . Since A = 0, we get that the spectral radius of A is r(A) = lim A2 n
n+1
n+1
1
1
2n+1 ≥ lim c 2n+1 A 2n+1 = 1 n
Since we already have that r(A) ≤ 1 we get r(A) = 1. Now assume that A is invertible. Using the notation from the preceding paragraph we have that A−1 ⊕ A−1 = RA−2 R−1 so that from the first part r(A−1 ) = 1, Thus σ(A) ⊆ ∂D. 2.8. Proposition. If A is a compact operator and A satisfies Condition S, then A = 0. Proof. Assume A satisfies Condition S and A = 0. First observe that A has no non-zero eigenvalues in the open disk. Indeed, if λ is an eigenvalue with 0 < |λ| < 1, then, because A satisfies Condition S, −k for every k in N there is a λk in D ∩ σp (A) with λ2k = λ. But then |λk | → 1, implying that {λk } has an accumulation point that is not 0, contradicting the compactness of A. By the preceding proposition A must have an eigenvalue λ on the unit circle. Again using the compactness of A there can only be a finite number of these and the span of the generalized eigenspaces of all of these has finite dimension n. Thus there is a decomposition of H into possibly non-orthognal invariant subspaces H0 and H1 , where A0 = A|H0 is compact with σ(A0 ) = {0}, n = dim H1 < ∞, and A1 = A|H1 has σ(A1 ) = σ(A) ∩ ∂D. But A20 A21 = A2 ≈ A ⊕ A = (A0 ⊕ A0 ) (A1 ⊕ A1 ). But by matching the spectra of these operators and noting that the span of the generalized eigenspaces for A ⊕ A corresponding to the non-zero eigenvalues is 2n while the corresponding span for A2 is n, we arrive at a contradiction. The proof of the next result is straightforward and provides a rich set of examples.
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2.9. Proposition. Let 0 < r < 1 and let T be any operator with σ(T ) ⊆ {z ∈ C : r2 ≤ |z| ≤ r} such that T has roots of all orders. If ∞ ∞ 1 2n T ⊕ T 2n A0 = n=0
then A =
(∞) A0
n=1
satisfies Condition U.
In fact A20 ∼ = A0 , from which the assertion follows. 2.10. Proposition. Assume that A satisfies Condition S, λ ∈ σ(A), and ±μ are the two square roots of λ. If A − λ is left (respectively, right) semi-Fredholm, then A + μ and A − μ are left (respectively, right) semi-Fredholm and 2 ind (A − λ) = ind (A − μ) + ind (A + μ) ¯ and A − λ is Proof. Note that since ±¯ μ are the two square roots of λ ∗ ¯ right semi-Fredholm if and only if A − λ is left semi-Fredolm; thus it suffices to prove this proposition only in the case that A − λ is left semi-Fredholm, an assumption we now make. It follows that A ⊕ A − λ is also left semi-Fredholm. Because A satisfies Condition S, we have that A2 − λ = (A − μ)(A + μ) is left semi-Fredholm. Recall ([1], p 350) that A − λ is left semi-Fredholm if and only if there is no orthonormal sequence {en } such that (A − λ)en → 0. It easily follows that there is no orthonormal sequence {en } with (A ± μ)en converging to 0 and so both A + μ and A − μ are left semi-Fredholm. Since A ⊕ A − λ and (A − μ)(A + μ) are similar, the statement on the index is immediate from its properties. Assume A satisfies Condition S and let P (A) be the semi-Fredholm domain of A; that is, P (A) = {λ ∈ σ(A) : A − λ is semi-Fredholm}. Observe two things about the preceding proposition. First if λ ∈ P (A) and ±μ ∈ σ(A), it must be that ±μ ∈ P (A); that is, ±μ cannot lie in some other part of the spectrum of A. Second, while the condition that σ(A) = σ(A)2 must hold when A satisfies Condition S, this only implies that at least one of ±μ belongs to σ(A). So it could be that μ ∈ P (A) while A + μ is invertible. This last observation leads to the next corollary. 2.11. Corollary. Maintaining the notation of the preceding proposition, if −μ ∈ / σ(A), then ind (A − μ) = 2 ind (A − λ). 2.12. Corollary. If A satisfies Condition S, P (A) = −P (A), and ind (A − λ) is constant for λ in P (A), then ind (A − λ) = 0 or ±∞ for all λ in P (A).
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 6 A,
3. Normal operators As already noted, two normal operators satisfy Condition S exactly when they satisfy Condition U ([1], Corollary IX.6.11). So throughout this section we will confine our attention to Condition U. In this discussion we will rely heavily on §IX.10 in [1]. If μ is the scalar-valued spectral measure for N , the scalar-valued spectral measure for N 2 is ν(Δ) = μ ◦ (z 2 )−1 (Δ) = μ({z : z 2 ∈ Δ}). Hence the next proposition is a direct consequence of Theorem 10.21 in [1]. 3.1. Proposition. If N is a normal operator with σ(N ) ⊆ cl D, scalarvalued spectral measure μ, and multiplicity function mN , then N satisfies Condition U if and only if: (a) μ is equivalent to the measure ν = μ ◦ (z 2 )−1 ; (b) the multiplicity function for N 2 is 2mN . 3.2. Corollary. If N is a normal operator that satisfies Condition U and ker N = (0), then ker N is infinite dimensional. Proof. In fact ker N 2 = ker N ; hence mN (0) = mN 2 (0) = 2mN (0), and this implies mN (0) = ∞. In addition to Example 1.3, note that the bilateral shift of any multiplicity satisfies condition U. For any measure μ, Nμ is the normal operator defined on L2 (μ) as multiplication by the independent variable; and for any extended integer n and any operator T , T (n) denotes the direct sum of T with itself n times. Recall (see [1], Theorem IX.10.16) that if N is a normal operator on a separable Hilbert space, then there are mutually singular measures μ∞ , μ1 , μ2 , . . . such that 3.3 N∼ = N (∞) ⊕ Nμ ⊕ N (2) ⊕ · · · μ∞
1
μ2
Since the measures μ∞ , μ1 , . . . in (3.3) are mutually singular, there are pairwise disjoint Borel sets Δ∞ , Δ1 , . . . that carry these measures. Maintaining the notation of (3.3), the scalar-valued spectral measure μ of N is the sum of all the measures μ∞ , μ1 , . . . (we will assume that n μn < ∞ so that this sum converges) and mN (λ) = n if and only if λ ∈ Δn a.e. [μ]. It follows that N2 ∼ = N (∞) ⊕ N 2 ⊕ [N 2 ](2) ⊕ · · · μ∞
and
μ1
μ2
⊕ Nμ(2) ⊕ Nμ(4) ⊕ · · · ⊕ Nμ(2n) ⊕ ··· N ⊕N ∼ = Nμ(∞) ∞ n 1 2
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However the fact that μi ⊥ μj for i = j implies that any unitary (∞) (∞) (2) (4) intertwining Nμ∞ ⊕ Nμ21 ⊕ [Nμ22 ](2) ⊕ · · · and Nμ∞ ⊕ Nμ1 ⊕ Nμ2 ⊕ · · · (n) (n) (n) must map each L2 (μn ) onto L2 (μn ) ⊕ L2 (μn ). (See, for example, [1], Lemma IX.10.19.) This gives us the following. 3.4. Lemma. Maintaining the notation in (3.3), the normal operator (n) N satisfies Condition U if and only if Nμn satisfies Condition U for 1 ≤ n ≤ ∞. We give here the computation of the multiplicity function for the square of the normal operator A = Nμ operating on L2 (μ). Temporarily assume that ker A = (0) and put ν = μ ◦ (z 2 )−1 . Let F be a carrier of ν that does not contain 0. Since there are an uncountable number of rays from the origin √ to ∞, there is one with ν-measure 0; fix such a ray. Denote by z the branch of the square √ root function defined on the complement of this ray. Let E± = ± F ; so E+ ∩ E− = ∅ and E+ = −E− . Put E = E+ ∪ E− ; it is easy to check that E is a carrier of μ and F = E 2 . Put μ± = μ|E± and note that μ+ ⊥ μ− . Since E is a carrier of μ, μ+ + μ− = μ and mA , the multiplicity function for A, is equal a.e.[μ] to the characteristic function of E. Now√let F± = E±2 . It follows that F+ ∪ F− = F and F+ ∩ F− = {λ ∈ F : ± λ ∈ E}. √ Claim. z 2 : E± → F± is a bijection with inverse mapping ± z. In fact this is√straightforward from the definitions and the fact that the functions ± z have disjoint ranges. Let ν± be the measure defined on F± by ν± = ν|F± . Note that √ z 2 : (E± , μ± ) → (F± , ν± ) is a measure isomorphism with inverse ± z, and so √ 3.5 ν± (Λ) = μ± (± Λ) for any Borel set Λ contained in F± . Now define U± : L2 (F± , ν± ) → L2 (E± , μ± ) by U± g = g ◦ z 2 . It is easy to check that U± is a unitary. Now define A± : L2 (E± , μ± ) → L2 (E± , μ± ) and B± : L2 (F± , ν± ) → L2 (F± , ν± ) as multiplication by the independent variable. Since μ+ ⊥ μ− and μ = μ+ + μ− , we have that A+ ⊕ A− ∼ = Nμ = A. It is easy to compute that U± B± = A2± U± . That is, A2± ∼ = B± . Therefore mA2± = mB± = 1 a.e.[ν± ] on F± . Note that this says that mA2 (λ) = 2 a.e.[ν] on F+ ∩
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 8 A,
F− and mA2 (λ) = 1 a.e.[ν] on the symmetric difference of F+ and F− . Also observe that the original assumption that ker A = (0) was unnecessary as we can now attach 0 to either F+ or F− (but not both). We summarize this as follows. 3.6. Proposition. Let μ be a compactly supported measure on the plane and put ν = μ ◦ (z 2 )−1 , the scalar-valued spectral measure for Nμ2 . If √ F is a carrier of ν, z a branch of the square root defined a.e.[ν], √ E± = ± F , and F± = E±2 , then the multiplicity function for A2 is given a.e.[ν] by 2 if λ ∈ F+ ∩ F− mA2 (λ) = 1 if λ ∈ (F+ \F− ) ∪ (F− \F+ ) This puts us in a position to characterize the normal operators satisfying Condition U. Indeed in light of Lemma 3.4 the next result accomplishes this. 3.7. Theorem. Let μ be a compactly supported measure on the plane, (n) let n be a positive extended integer, and let A = Nμ . Put ν = μ ◦ (z 2 )−1 . (a) If n = ∞, then A satisfies Condition U if and only if μ and ν are mutually absolutely continuous. (b) If n is finite, then A satisfies Condition U if and only if ker A = (0) and the measures μ, ν, and Δ → μ(−Δ) are all mutually absolutely continuous. Proof. (a) Since A has infinite uniform multiplicity, so do A2 and A⊕A. Hence part (a) follows from Proposition 3.1. (b) As was observed before, if A has a finite dimensional kernel, it cannot satisfy Condition U since mA2 (0) = dim ker A2 = mA (0) = n while mA⊕A (0) = 2n. Assume A satisfies Condition U. We thus have that ker A = (0) and, by Proposition 3.1, μ and ν are mutually absolutely continuous and mA2 = 2n. But using the notation of Proposition 3.6 we have that mA2 (λ) = n a.e.[ν] on (F+ \F− ) ∪ (F− \F+ ). This implies that ν[(F+ \F− ) ∪ (F− \F+ )] = 0. Thus we may assume that F = F+ = F− 2 so that ν = √ ν± . Assume that Δ ⊆ E+ and 0 = μ(Δ). So Δ ⊆ F+ = F 2 2 and Δ = Δ . By (3.5) 0 = μ(Δ) = μ+ (Δ) = ν+ (Δ ) = ν(Δ2 ). Also −Δ ⊆ E− and therefore μ(−Δ) = ν− (Δ2 ) = ν(Δ2 ) = 0. Similarly if Δ ⊆ E− and μ(Δ) = 0, then μ(−Δ) = 0. From here it follows that μ and the measure Δ → μ(−Δ) are mutually absolutely continuous. Conversely, assume the conditions stated in part (b) are true. It follows that ν[(F+ \F− ) ∪ (F− \F+ )] = 0. Indeed, if there is a Borel
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√
√ set Λ contained in F \F with ν(Λ) > 0, then μ( Λ) = μ ( Λ) > 0 + − + √ while μ(− Λ) = 0. Thus mA2 (λ) = 2n a.e.[ν]. By Proposition 3.1, A satisfies Condition U. 3.8. Corollary. If G is an open subset of D, A is the operator defined as multiplication by the independent variable on L2 (G), the L2 -space of area measure on G, then the following hold. (a) A(∞) satisfies condition U if and only if G = G2 . (b) If 1 ≤ n < ∞, then A(n) satisfies condition U if and only if G = G2 = −G. It is worth singling out what happens with hermitian operators. As noted before, the zero operator and the identity operator on infinite dimensional spaces satisfy Condition U, and it is convenient in stating a characterization of the hermitian operators that satisfy this condition to exclude these cases. Also if A satisfies Condition U, we have that R ⊇ σ(A) = σ(A)2 , so σ(A) ⊆ [0, 1]. 3.9. Theorem. If A is a hermitian operator with σ(A) ⊆ [0, 1] but σ(A) is not contained in the two-point set {0, 1} and μ is its scalar-valued spectral measure, then the following statements are equivalent. (a) A satisfies Condition U. (b) The measures μ and Δ → μ({x : x2 ∈ Δ}) are mutually absolutely continuous and A has uniform infinite multiplicity. (c) There is a hermitian operator T with uniform infinite multiplicity and σ(T ) ⊆ [ 14 , 12 ] such that one of the following hold: (i) A ∼ = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ; (ii) A ∼ = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ⊕ I (∞) ⊕ 0(∞) ; (iii) A ∼ = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ⊕ I (∞) ; ∼ (iv) A = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ⊕ 0(∞) ; where I (∞) and 0(∞) are the identity and zero operators on infinite dimensional spaces. Proof. (a) implies (b). From Theorem 3.7 we know that μ and μ ◦ (z 2 )−1 are mutually absolutely continuous. Also since σ(A) ⊆ [0, 1], the measures μ and μ ◦ (−z) are mutually singular. Hence Theorem 3.7 implies that mA (x) = ∞ a.e. [μ]. (b) implies (c). First assume that neither 0 nor 1 is an eigenvalue of A. So μ is carried by the open unit interval. Let A = tdP (t) be the spectral representation of A and let T = A|P ( 14 , 12 ]H. Note that the scalar-valued spectral measure for T is μ|( 14 , 12 ] and its multiplicity function is mT = mA |( 14 , 12 ]. Now T 2 = A2 |P ( 14 , 12 ]H and by examining
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 10 A,
the scalar-valued spectral measures and multiplicity functions we see that (b) implies that T 2 ∼ = A|P (4−2 , 2−2 ]H. Continuing we conclude that for all n ≥ 0, n n n T2 ∼ = A|P 4−2 , 2−2 H By similar reasoning we get that for n ≥ 1,
−n −n −n T2 ∼ H = A|P 4−2 , 2−2 This says that part (i) of (d) holds when A has neither 0 nor 1 as an eigenvalue. The remaining parts will hold when either 0, 1, or both are eigenvalues. (c) implies (a). This is straightforward using Proposition 2.9 or Theorem 3.7 or it can be done directly. 3.10. Corollary. If A is a self-adjoint operator whose scalar-valued spectral measure is equivalent to Lebesgue measure on [0, 1], then A satisfies Condition U if and only if it has uniform infinite multiplicity. 4. Weighted shifts Let {en : n ≥ 0} be an orthonormal basis for the Hilbert space H and let {an } be a bounded sequence of strictly positive scalars with supn an = 1, so that Aen = an en+1 defines a unilateral shift on H of norm 1. This notation will remain fixed in this section, where we will characterize the unilateral weighted shifts that satisfy condition S. Clearly A2 en = an an+1 en+2 . This leads to the following result that is the starting point for this section. 4.1. Lemma. If A is a weighted shift with weight sequence {an }, then A2 ∼ = S0 ⊕S1 , where S0 and S1 are weighted shifts with weight sequences {a2n a2n +1 } and {a2n+1 a2n+2 }, respectively. Proof. If H0 = {e2n : n ≥ 0} and H1 = {e2n+1 : n ≥ 0}, then H0 and H1 are reducing subspaces for A2 with H = H0 ⊕H1 . If S0 = A2 |H0 and S1 = A2 |H1 . These operators satisfy the lemma. This leads us to a consideration of operator-valued weighted shifts where the operators are 2 × 2 matrices. The next lemma is just a rephrasing of the preceding one. 4.2. Lemma. If for each n ≥ 0, Bn and Cn are the 2 × 2 diagonal matrices Bn = diag(an , an )
Cn = diag(a2n a2n+1 , a2n+1 a2n+2 )
and B and C are the operator-valued weighted shifts with weight sequences {Bn } and {Cn }, then A ⊕ A ∼ = C. = B and A2 ∼
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So to discuss Condition S and Condition U for weighted shifts we are led to consider when two operator-valued weighted shifts with invertible matrix weights are similar (or unitarily equivalent). This first result is undoubtedly known by experts, but a handy reference is not available and so we sketch a proof. Indeed it is an extension to operator-valued weighted shifts of a well known result for scalar-valued weighted shifts ([2], Problem 90, or [3], page 54). 4.3. Proposition. If {Bn } and {Cn } are two sequences of invertible matrices on C2 , then the operator weighted shifts with these weights are similar if and only if there is an invertible matrix X on C2 and a finite constant M such that Cn · · · C0 XB0−1 · · · Bn−1 ≤ M and
Bn · · · B0 X −1 C0−1 · · · Cn−1 ≤ M
for all n ≥ 0.
Proof. Let B be the operator weighted shift with weights {Bn } and C the operator weighted shift with weights {Cn }. We will show first that the condition is sufficient. Let R be the diagonal operator with entries X, C0 XB0−1 , C1 C0 XB0−1 B1−1 , . . . , Cn Cn−1 . . . C0 XB0−1 B2−1 . . . Bn−1 , . . . . It is easy to see that the condition implies that R is bounded and invertible. A straightforward computation shows that RB = CR. Conversely, if there is an invertible operator R = (Xij ) such that RB = CR, writing the equality of the diagonal entries in these two ma−1 trix products we obtain exactly that for n ≥ 1, Xnn = Cn−1 . . . C0 X00 B0−1 B1−1 . . . Bn−1 and that X0j = 0 for every j ≥ 1. Since Xnn ≤ R = M , putting X = X00 we have the first desired upper bound. If R−1 = (Yij ), then a similar computation using R−1 C = BR−1 shows that Y00 is invertible −1 and Y00 = X00 . So the same computations gives that sup Bn Bn−1 . . . B0 X −1 C0−1 C1−1 . . . Cn−1 < ∞ n
giving the desired second upper bound. The proof of the next lemma is elementary. 4.4. Lemma. If
x11 x12 X= x21 x22
is an invertible matrix and X
−1
=
y11 y12 y21 y22
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 12 A,
then either x11 x22 y11 y22 = 0 or x12 x21 y12 y21 = 0. 4.5. Proposition. Let S1 , S2 , T1 , T2 be weighted shifts with weight se(n) (n) (n) (n) quences {s1 }, {s2 } and {t1 }, {t2 }, respectively. S1 ⊕ S2 ≈ T1 ⊕ T2 if and only if there are constants 0 < m ≤ M < ∞ such that for all n ≥ 1 either (0)
4.6 m ≤
s1
(0)
t1
(1)
·
s1
·
s1
(n)
(1)
t1
···
s1
···
s1
(n)
t1
(0)
≤M
and
≤M
and
m≤
s2
m≤
s2
(0)
t2
(1)
·
s2
·
s2
(1)
t2
(n)
···
s2
···
s2
(n)
t2
≤M
or (0)
4.7 m ≤
s1
(0)
t2
(1)
(n)
(1)
t2
(n)
t2
(0)
(0)
t1
(1)
(1)
t1
(n)
(n)
t1
≤M
for all n ≥ 0. If this is the case, then we have that S1 ⊕ S2 ≈ T1 ⊕ T2 if and only if either S1 ≈ T1 and S2 ≈ T2 or S1 ≈ T2 and S2 ≈ T1 . Proof. Let’s begin by pointing out that (4.6) is equivalent to the condition that S1 ≈ T1 and S2 ≈ T2 . (See [2], Problem 90 or [3], page 54.) Similarly (4.7) is equivalent to the condition that S1 ≈ T2 and S2 ≈ T1 . So it suffices to prove the first part of the proposition. Assume S1 ⊕ S2 ≈ T1 ⊕ T2 . Let σn and τn be diagonal 2 × 2 matrices (n) (n) (n) (n) with entries {s1 } and {s2 } and {t1 } and {t2 }; and let S and T be the operator weighted shifts with weight sequences {σn } and {τn }. So S ∼ = S1 ⊕ S2 and T ∼ = T1 ⊕ T2 , and therefore S ≈ T . By Proposition 4.3, there is an invertible matrix X such that sup τn τn−1 . . . τ0 Xσ0−1 σ1−1 . . . σn−1 < ∞ n
and sup σn σn−1 . . . σ0 X −1 τ0−1 τ1−1 . . . τn−1 < ∞. n
Let
X=
x11 x12 x21 x22
y11 y12 . y21 y22
and X
−1
=
Then ⎛ ⎜ τn τn−1 . . . τ0 Xσ0−1 σ1−1 . . . σn−1 = ⎜ ⎝
(0) (1)
(n)
t1 ·t1 ···t1
(0) (1) (n) s1 ·s1 ···s1 (0) (1)
(n)
t2 ·t2 ···t2
(0) (1) (n) s1 ·s1 ···s1
x11 x21
(0) (1)
(n)
t1 ·t1 ···t1
(0) (1) (n) s2 ·s2 ···s2 (0) (1)
(n)
t2 ·t2 ···t2
(0) (1) (n) s2 ·s2 ···s2
x12 x22
⎞ ⎟ ⎟ ⎠
POWERS AND DIRECT SUMS
and
⎛
⎜ σn σn−1 . . . σ0 X −1 τ0−1 τ1−1 . . . τn−1 = ⎜ ⎝
(0)
(1)
(n)
s1 ·s1 ···s1
(0) (1) (n) t1 ·t1 ···t1 (0)
(1)
(n)
s2 ·s2 ···s2
(0) (1) (n) t1 ·t1 ···t1
13
(0)
(1)
(n)
s1 ·s1 ···s1
y11
(0) (1) (n) t2 ·t2 ···t2 (0)
(1)
(n)
s2 ·s2 ···s2
y21
(0) (1) (n) t2 ·t2 ···t2
y12
⎞ ⎟ ⎟. ⎠
y22
We will assume that x11 x22 y11 y22 = 0 (4.4). Since the absolute value of an entry in a matrix is less than or equal to the norm of the matrix, we get that the sequences (0) (1) (n) (0) (1) (n) s1 s2 s1 s1 s2 s2 · (1) · · · (n) · (1) · · · (n) and (0) (0) t1 t1 t1 t2 t2 t2 as well as the sequences of their reciprocals are bounded. Thus there are constants 0 < c ≤ C < ∞ with (0)
m≤
s1
(0)
t1
(1)
·
s1
(1)
t1
(n)
···
s1
(n)
t1
(0)
≤M
and m ≤
s2
(0)
t2
(1)
·
s2
(1)
t2
(n)
···
s2
(n)
t2
≤M
for all n ≥ 1. The second option in the Lemma 4.3 leads to the second option in the conclusion of this proposition. The proof of the converse follows the same lines. We can now focus on the characterization of the weighted shifts satisfying Condition S. Recall the notation introduced at the start of this section. 4.8. Theorem. If A is a unilateral weighted shift with weight sequence {an } satisfying 0 < an ≤ 1 for all n, then A satisfies Condition S if and only if there is a positive constant c with c ≤ an an+1 · · · a2n for all n ≥ 0. Proof. We use Proposition 4.5 with S1 and S2 defined as in the preceding lemma and T1 = T2 = A. We need only interpret (4.6) and (4.7) for these shifts and show that this is equivalent to the condition given in this theorem. Since T1 = T2 here, the inequalities in Proposition 4.5 are the same, so we need only work with (4.6). Note that here the first set of inequalities in (4.6) becomes c≤
(a1 a2 )(a3 a4 ) · · · (a2n+1 a2n+2 ) ≤ C. a0 a1 · · · an
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 14 A,
This becomes
an+1 · · · a2n+2 ≤ C; a0 Since an ≤ 1 for all n ≥ 0, we can replace the constant C by 1; replacing c by a smaller constant if necessary, we get c≤
4.9
c ≤ an+1 · · · a2n+2
for n ≥ 0
The second set of inequalities in (4.6) becomes in this case c≤
(a0 a1 )(a2 a3 ) · · · (a2n a2n+1 ) ≤ C. a0 a1 · · · an
This reduces to 4.10
c ≤ an · · · a2n
for n ≥ 0.
Therefore both of these inequalities are covered by (4.10), which is the inequality in the statement of the theorem. 4.11. Corollary. The Bergman shift satisfies Condition S. Proof. The weight sequence for the Bergman shift is given by n+1 an = n+2 It follows that n+1 1 = an an+1 · · · a2n = 2n + 2 2 4.12. Corollary. If A is a hyponormal weighted shift with weight sequence {an }, then A satisfies Condition S if and only if there is a 1 constant c with 0 < c < 1 and an ≥ c n for all n ≥ 1. Proof. It is well known that a weighted shift is hyponormal if and only if the weight sequence is increasing. In these circumstances an · · · a2n ≥ 1 ≥ cc n ≥ c2 . an+1 n Now we examine Condition U for unilateral weighted shifts. 4.13. Theorem. Let A be a unilateral weighted shift with weight sequence of positive numbers {an }, then A satisfies condition U if and only if there is a constant a ≥ 1 such that: ⎧ ⎪ ⎨a when n is even 4.14 an = 1 when n = 1 ⎪ ⎩ ak if n = 2k + 1 and k ≥ 1 a
POWERS AND DIRECT SUMS
15
Proof. Assume a ≥ 1 and {an } is defined as in (4.14). It is clear that A is a bounded operator with A = a. As in Lemma 4.1, let H0 and H1 be the closed linear spans of the even and odd basis vectors, respectively. Define U : H → H ⊕ H by letting U e2n = en ⊕ 0 and U e2n+1 = 0⊕en for all n ≥ 0. Clearly U is a unitary that maps H0 onto H ⊕ (0) and H1 onto (0) ⊕ H. We claim that U A2 = (A ⊕ A)U . In fact when n = 2k for some k ≥ 0, U A2 en = aa2k+1 U e2k+2 = ak (ek+1 ⊕ 0) = Aek ⊕ 0 = (A ⊕ A)U en . A similar computation shows that U A2 en = (A ⊕ A)U en when n is odd. This finishes this part of the proof. Now let us assume that A is a weighted shift operator with weight sequence {an } that satisfies condition U . As in the proof of Theorem 4.8 we regard A ⊕ A as the operator-valued weighted shift with weight sequence {Bn }n≥0 , where each Bn is a 2 × 2 diagonal matrix Bn = diag(an , an ). Recall from Lemma 4.1 that A2 is unitarily equivalent to the operator-valued weighted shift with weight sequence {Cn }n≥0 , where each Cn is a 2 × 2 diagonal matrix Cn = diag(a2n a2n+1 , a2n+1 a2n+2 ). We regard B and C as operating on the 2 same space K = ⊕∞ n=0 Kn where each Kn = C . ∼ Since A satisfies Condition U, B = C. Let W be a unitary on K such that B = W ∗ CW ; so B ∗n = W ∗ C ∗n W for n ≥ 1. Thus W must map ker B ∗n isometrically onto ker C ∗n . It follows that each Kn reduces W so that W = ⊕∞ n=1 Wn , where each Wn is a 2 × 2 unitary matrix operating on Kn . Moreover we have that 4.15
Wn+1 Bn = Cn Wn
Letting W = [wijn ], multiplying the matrices in (4.15), and comparing entries we get the equations n+1 n w11 an = w11 a2n a2n+1 ,
n+1 n w12 an = w12 a2n a2n+1
n+1 n n+1 n w21 an = w21 a2n+1 a2n+2 , w22 an = w22 a2n+1 a2n+2 Using the fact that both the rows and the columns of each Wn are n n n n | = |w22 | and |w12 | = |w21 | for each orthonormal pairs, we have that |w11 n n ≥ 0. Fix some n ≥ 0 and suppose w11 = 0. Using the above equations n n and w22 and the fact that the weights are all positive, that involve w11 n+1 n+1 we see that w11 = 0 = w22 . Therefore solving for an and taking the modulus of the resulting equations yields that a2n+1 a2n+2 = a2n a2n+1 . Canceling we get
4.16
a2n = a2n+2
n for all n ≥ 0. This was under the assumption that w11 = 0; now assume n n n n that w11 = 0. So it follows that w22 = 0 and |w12 | = |w21 |. Arguing just as in the previous case but using the other two equations above yields
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 16 A,
the same conclusion (4.16). Putting a = a0 , (4.16) implies a2n = a for all n ≥ 0. Taking the absolute values of the determinants of both sides of (4.15) shows that a2n = det Bn = det Cn = a2n a2n+2 a22n+1 = a2 a22n+1 ; hence a2n+1 = an /a and we have established that (4.14) is valid. 4.17. Corollary. The only hyponormal weighted shift that satisfies Condition U is the shift with weight sequence 1, 1, . . . . Proof. If A is such a weighted shift with weights as in the preceding theorem, the fact that A is hyponormal implies an ≤ an+1 for all n. A moment’s thought shows that this implies that an = 1 for all n ≥ 0.
POWERS AND DIRECT SUMS
17
References [1] J B Conway, A Course in Functional Analysis, Second Edition, SpringerVerlag, New York (1990). [2] P R Halmos, A Hilbert Space Problem Book, Springer-Verlag, New York (1982). [3] A L Shields Weighted shift operators and analytic function theory, Amer Math Soc Math Surveys, 13 (1974) 49–128. The George Washington University E-mail address: [email protected] SUNY Brockport E-mail address: [email protected] Khalifa University of Science, Technology and Research, Abu Dhabi, UAE E-mail address: [email protected]
John B. Conway, Gabriel Prˇajiturˇa, Alejandro Rodríguez-Martínez
PII: DOI: Reference:
S0022-247X(13)01086-X 10.1016/j.jmaa.2013.12.007 YJMAA 18111
To appear in:
Journal of Mathematical Analysis and Applications
Received date: 20 May 2010
Please cite this article in press as: J.B. Conway et al., Powers and direct sums, J. Math. Anal. Appl. (2014), http://dx.doi.org/10.1016/j.jmaa.2013.12.007
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POWERS AND DIRECT SUMS ˇ ˇ JOHN B. CONWAY, GABRIEL PRAJITUR A, ´ ´ AND ALEJANDRO RODRIGUEZ-MARTINEZ
1. Introduction If a general theory of multiplicity ever comes to be, it will likely be the case that for any operator A, the operator A(2) = A ⊕ A has twice the multiplicity of A. As in the case of hermitian operators, it might sometimes be the case that A2 has twice the multiplicity of A and sometimes that it does not. Though this paper will not try to begin a general theory of multiplicity, it does explore the relationship between A ⊕ A and A2 . Specifically it studies operators A such that A ⊕ A and A2 are similar, a question that has an intrinsic interest independent of any attempt at multiplicity theory. 1.1. Definition. If A ∈ B(H), A satisfies Condition S if A2 ≈ A ⊕ A. Say that A satisfies Condition U if A2 ∼ = A ⊕ A. (≈ means similar and ∼ = means unitarily equivalent.) Realize that for normal operators Condition S and Condition U are the same since two normal operators are similar if and only if they are unitarily equivalent ([1], Corollary IX.6.11). We will shortly see several examples of operators satisfying Condition S, but for the time being we present only two. 1.2. Example. If S is the unilateral shift, then S satisfies Condition U. 1.3. Example. If A denotes multiplication by the independent variable on L2 [0, 1], then A does not satisfy Condition S. In fact since x2 is oneto-one on [0, 1], A2 is unitarily equivalent to A; however A ⊕ A has uniform multiplicity 2. On the other hand, A(∞) , the direct sum of A with itself an infinite number of times, does satisfy Condition S (and therefore Condition U). The paper is organized as follows. §2 presents some spectral properties of operators that satisfy Condition S. In particular it is shown that such operators have spectral radius 1 and that the only compact 1
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 2 A,
operator satisfying Condition S is the zero operator. §3 gives a complete characterization of the normal operators satisfying Condition S. In §4 the unilateral weighted shifts satisfying Condition S as well as those satisfying Condition U are characterized. This results in a rather simple criterion for a hyponormal weighted shift to satisfy Condition S and the fact that the isometric weighted shift is the only hyponormal weighted shift that satisfies Condition U. 2. Spectral results The proofs of the next two results are straightforward. Recall that σe (A) denotes the essential spectrum of A; that is, the spectrum of the image of A in the Calkin algebra. 2.1. Proposition. If A satisfies Condition S, then σ(A) = σ(A)2 and σe (A) = σe (A)2 . 2.2. Proposition. (a) If A and B satisfy Condition S (or U ), then so does A ⊕ B. (b) A satisfies Condition S (or U) if and only if A∗ does. n (c) If A satisfies Condition S (or U ), then so does A2 for all n ∈ N. Note that a scalar multiple of an operator satisfying Condition S (or U) satisfies the same condition only if the scalar is either 0 or 1. In light of Proposition 2.1 it becomes important for the problem of characterizing the operators satisfying Condition S to characterize the compact subsets K of C that satisfy K = K 2 . At present we cannot do this, but we can make such a characterization when K is a subset of R. 2.3. Proposition. A compact subset K of R satisfies K = K 2 if and only if one of the following holds: (a) K = {0}; (b) K = {1}; (c) K = {0, 1}; (d) for any r in the open (0, 1) there is a compact subset D of interval n=∞ 2n 1/2n D ∪ ∪ {0, 1} [r2 , r] such that K = n=∞ n=0 n=1 D Proof. It is clear that any set K that has the form in parts (a) through (d) must satisfy K = K 2 , so we look at the converse. Assume K = K 2 and assume that none of the conditions (a) through (c) is true; let 0 < r < 1. We first note that K ∩ [r2 , r] = ∅. In fact if s ∈ K and n s = 0, 1, let n be the smallest natural number such that s2 ≤ r. If n n−1 it were the case that s2 ≤ r2 , then we would have that s2 ≤ r,
POWERS AND DIRECT SUMS
3
2n
a contradiction; s > r2 . That is D = K ∩ [r2 , r] = ∅. Let n=∞ n=∞ 2n so n L = n=0 D ∪ n=1 D1/2 ∪ {0, 1}. Clearly L ⊆ K and L = L2 . Using an argument similar to the one used to show that K ∩ [r2 , r] = ∅ we can establish the reverse inclusion and thus the equality of K and L. The next result is straightforward. 2.4. Proposition. Assume E is a non-empty, bounded subset of C that satisfies E = E 2 . (a) E ⊆ cl D. (b) If E ∩ D = ∅, then 0 ∈ cl E and cl E ∩ ∂D = ∅. (c) If int E = ∅ and E = −E, then (int E)2 = int E. (d) (cl E)2 = cl E. The next result illustrates the utility of an extra assumption on sets E such that E = E 2 . The authors wish to thank the referee for this suggestion. 2.5. Lemma. If E ⊆ cl D and E = E 2 = −E, then whenever a ∈ E, E contains a dense set of the circle {z : |z| = |a|}. Proof. When a = |a|eiα ∈ E, let Ta = {θ ∈ [0, 2π] : aeiθ ∈ E}. The lemma will follow by establishing the following. Claim. If θ ∈ Ta , then θ + addition is modulo 2π.
kπ 2n
∈ Ta for 0 ≤ k ≤ 2n+1 , where the
This is established by induction. Assume n = 1. If θ ∈ Ta , then |a|ei(α+θ) ∈ E; thus −|a|2 ei(2α+2θ) ∈ E. Taking square roots it follows that i|a|ei(α+θ) = |a|ei(α+θ+π/2) ∈ E and |a|ei(α+θ+3π/2) ∈ E. That is, θ + π2 , θ + 3π ∈ Ta . This says that θ + kπ ∈ Ta for 0 ≤ k ≤ 4 since the 2 2 case k = 2, 4 are trivial. The proof of the induction step is similar. Note that if E is as in the preceding lemma, then so is cl D\E. 2.6. Proposition. If G is a non-empty open subset of D that satisfies G = G2 = −G and G = D, then G is the union of a countable number of pairwise disjoint open annuli. Proof. Since E = cl D\G satisfies E = E 2 = −E and E is closed, we get that when a ∈ G, then the circle {z : |z| = |a|} ⊆ E. This implies that each component of G is an annulus. If K is a compact subset of cl D that satisfies K = K 2 = −K, then we have a characterization of K since G = D\K satisfies the preceding proposition.
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 4 A,
We now turn to some basic spectral properties of operators satisfying Condition S. 2.7. Proposition. If A satisfies Condition S and A = 0, then the spectral radius of A is 1; consequently, A ≥ 1. Moreover if A is an invertible operator satisfying Condition S, then σ(A) ⊆ ∂D. Proof. First observe that Proposition 2.1 implies that σ(A) ⊆ cl D or r(A) ≤ 1. Assume A = 0 and there is an invertible operator R such that A ⊕ A = RA2 R−1 . If c = ( R R−1 )−1 , then A ≤ 1c A2 , so that A2 ≥ c A . Similarly A2 ⊕A2 = RA4 R−1 and so A4 ≥ c A2 . n n n+1 n+1 n By induction we have A2 ⊕A2 = RA2 R−1 and so A 2 ≥ c A2 n+1 for all n ≥ 1. This implies A2 ≥ cn+1 A . Since A = 0, we get that the spectral radius of A is r(A) = lim A2 n
n+1
n+1
1
1
2n+1 ≥ lim c 2n+1 A 2n+1 = 1 n
Since we already have that r(A) ≤ 1 we get r(A) = 1. Now assume that A is invertible. Using the notation from the preceding paragraph we have that A−1 ⊕ A−1 = RA−2 R−1 so that from the first part r(A−1 ) = 1, Thus σ(A) ⊆ ∂D. 2.8. Proposition. If A is a compact operator and A satisfies Condition S, then A = 0. Proof. Assume A satisfies Condition S and A = 0. First observe that A has no non-zero eigenvalues in the open disk. Indeed, if λ is an eigenvalue with 0 < |λ| < 1, then, because A satisfies Condition S, −k for every k in N there is a λk in D ∩ σp (A) with λ2k = λ. But then |λk | → 1, implying that {λk } has an accumulation point that is not 0, contradicting the compactness of A. By the preceding proposition A must have an eigenvalue λ on the unit circle. Again using the compactness of A there can only be a finite number of these and the span of the generalized eigenspaces of all of these has finite dimension n. Thus there is a decomposition of H into possibly non-orthognal invariant subspaces H0 and H1 , where A0 = A|H0 is compact with σ(A0 ) = {0}, n = dim H1 < ∞, and A1 = A|H1 has σ(A1 ) = σ(A) ∩ ∂D. But A20 A21 = A2 ≈ A ⊕ A = (A0 ⊕ A0 ) (A1 ⊕ A1 ). But by matching the spectra of these operators and noting that the span of the generalized eigenspaces for A ⊕ A corresponding to the non-zero eigenvalues is 2n while the corresponding span for A2 is n, we arrive at a contradiction. The proof of the next result is straightforward and provides a rich set of examples.
POWERS AND DIRECT SUMS
5
2.9. Proposition. Let 0 < r < 1 and let T be any operator with σ(T ) ⊆ {z ∈ C : r2 ≤ |z| ≤ r} such that T has roots of all orders. If ∞ ∞ 1 2n T ⊕ T 2n A0 = n=0
then A =
(∞) A0
n=1
satisfies Condition U.
In fact A20 ∼ = A0 , from which the assertion follows. 2.10. Proposition. Assume that A satisfies Condition S, λ ∈ σ(A), and ±μ are the two square roots of λ. If A − λ is left (respectively, right) semi-Fredholm, then A + μ and A − μ are left (respectively, right) semi-Fredholm and 2 ind (A − λ) = ind (A − μ) + ind (A + μ) ¯ and A − λ is Proof. Note that since ±¯ μ are the two square roots of λ ∗ ¯ right semi-Fredholm if and only if A − λ is left semi-Fredolm; thus it suffices to prove this proposition only in the case that A − λ is left semi-Fredholm, an assumption we now make. It follows that A ⊕ A − λ is also left semi-Fredholm. Because A satisfies Condition S, we have that A2 − λ = (A − μ)(A + μ) is left semi-Fredholm. Recall ([1], p 350) that A − λ is left semi-Fredholm if and only if there is no orthonormal sequence {en } such that (A − λ)en → 0. It easily follows that there is no orthonormal sequence {en } with (A ± μ)en converging to 0 and so both A + μ and A − μ are left semi-Fredholm. Since A ⊕ A − λ and (A − μ)(A + μ) are similar, the statement on the index is immediate from its properties. Assume A satisfies Condition S and let P (A) be the semi-Fredholm domain of A; that is, P (A) = {λ ∈ σ(A) : A − λ is semi-Fredholm}. Observe two things about the preceding proposition. First if λ ∈ P (A) and ±μ ∈ σ(A), it must be that ±μ ∈ P (A); that is, ±μ cannot lie in some other part of the spectrum of A. Second, while the condition that σ(A) = σ(A)2 must hold when A satisfies Condition S, this only implies that at least one of ±μ belongs to σ(A). So it could be that μ ∈ P (A) while A + μ is invertible. This last observation leads to the next corollary. 2.11. Corollary. Maintaining the notation of the preceding proposition, if −μ ∈ / σ(A), then ind (A − μ) = 2 ind (A − λ). 2.12. Corollary. If A satisfies Condition S, P (A) = −P (A), and ind (A − λ) is constant for λ in P (A), then ind (A − λ) = 0 or ±∞ for all λ in P (A).
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 6 A,
3. Normal operators As already noted, two normal operators satisfy Condition S exactly when they satisfy Condition U ([1], Corollary IX.6.11). So throughout this section we will confine our attention to Condition U. In this discussion we will rely heavily on §IX.10 in [1]. If μ is the scalar-valued spectral measure for N , the scalar-valued spectral measure for N 2 is ν(Δ) = μ ◦ (z 2 )−1 (Δ) = μ({z : z 2 ∈ Δ}). Hence the next proposition is a direct consequence of Theorem 10.21 in [1]. 3.1. Proposition. If N is a normal operator with σ(N ) ⊆ cl D, scalarvalued spectral measure μ, and multiplicity function mN , then N satisfies Condition U if and only if: (a) μ is equivalent to the measure ν = μ ◦ (z 2 )−1 ; (b) the multiplicity function for N 2 is 2mN . 3.2. Corollary. If N is a normal operator that satisfies Condition U and ker N = (0), then ker N is infinite dimensional. Proof. In fact ker N 2 = ker N ; hence mN (0) = mN 2 (0) = 2mN (0), and this implies mN (0) = ∞. In addition to Example 1.3, note that the bilateral shift of any multiplicity satisfies condition U. For any measure μ, Nμ is the normal operator defined on L2 (μ) as multiplication by the independent variable; and for any extended integer n and any operator T , T (n) denotes the direct sum of T with itself n times. Recall (see [1], Theorem IX.10.16) that if N is a normal operator on a separable Hilbert space, then there are mutually singular measures μ∞ , μ1 , μ2 , . . . such that 3.3 N∼ = N (∞) ⊕ Nμ ⊕ N (2) ⊕ · · · μ∞
1
μ2
Since the measures μ∞ , μ1 , . . . in (3.3) are mutually singular, there are pairwise disjoint Borel sets Δ∞ , Δ1 , . . . that carry these measures. Maintaining the notation of (3.3), the scalar-valued spectral measure μ of N is the sum of all the measures μ∞ , μ1 , . . . (we will assume that n μn < ∞ so that this sum converges) and mN (λ) = n if and only if λ ∈ Δn a.e. [μ]. It follows that N2 ∼ = N (∞) ⊕ N 2 ⊕ [N 2 ](2) ⊕ · · · μ∞
and
μ1
μ2
⊕ Nμ(2) ⊕ Nμ(4) ⊕ · · · ⊕ Nμ(2n) ⊕ ··· N ⊕N ∼ = Nμ(∞) ∞ n 1 2
POWERS AND DIRECT SUMS
7
However the fact that μi ⊥ μj for i = j implies that any unitary (∞) (∞) (2) (4) intertwining Nμ∞ ⊕ Nμ21 ⊕ [Nμ22 ](2) ⊕ · · · and Nμ∞ ⊕ Nμ1 ⊕ Nμ2 ⊕ · · · (n) (n) (n) must map each L2 (μn ) onto L2 (μn ) ⊕ L2 (μn ). (See, for example, [1], Lemma IX.10.19.) This gives us the following. 3.4. Lemma. Maintaining the notation in (3.3), the normal operator (n) N satisfies Condition U if and only if Nμn satisfies Condition U for 1 ≤ n ≤ ∞. We give here the computation of the multiplicity function for the square of the normal operator A = Nμ operating on L2 (μ). Temporarily assume that ker A = (0) and put ν = μ ◦ (z 2 )−1 . Let F be a carrier of ν that does not contain 0. Since there are an uncountable number of rays from the origin √ to ∞, there is one with ν-measure 0; fix such a ray. Denote by z the branch of the square √ root function defined on the complement of this ray. Let E± = ± F ; so E+ ∩ E− = ∅ and E+ = −E− . Put E = E+ ∪ E− ; it is easy to check that E is a carrier of μ and F = E 2 . Put μ± = μ|E± and note that μ+ ⊥ μ− . Since E is a carrier of μ, μ+ + μ− = μ and mA , the multiplicity function for A, is equal a.e.[μ] to the characteristic function of E. Now√let F± = E±2 . It follows that F+ ∪ F− = F and F+ ∩ F− = {λ ∈ F : ± λ ∈ E}. √ Claim. z 2 : E± → F± is a bijection with inverse mapping ± z. In fact this is√straightforward from the definitions and the fact that the functions ± z have disjoint ranges. Let ν± be the measure defined on F± by ν± = ν|F± . Note that √ z 2 : (E± , μ± ) → (F± , ν± ) is a measure isomorphism with inverse ± z, and so √ 3.5 ν± (Λ) = μ± (± Λ) for any Borel set Λ contained in F± . Now define U± : L2 (F± , ν± ) → L2 (E± , μ± ) by U± g = g ◦ z 2 . It is easy to check that U± is a unitary. Now define A± : L2 (E± , μ± ) → L2 (E± , μ± ) and B± : L2 (F± , ν± ) → L2 (F± , ν± ) as multiplication by the independent variable. Since μ+ ⊥ μ− and μ = μ+ + μ− , we have that A+ ⊕ A− ∼ = Nμ = A. It is easy to compute that U± B± = A2± U± . That is, A2± ∼ = B± . Therefore mA2± = mB± = 1 a.e.[ν± ] on F± . Note that this says that mA2 (λ) = 2 a.e.[ν] on F+ ∩
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 8 A,
F− and mA2 (λ) = 1 a.e.[ν] on the symmetric difference of F+ and F− . Also observe that the original assumption that ker A = (0) was unnecessary as we can now attach 0 to either F+ or F− (but not both). We summarize this as follows. 3.6. Proposition. Let μ be a compactly supported measure on the plane and put ν = μ ◦ (z 2 )−1 , the scalar-valued spectral measure for Nμ2 . If √ F is a carrier of ν, z a branch of the square root defined a.e.[ν], √ E± = ± F , and F± = E±2 , then the multiplicity function for A2 is given a.e.[ν] by 2 if λ ∈ F+ ∩ F− mA2 (λ) = 1 if λ ∈ (F+ \F− ) ∪ (F− \F+ ) This puts us in a position to characterize the normal operators satisfying Condition U. Indeed in light of Lemma 3.4 the next result accomplishes this. 3.7. Theorem. Let μ be a compactly supported measure on the plane, (n) let n be a positive extended integer, and let A = Nμ . Put ν = μ ◦ (z 2 )−1 . (a) If n = ∞, then A satisfies Condition U if and only if μ and ν are mutually absolutely continuous. (b) If n is finite, then A satisfies Condition U if and only if ker A = (0) and the measures μ, ν, and Δ → μ(−Δ) are all mutually absolutely continuous. Proof. (a) Since A has infinite uniform multiplicity, so do A2 and A⊕A. Hence part (a) follows from Proposition 3.1. (b) As was observed before, if A has a finite dimensional kernel, it cannot satisfy Condition U since mA2 (0) = dim ker A2 = mA (0) = n while mA⊕A (0) = 2n. Assume A satisfies Condition U. We thus have that ker A = (0) and, by Proposition 3.1, μ and ν are mutually absolutely continuous and mA2 = 2n. But using the notation of Proposition 3.6 we have that mA2 (λ) = n a.e.[ν] on (F+ \F− ) ∪ (F− \F+ ). This implies that ν[(F+ \F− ) ∪ (F− \F+ )] = 0. Thus we may assume that F = F+ = F− 2 so that ν = √ ν± . Assume that Δ ⊆ E+ and 0 = μ(Δ). So Δ ⊆ F+ = F 2 2 and Δ = Δ . By (3.5) 0 = μ(Δ) = μ+ (Δ) = ν+ (Δ ) = ν(Δ2 ). Also −Δ ⊆ E− and therefore μ(−Δ) = ν− (Δ2 ) = ν(Δ2 ) = 0. Similarly if Δ ⊆ E− and μ(Δ) = 0, then μ(−Δ) = 0. From here it follows that μ and the measure Δ → μ(−Δ) are mutually absolutely continuous. Conversely, assume the conditions stated in part (b) are true. It follows that ν[(F+ \F− ) ∪ (F− \F+ )] = 0. Indeed, if there is a Borel
POWERS AND DIRECT SUMS
9
√
√ set Λ contained in F \F with ν(Λ) > 0, then μ( Λ) = μ ( Λ) > 0 + − + √ while μ(− Λ) = 0. Thus mA2 (λ) = 2n a.e.[ν]. By Proposition 3.1, A satisfies Condition U. 3.8. Corollary. If G is an open subset of D, A is the operator defined as multiplication by the independent variable on L2 (G), the L2 -space of area measure on G, then the following hold. (a) A(∞) satisfies condition U if and only if G = G2 . (b) If 1 ≤ n < ∞, then A(n) satisfies condition U if and only if G = G2 = −G. It is worth singling out what happens with hermitian operators. As noted before, the zero operator and the identity operator on infinite dimensional spaces satisfy Condition U, and it is convenient in stating a characterization of the hermitian operators that satisfy this condition to exclude these cases. Also if A satisfies Condition U, we have that R ⊇ σ(A) = σ(A)2 , so σ(A) ⊆ [0, 1]. 3.9. Theorem. If A is a hermitian operator with σ(A) ⊆ [0, 1] but σ(A) is not contained in the two-point set {0, 1} and μ is its scalar-valued spectral measure, then the following statements are equivalent. (a) A satisfies Condition U. (b) The measures μ and Δ → μ({x : x2 ∈ Δ}) are mutually absolutely continuous and A has uniform infinite multiplicity. (c) There is a hermitian operator T with uniform infinite multiplicity and σ(T ) ⊆ [ 14 , 12 ] such that one of the following hold: (i) A ∼ = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ; (ii) A ∼ = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ⊕ I (∞) ⊕ 0(∞) ; (iii) A ∼ = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ⊕ I (∞) ; ∼ (iv) A = · · · ⊕ T 1/4 ⊕ T 1/2 ⊕ T ⊕ T 2 ⊕ T 4 ⊕ · · · ⊕ 0(∞) ; where I (∞) and 0(∞) are the identity and zero operators on infinite dimensional spaces. Proof. (a) implies (b). From Theorem 3.7 we know that μ and μ ◦ (z 2 )−1 are mutually absolutely continuous. Also since σ(A) ⊆ [0, 1], the measures μ and μ ◦ (−z) are mutually singular. Hence Theorem 3.7 implies that mA (x) = ∞ a.e. [μ]. (b) implies (c). First assume that neither 0 nor 1 is an eigenvalue of A. So μ is carried by the open unit interval. Let A = tdP (t) be the spectral representation of A and let T = A|P ( 14 , 12 ]H. Note that the scalar-valued spectral measure for T is μ|( 14 , 12 ] and its multiplicity function is mT = mA |( 14 , 12 ]. Now T 2 = A2 |P ( 14 , 12 ]H and by examining
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 10 A,
the scalar-valued spectral measures and multiplicity functions we see that (b) implies that T 2 ∼ = A|P (4−2 , 2−2 ]H. Continuing we conclude that for all n ≥ 0, n n n T2 ∼ = A|P 4−2 , 2−2 H By similar reasoning we get that for n ≥ 1,
−n −n −n T2 ∼ H = A|P 4−2 , 2−2 This says that part (i) of (d) holds when A has neither 0 nor 1 as an eigenvalue. The remaining parts will hold when either 0, 1, or both are eigenvalues. (c) implies (a). This is straightforward using Proposition 2.9 or Theorem 3.7 or it can be done directly. 3.10. Corollary. If A is a self-adjoint operator whose scalar-valued spectral measure is equivalent to Lebesgue measure on [0, 1], then A satisfies Condition U if and only if it has uniform infinite multiplicity. 4. Weighted shifts Let {en : n ≥ 0} be an orthonormal basis for the Hilbert space H and let {an } be a bounded sequence of strictly positive scalars with supn an = 1, so that Aen = an en+1 defines a unilateral shift on H of norm 1. This notation will remain fixed in this section, where we will characterize the unilateral weighted shifts that satisfy condition S. Clearly A2 en = an an+1 en+2 . This leads to the following result that is the starting point for this section. 4.1. Lemma. If A is a weighted shift with weight sequence {an }, then A2 ∼ = S0 ⊕S1 , where S0 and S1 are weighted shifts with weight sequences {a2n a2n +1 } and {a2n+1 a2n+2 }, respectively. Proof. If H0 = {e2n : n ≥ 0} and H1 = {e2n+1 : n ≥ 0}, then H0 and H1 are reducing subspaces for A2 with H = H0 ⊕H1 . If S0 = A2 |H0 and S1 = A2 |H1 . These operators satisfy the lemma. This leads us to a consideration of operator-valued weighted shifts where the operators are 2 × 2 matrices. The next lemma is just a rephrasing of the preceding one. 4.2. Lemma. If for each n ≥ 0, Bn and Cn are the 2 × 2 diagonal matrices Bn = diag(an , an )
Cn = diag(a2n a2n+1 , a2n+1 a2n+2 )
and B and C are the operator-valued weighted shifts with weight sequences {Bn } and {Cn }, then A ⊕ A ∼ = C. = B and A2 ∼
POWERS AND DIRECT SUMS
11
So to discuss Condition S and Condition U for weighted shifts we are led to consider when two operator-valued weighted shifts with invertible matrix weights are similar (or unitarily equivalent). This first result is undoubtedly known by experts, but a handy reference is not available and so we sketch a proof. Indeed it is an extension to operator-valued weighted shifts of a well known result for scalar-valued weighted shifts ([2], Problem 90, or [3], page 54). 4.3. Proposition. If {Bn } and {Cn } are two sequences of invertible matrices on C2 , then the operator weighted shifts with these weights are similar if and only if there is an invertible matrix X on C2 and a finite constant M such that Cn · · · C0 XB0−1 · · · Bn−1 ≤ M and
Bn · · · B0 X −1 C0−1 · · · Cn−1 ≤ M
for all n ≥ 0.
Proof. Let B be the operator weighted shift with weights {Bn } and C the operator weighted shift with weights {Cn }. We will show first that the condition is sufficient. Let R be the diagonal operator with entries X, C0 XB0−1 , C1 C0 XB0−1 B1−1 , . . . , Cn Cn−1 . . . C0 XB0−1 B2−1 . . . Bn−1 , . . . . It is easy to see that the condition implies that R is bounded and invertible. A straightforward computation shows that RB = CR. Conversely, if there is an invertible operator R = (Xij ) such that RB = CR, writing the equality of the diagonal entries in these two ma−1 trix products we obtain exactly that for n ≥ 1, Xnn = Cn−1 . . . C0 X00 B0−1 B1−1 . . . Bn−1 and that X0j = 0 for every j ≥ 1. Since Xnn ≤ R = M , putting X = X00 we have the first desired upper bound. If R−1 = (Yij ), then a similar computation using R−1 C = BR−1 shows that Y00 is invertible −1 and Y00 = X00 . So the same computations gives that sup Bn Bn−1 . . . B0 X −1 C0−1 C1−1 . . . Cn−1 < ∞ n
giving the desired second upper bound. The proof of the next lemma is elementary. 4.4. Lemma. If
x11 x12 X= x21 x22
is an invertible matrix and X
−1
=
y11 y12 y21 y22
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 12 A,
then either x11 x22 y11 y22 = 0 or x12 x21 y12 y21 = 0. 4.5. Proposition. Let S1 , S2 , T1 , T2 be weighted shifts with weight se(n) (n) (n) (n) quences {s1 }, {s2 } and {t1 }, {t2 }, respectively. S1 ⊕ S2 ≈ T1 ⊕ T2 if and only if there are constants 0 < m ≤ M < ∞ such that for all n ≥ 1 either (0)
4.6 m ≤
s1
(0)
t1
(1)
·
s1
·
s1
(n)
(1)
t1
···
s1
···
s1
(n)
t1
(0)
≤M
and
≤M
and
m≤
s2
m≤
s2
(0)
t2
(1)
·
s2
·
s2
(1)
t2
(n)
···
s2
···
s2
(n)
t2
≤M
or (0)
4.7 m ≤
s1
(0)
t2
(1)
(n)
(1)
t2
(n)
t2
(0)
(0)
t1
(1)
(1)
t1
(n)
(n)
t1
≤M
for all n ≥ 0. If this is the case, then we have that S1 ⊕ S2 ≈ T1 ⊕ T2 if and only if either S1 ≈ T1 and S2 ≈ T2 or S1 ≈ T2 and S2 ≈ T1 . Proof. Let’s begin by pointing out that (4.6) is equivalent to the condition that S1 ≈ T1 and S2 ≈ T2 . (See [2], Problem 90 or [3], page 54.) Similarly (4.7) is equivalent to the condition that S1 ≈ T2 and S2 ≈ T1 . So it suffices to prove the first part of the proposition. Assume S1 ⊕ S2 ≈ T1 ⊕ T2 . Let σn and τn be diagonal 2 × 2 matrices (n) (n) (n) (n) with entries {s1 } and {s2 } and {t1 } and {t2 }; and let S and T be the operator weighted shifts with weight sequences {σn } and {τn }. So S ∼ = S1 ⊕ S2 and T ∼ = T1 ⊕ T2 , and therefore S ≈ T . By Proposition 4.3, there is an invertible matrix X such that sup τn τn−1 . . . τ0 Xσ0−1 σ1−1 . . . σn−1 < ∞ n
and sup σn σn−1 . . . σ0 X −1 τ0−1 τ1−1 . . . τn−1 < ∞. n
Let
X=
x11 x12 x21 x22
y11 y12 . y21 y22
and X
−1
=
Then ⎛ ⎜ τn τn−1 . . . τ0 Xσ0−1 σ1−1 . . . σn−1 = ⎜ ⎝
(0) (1)
(n)
t1 ·t1 ···t1
(0) (1) (n) s1 ·s1 ···s1 (0) (1)
(n)
t2 ·t2 ···t2
(0) (1) (n) s1 ·s1 ···s1
x11 x21
(0) (1)
(n)
t1 ·t1 ···t1
(0) (1) (n) s2 ·s2 ···s2 (0) (1)
(n)
t2 ·t2 ···t2
(0) (1) (n) s2 ·s2 ···s2
x12 x22
⎞ ⎟ ⎟ ⎠
POWERS AND DIRECT SUMS
and
⎛
⎜ σn σn−1 . . . σ0 X −1 τ0−1 τ1−1 . . . τn−1 = ⎜ ⎝
(0)
(1)
(n)
s1 ·s1 ···s1
(0) (1) (n) t1 ·t1 ···t1 (0)
(1)
(n)
s2 ·s2 ···s2
(0) (1) (n) t1 ·t1 ···t1
13
(0)
(1)
(n)
s1 ·s1 ···s1
y11
(0) (1) (n) t2 ·t2 ···t2 (0)
(1)
(n)
s2 ·s2 ···s2
y21
(0) (1) (n) t2 ·t2 ···t2
y12
⎞ ⎟ ⎟. ⎠
y22
We will assume that x11 x22 y11 y22 = 0 (4.4). Since the absolute value of an entry in a matrix is less than or equal to the norm of the matrix, we get that the sequences (0) (1) (n) (0) (1) (n) s1 s2 s1 s1 s2 s2 · (1) · · · (n) · (1) · · · (n) and (0) (0) t1 t1 t1 t2 t2 t2 as well as the sequences of their reciprocals are bounded. Thus there are constants 0 < c ≤ C < ∞ with (0)
m≤
s1
(0)
t1
(1)
·
s1
(1)
t1
(n)
···
s1
(n)
t1
(0)
≤M
and m ≤
s2
(0)
t2
(1)
·
s2
(1)
t2
(n)
···
s2
(n)
t2
≤M
for all n ≥ 1. The second option in the Lemma 4.3 leads to the second option in the conclusion of this proposition. The proof of the converse follows the same lines. We can now focus on the characterization of the weighted shifts satisfying Condition S. Recall the notation introduced at the start of this section. 4.8. Theorem. If A is a unilateral weighted shift with weight sequence {an } satisfying 0 < an ≤ 1 for all n, then A satisfies Condition S if and only if there is a positive constant c with c ≤ an an+1 · · · a2n for all n ≥ 0. Proof. We use Proposition 4.5 with S1 and S2 defined as in the preceding lemma and T1 = T2 = A. We need only interpret (4.6) and (4.7) for these shifts and show that this is equivalent to the condition given in this theorem. Since T1 = T2 here, the inequalities in Proposition 4.5 are the same, so we need only work with (4.6). Note that here the first set of inequalities in (4.6) becomes c≤
(a1 a2 )(a3 a4 ) · · · (a2n+1 a2n+2 ) ≤ C. a0 a1 · · · an
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 14 A,
This becomes
an+1 · · · a2n+2 ≤ C; a0 Since an ≤ 1 for all n ≥ 0, we can replace the constant C by 1; replacing c by a smaller constant if necessary, we get c≤
4.9
c ≤ an+1 · · · a2n+2
for n ≥ 0
The second set of inequalities in (4.6) becomes in this case c≤
(a0 a1 )(a2 a3 ) · · · (a2n a2n+1 ) ≤ C. a0 a1 · · · an
This reduces to 4.10
c ≤ an · · · a2n
for n ≥ 0.
Therefore both of these inequalities are covered by (4.10), which is the inequality in the statement of the theorem. 4.11. Corollary. The Bergman shift satisfies Condition S. Proof. The weight sequence for the Bergman shift is given by n+1 an = n+2 It follows that n+1 1 = an an+1 · · · a2n = 2n + 2 2 4.12. Corollary. If A is a hyponormal weighted shift with weight sequence {an }, then A satisfies Condition S if and only if there is a 1 constant c with 0 < c < 1 and an ≥ c n for all n ≥ 1. Proof. It is well known that a weighted shift is hyponormal if and only if the weight sequence is increasing. In these circumstances an · · · a2n ≥ 1 ≥ cc n ≥ c2 . an+1 n Now we examine Condition U for unilateral weighted shifts. 4.13. Theorem. Let A be a unilateral weighted shift with weight sequence of positive numbers {an }, then A satisfies condition U if and only if there is a constant a ≥ 1 such that: ⎧ ⎪ ⎨a when n is even 4.14 an = 1 when n = 1 ⎪ ⎩ ak if n = 2k + 1 and k ≥ 1 a
POWERS AND DIRECT SUMS
15
Proof. Assume a ≥ 1 and {an } is defined as in (4.14). It is clear that A is a bounded operator with A = a. As in Lemma 4.1, let H0 and H1 be the closed linear spans of the even and odd basis vectors, respectively. Define U : H → H ⊕ H by letting U e2n = en ⊕ 0 and U e2n+1 = 0⊕en for all n ≥ 0. Clearly U is a unitary that maps H0 onto H ⊕ (0) and H1 onto (0) ⊕ H. We claim that U A2 = (A ⊕ A)U . In fact when n = 2k for some k ≥ 0, U A2 en = aa2k+1 U e2k+2 = ak (ek+1 ⊕ 0) = Aek ⊕ 0 = (A ⊕ A)U en . A similar computation shows that U A2 en = (A ⊕ A)U en when n is odd. This finishes this part of the proof. Now let us assume that A is a weighted shift operator with weight sequence {an } that satisfies condition U . As in the proof of Theorem 4.8 we regard A ⊕ A as the operator-valued weighted shift with weight sequence {Bn }n≥0 , where each Bn is a 2 × 2 diagonal matrix Bn = diag(an , an ). Recall from Lemma 4.1 that A2 is unitarily equivalent to the operator-valued weighted shift with weight sequence {Cn }n≥0 , where each Cn is a 2 × 2 diagonal matrix Cn = diag(a2n a2n+1 , a2n+1 a2n+2 ). We regard B and C as operating on the 2 same space K = ⊕∞ n=0 Kn where each Kn = C . ∼ Since A satisfies Condition U, B = C. Let W be a unitary on K such that B = W ∗ CW ; so B ∗n = W ∗ C ∗n W for n ≥ 1. Thus W must map ker B ∗n isometrically onto ker C ∗n . It follows that each Kn reduces W so that W = ⊕∞ n=1 Wn , where each Wn is a 2 × 2 unitary matrix operating on Kn . Moreover we have that 4.15
Wn+1 Bn = Cn Wn
Letting W = [wijn ], multiplying the matrices in (4.15), and comparing entries we get the equations n+1 n w11 an = w11 a2n a2n+1 ,
n+1 n w12 an = w12 a2n a2n+1
n+1 n n+1 n w21 an = w21 a2n+1 a2n+2 , w22 an = w22 a2n+1 a2n+2 Using the fact that both the rows and the columns of each Wn are n n n n | = |w22 | and |w12 | = |w21 | for each orthonormal pairs, we have that |w11 n n ≥ 0. Fix some n ≥ 0 and suppose w11 = 0. Using the above equations n n and w22 and the fact that the weights are all positive, that involve w11 n+1 n+1 we see that w11 = 0 = w22 . Therefore solving for an and taking the modulus of the resulting equations yields that a2n+1 a2n+2 = a2n a2n+1 . Canceling we get
4.16
a2n = a2n+2
n for all n ≥ 0. This was under the assumption that w11 = 0; now assume n n n n that w11 = 0. So it follows that w22 = 0 and |w12 | = |w21 |. Arguing just as in the previous case but using the other two equations above yields
ˇ ˇ AND ALEJANDRO RODR´IGUEZ-MART´INEZ JOHN B. CONWAY, GABRIEL PRAJITUR 16 A,
the same conclusion (4.16). Putting a = a0 , (4.16) implies a2n = a for all n ≥ 0. Taking the absolute values of the determinants of both sides of (4.15) shows that a2n = det Bn = det Cn = a2n a2n+2 a22n+1 = a2 a22n+1 ; hence a2n+1 = an /a and we have established that (4.14) is valid. 4.17. Corollary. The only hyponormal weighted shift that satisfies Condition U is the shift with weight sequence 1, 1, . . . . Proof. If A is such a weighted shift with weights as in the preceding theorem, the fact that A is hyponormal implies an ≤ an+1 for all n. A moment’s thought shows that this implies that an = 1 for all n ≥ 0.
POWERS AND DIRECT SUMS
17
References [1] J B Conway, A Course in Functional Analysis, Second Edition, SpringerVerlag, New York (1990). [2] P R Halmos, A Hilbert Space Problem Book, Springer-Verlag, New York (1982). [3] A L Shields Weighted shift operators and analytic function theory, Amer Math Soc Math Surveys, 13 (1974) 49–128. The George Washington University E-mail address: [email protected] SUNY Brockport E-mail address: [email protected] Khalifa University of Science, Technology and Research, Abu Dhabi, UAE E-mail address: [email protected]