Patryk Pagacz Uniwersytet Jagielloński Characterization of strong stability of power-bounded operators Praca semestralna nr 3 (semestr zimowy 2011/12) Opiekun pracy: Jaroslav Zemanek
CHARACTERIZATION OF STRONG STABILITY OF POWER-BOUNDED OPERATORS PATRYK PAGACZ 1 Abstract. By a bounded backward sequence of the operator T we mean a bounded sequence {x n } satisfying T x n+1 = x n. In [8] we have characterized contractions with strongly stable nonunitary part in terms of bounded backward sequences. The main purpose of this work is to extend that result to powerbounded operators. Additionally, we show that a power-bounded operator is strongly stable (C 0 ) if and only if its adjoint does not have any nonzero bounded backward sequence. Similarly, a power-bounded operator is non-vanishing (C 1 ) if and only if its adjoint has a lot of bounded backward sequences. 1. Preliminaries Let H be a complex, separable Hilbert space. We denote by B(H) the space of bounded linear transformations acting on H. By a contraction we mean T B(H) such that T x x for each x H. By a powerbounded operator we mean T B(H) such that T n is uniformly bounded for all n = 1, 2, 3,... An operator T is said to be completely nonunitary (abbreviated cnu) if T restricted to every reducing subspace of H is nonunitary. As usual, by T we mean the adjoint of T. We define as usually: Definition 1.1. An operator T B(H) is said to be of class C 0 if for each x H. lim inf n T n x = 0 Note that a power-bounded operator T is of class C 0 if and only if it is strongly stable (T n 0, SOT). Indeed, let T be C 0. If we fix x H then for each ɛ > 0 there is k N such that T k x < ɛ, so for all m > k we have T m x = T m k T k x T m k T k x ɛ sup T n. 1 Key words and phrases: power-bounded operators, C 0 operators, C 0 operators, strongly stable operators. AMS(MOS) subject classifcation (2010): 47A05, 47A45, 47B37, 47G10. 1
2 PATRYK PAGACZ In general, C 0 operators can be extremely different from strongly stable operators. The following example shows a bounded operator of class C 0, which is not strongly stable at any (nonzero) point. Example 1.2. Let {N k } k be the sequence such that { N1 = 1 N k+1 = 3N k + 2Nk 2, for k = 1, 2,... Then let us define the operator S as the unilateral shift with weights w 1, w 2,..., i.e., S : l 2 (x 1, x 2,...) (0, w 1 x 1, w 2 x 2,...) l 2, where w 1 = 1 w i = 1, for i = N 2 k + 1, N k + 2,..., 3N k w i = 2 1 N k, for i = 3N k + 1, 3N k + 2,..., N k+1. For nonzero x = (x 1, x 2,...) l 2 there is i 0 such that x i0 0. But by definition of S we have S N k e1 = e Nk +1 for all k N, thus S N k i 0 x S N k i 0 1 x i0 e i0 = w 1 w 2... w i0 x i0. So S n x 0 for all nonzero x l 2. Now we show that S is of class C 0. Fix x = (x 1, x 2,...) l 2 and ɛ > 0. We can assume x = 1. Since {N k } k increases, then there is N {N k k = 1, 2,...} such that x i 2 < ɛ and ( ) 1 2 32 2 < ɛ. By that we obtain: N 2 i=n+1 = + S 2N x 2 = x i 2 S 2N e i 2 + x i 2 S N (w i w i+1...w i+n 1 e i+n ) 2 + j=n+1 x i 2 w i w i+1... w N x j 2 2 1 1 1 N 2 N... 2 N }{{} 2 = x i 2 2N ( 1 2 j=n+1 j=n+1 ( 1 2 x j 2 S 2N e j 2 = x j 2 w j w j+1...w j+2n 1 2 ) i 1 S N e i+n 2 + x i 2 S N e i+n 2 + j=n+1 ) N e i+2n 2 + ɛ 32 16 x ɛ 2 + ɛ 2 = ɛ. In contrast to the above notion we have: x j 2 4 2 = Definition 1.3. An operator T B(H) is said to be of class C 1 if for each nonzero x H. lim inf n T n x > 0
CHARACTERIZATION OF STRONG STABILITY OF POWER-BOUNDED OPERATORS3 Operators of class C 1 are also called non-vanishing. We also say that T is of class C 0 or C 1 if its adjoint is of class C 0 or C 1, respectively. Let us define M(T ) := {x H {x n } : x = x 0, T x n+1 = x n and {x n } is bounded }. Naturally, such a sequence {x n } can be called as the bounded backward sequence. 2. Introduction In the paper [8] we have presented the following theorem with some applications. Theorem 2.1. Let T be a contraction. The following conditions are equivalent: for any bounded backward sequence {x n } of T, the sequence of norms { x n } is constant, the nonunitary part of T is of class C 0. We have been asked the natural question about a possible generalization to power-bounded operators. In this work we will try to answer this question. The easy extension of the above theorem for power-bounded operators is not true. To see this, let us consider the following: Example 2.2. Let T : l 2 (x 1, x 2, x 3,...) (0, x 1 + x 2, 0, x 3 + x 4, 0,...) l 2. It is clear that T is power-bounded, in fact T = T 2. Additionally, if x = (a 1, a 2, a 3,...) M(T ) T (H), then a 2k+1 = 0 for all k N. Hence T 1 ({x}) = {x}. Thus, even any (not necessary bounded) backward sequence of T must be constant. On the other hand, T has trivial unitary part and is not C 0, since T = T 2. 3. Characterization of C 1 and C 0 power-bounded operators To introduce the next theorem, let us recall the construction of isometric asymptotes (see [7]). Let us define a new semi-inner product on H: [x, y] := glim{ T n x, T n y }, where glim denote a Banach limit. Thus, the factor space H/H 0, where H 0 stands for the linear manifold H 0 := {x H [x, x] = 0}, endowed with the inner product [x + H 0, y + H 0 ] = [x, y], is an inner product space. Let K denote the resulting Hilbert space obtained by completion. Let X denote the natural embedding of Hilbert space H into K i.e. X : H x x+h 0 K.
4 PATRYK PAGACZ We can see that: XT x = Xx. So there is an isometry V : K K such that XT = V X. The isometry V is called isometric asymptote. Lemma 3.1. For any power-bounded operator T B(H) the corresponding X from the construction above satisfies: X (K) = M(T ). Proof. By definition of V and X, we have T X = X V. Let x n := X V n x, then T x n+1 = T X V n+1 x = X V V n+1 x = x n. Moreover x n X x, for all n N. Thus X x = x 0 M(T ), where x H. Hence X (H) M(T ). To prove the converse, let us fix x M(T ). By definition of M(T ), there exists {x n } H, a bounded backward sequence of x. Let y H, then (1) x, y = T n x n, y = x n, T n y x n T n y. So x, y sup x n lim inf T n y sup x n Xy. So by Theorem n 1 in [9], we have x X (K). At the begining, we have observed that if lim inf n some x H, then lim T n x = 0. So we have: n Now by Lemma 3.1 we obtain: {x H T n x 0} = N (X). Corollary 3.2. Let T be a power-bounded operator, then We also have: H = {x H T n x 0} M(T ). T n x = 0, for Theorem 3.3. A power-bounded operator T is C 1 if and only if M(T ) = H. It is trivial that M(T ) is included in the set of origins of all backward sequences, that is, T (H) := T n (H). But in general, the converse inclusion does not hold, even for C 1 contractions. To see this let us consider the following example. Example 3.4. Let H = l 2. Then H := l 2 (H) = {{x n } H x n 2 < } is a separable Hilbert space, with the norm {x n } := x n 2. For the element {x n } H sometimes we write x n. Let S w be the backward unilateral shift with weights w = (w 1, w 2,...),
CHARACTERIZATION OF STRONG STABILITY OF POWER-BOUNDED OPERATORS5 i.e., S w : H (x 1, x 2,...) (w 1 x 2, w 2 x 3...) H. If we put wi n = ( 1 ) 1 i 1 1 i for all n N and i > 2, and w n n 1 = w2 n = 1 for all n N, then T = S w n is a C 1 contraction. Indeed, T m = S m w n, where Sm w n(x 1, x 2, x 3,...) = (w n 1 w n 2... w n mx m+1, w n 2 w n 3... wm+1x n m+2,...) and lim m wn 1 w2 n... wm n = lim ( 1 ) 1 2 1 m = 1 > 0. m n n 2 Now, let us consider x = ( 1, 0, 0,...) H. n For m N we have x = T m a m, where a m = (0, 0,..., 0, ( }{{} 1 ) 1 2 + 1 m, 0, 0,...) H. Thus x T (H). n m Now let {b m } m N H be a backward sequence for x, then x = T m b m and thus b m = (q 1, q 2,..., q m, ( 1 ) 1 2 + 1 m, 0, 0,...) for some n complex q 1, q 2,..., q m. So a m b m, but a m 2 = ( 1 n )2( 1 2 + 1 m ) ( for m ). Hence x M(T ). One more consequence of Corollary 3.2 is the following: Theorem 3.5. A power-bounded operator T is C 0 if and only if M(T ) = {0}. Another proof of this theorem (in the case of operators considered on Banach spaces) can be found in [10]. Corollary 3.6. If T is power-bounded and invertible, then T n x 0 for all x H if and only if T n x for all x H. Proof. If T is power-bounded, then T is power-bounded too. By Theorem 3.5 we obtain that T is strongly stable if and only if each nontrivial sequence such that T x n+1 = x n is unbounded. But we have x n = T n x 0. Thus the second condition means that sup T n x = for each nonzero x H. Now, if for some x H there is an increasing sequence {n k } k N such that sup T n k x < N, then for each n N we have k N T n x = T nk n T n k x T nk n T n k x N sup T n, since n k > n for some k N. Example 3.7. Let V be the classical integral Volterra operator defined, on the space L 2 [0, 1], by (V f)(x) := x 0 f(t)dt, for f L2 [0, 1]. It is easy to calculate that (V f)(x) = 1 f(t)dt. x Hence V + V = P, where P is the one-dimensional projection on
6 PATRYK PAGACZ subspace of constant functions. It is well-known that (I + V ) 1 = 1 (see Problem 150 in [4]). The Allan-Pedersen relation (see [1]) S 1 (I V )S = (I + V ) 1, where Sf(t) = e t f(t) show us that I V is similar to a contraction. So it is power-bounded. Furthermore, Proposition 3.3 from [5] yields to (2) lim n(i V ) n V f = 0, for all f L 2 [0, 1]. n But as we mentioned, I V is power-bounded. Moreover, V has dense range. Therefore I V is C 0. (To obtain this, instead of (2) we can use the Esterle-Katznelson-Tzafriri theorem (see [3], [6]), since σ(i V ) = {1}.) Now, by Corollary 3.6 we obtain (I + V P ) n f = ((I V ) ) n f for all f L 2 [0, 1]\{0}. Additionally, form (1) in the proof of Lemma 3.1 and (2) we have 1 n (I + V P ) n f for all f V (L 2 [0, 1])\{0}. Remark. To obtain the first part of this result we also can use Theorem 3.4 form [2] and observe that each local spectrum σ x (I +V P ) is equal {1}. (Because {1} = σ(i V ) = σ((i + V P ) ) = σ(i + V P ).) Example 3.8. According to the above example we see that the contraction (I + V ) 1 is of class C 0 and as before σ(i + V ) = {1}. So using Theorem 3.4 form [2] we obtain that (I + V ) n f for all nonzero f L 2 [0, 1]. Now by Corollary 3.6 we have (I V + P ) n f = (I + V ) n f 0 for all f L 2 [0, 1]. So the contraction (I + V ) 1 is of class C 00. 4. Main result To give a generalization of Theorem 2.1, we will need the following lemma(due to Kérchy, see [7]): Lemma 4.1. If T is power-bounded, then T can be represented by the matrix [ ] T11 T (3) 21, 0 T 22 where T 11, T 22 are power-bounded, T 11 is of class C 0 and T 22 is of class C 1. Proof. Let (3) be the matrix of T with respect to the orthogonal decomposition H = N N, where N := {x H T n x 0}. By definition N is invariant for T. So T N = T 11, thus T 11 is of class C 0 (and power-bounded). Moreover, we have: T 22 = P K T B(K), where K := N {0}.
CHARACTERIZATION OF STRONG STABILITY OF POWER-BOUNDED OPERATORS7 The subspace K is invariant for T. So we obtain T K = T22, thus T 22 is power-bounded. Now, we will show that T 22 is C 1. To see this, let us assume that T22f n 0 for some f K. For an arbitrary ɛ > 0, there is n 0 N such that T n 0 22 f < ɛ T n. 2M, where M := sup Let us suppose for a while that T 22 B(K, H). By definition of T 22 we have: (T T 22 )x = P N T x N for each x K. Hence for each k {1, 2, 3,..., n 0 } there exists m k T m +k 1 (T T 22 )T n 0 k 22 f ɛ 2n 0 for all m m k. Now, for m := max{m k k = 1, 2,..., n 0 } we have: N such that T m+n 0 f = T m (T n 0 T n0 1 T 22 + T n0 1 T 22 T n0 2 T22 2 +... + +T T n 0 1 22 T n 0 22 +T n 0 22 )f = n 0 T m+k 1 (T T 22 )T n 0 k 22 f +T m T n 0 22 f n 0 k=1 k=1 T m+k 1 (T T 22 )T n 0 k 22 f + T m T n 0 22 f n 0 ɛ 2n 0 + M ɛ 2M = ɛ. Thus T n f 0, contrary to f K. So T 22 is of class C 1. Now, we can give our generalization of Theorem 2.1: Theorem 4.2. Let T be a power-bounded operator. The following conditions are equivalent: for any bounded backward sequence {x n } of T, the sequence of norms { x n } is constant; [ ] T11 0 T can be decomposed as T =, where U is a unitary T 21 U and T 11 is of class C 0. [ ] T11 0 Proof. To the proof of the first implication, let T = be the T 21 T 22 matrix form Lemma 4.1, where T 11 B(H 1 ) is C 0 and T 22 B(H 2 ) is C 1. Now, H 2 is invariant for T, thus T 22 = T H2. Hence, each bounded backward sequence of T 22 is bounded backward sequence of T. So, by our assumption T 22 is an isometry on M(T 22 ). But by Theorem 3.3 we have M(T 22 ) = H 2. So T 22 is an isometry. Finally, it can be decomposed as T 22 = U S +, where U is unitary and S + is the unilateral shift. But T 22 is C 1. So we have T 22 = U. To prove the converse implication, let us assume that {x n } is the bounded backward sequence of T. Let x n = a n + b n, where a n H 1 and b n H 2. We have: T 11 a n+1 + (T 21 a n+1 + Ub n+1 ) = T a n+1 + T b n+1 = T x n+1 = x n = a n + b n.
8 PATRYK PAGACZ So T 11 a n+1 = a n and a n x n. It means that {a n } is a bounded backward sequence of T 11, but T 11 is of class C 0. So by Theorem 3.5 we obtain a n 0. Thus x n+1 = b n+1 = Ub n+1 = b n = x n. 5. Acknowledgements I am very grateful to professor J. Zemánek for his hospitality and discussions during my stay in the Institute of Mathematics of the Polish Academy of Sciences. References [1] G. R. Allan, Power-bounded elements and radical Banach algebras, Linear Operators (Warsaw, 1994), Banach Center Publ., vol. 38, Polish Acad. Sci., Warsaw, 1997, pp. 9-16. [2] B. Aupetit, D. Drissi Some spectral inequalities involving generalized scalar operators, Studia Math. 109(1994), 51-66. [3] J. Esterle, Quasimultipliers, representation of H, and the closed ideal problem for commutative Banach algebras. Radical Banach Algebras and Automatic Continuity., Lecture Notes in Math., Vol. 975. Springer, Berlin-Heidelberg- New York, 1983, pp. 66-162. [4] P. Halmos, A Hilbert Space Problem Book, Van Nostrand, Princeton, N. J., 1967. [5] Z. Léka, A note on the powers of Cesàro bounded operators, Czechoslovak Math. J. 60(2010), 1101-1107. [6] Y. Katznelson and Tzafriri, On power bounded operators, J. Funct. Anal. 68(1986), 313-328. [7] L. Kérchy, Isometric Asymptotes of Power Bounded Operators, Indiana Univ. Math. J. 38(1989), 173-188. [8] P. Pagacz, On Wold-type decomposition, Linear Algebra Appl. 436(2011), 3065-3071. [9] Z. Sebestyén, On range of adjoint operators in Hilbert space, Acta Sci. Math. (Szeged) 46(1983), 295-298. [10] Q. P. Vũ, Almost periodic and strongly stable semigroups of operators, Linear Operators (Warsaw, 1994), Banach Center Publ., vol. 38, Polish Acad. Sci., Warsaw, 1997, pp. 401-426. Instytut Matematyki, Uniwersytet Jagielloński, Łojasiewicza 6, 30-348, Kraków, Poland E-mail address: patryk.pagacz@im.uj.edu.pl