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Λ-Aluthge Iteration and Spectral Radius Integr. equ. oper. theory 99 (9999), 1{6 0378-620X/99000-0, DOI 10.1007/s00020-003-0000 Integral Equations °c 2008 BirkhÄauserVerlag Basel/Switzerland and Operator Theory ¸-Aluthge iteration and spectral radius Tin-Yau Tam Abstract. Let T 2 B(H) be an invertible operator on the complex Hilbert space H. For 0 < ¸ < 1, we extend Yamazaki's formula of the spectral radius ¸ 1¡¸ in terms of the ¸-Aluthge transform ¢¸(T ) := jT j UjT j where T = UjT j n is the polar decomposition of T . Namely, we prove that limn!1 jjj ¢¸(T ) jjj = r(T ) where r(T ) is the spectral radius of T and jjj ¢ jjj is a unitarily invariant norm such that (B(H); jjj ¢ jjj ) is a Banach algebra with jjj I jjj = 1. 1. Introduction Let H be a complex Hilbert space and let B(H) be the algebra of all bounded linear operators on H. For 0 < ¸ < 1, the ¸-Aluthge transform of T [2, 5, 11, 15] is ¸ 1¡¸ ¢¸(T ) := jT j UjT j ; where T = UjT j is the polar decomposition of T , that is, U is a partial isometry ¤ 1=2 n n¡1 0 and jT j = (T T ) . Set ¢¸(T ) := ¢¸(¢¸ (T )), n ¸ 1 and ¢¸(T ) := T . When 1 ¸ = 2 , it is called the Althuge transform [1] of T . See [3, 4, 7, 8, 11, 12, 13, 19]. Yamazaki [18] established the following interesting result n lim k¢ 1 (T )k = r(T ); (1.1) n!1 2 where r(T ) is the spectral radius of T and kT k is the spectral norm of T . Wang [17] then gave an elegant simple proof of (1.1) but apparently there is a gap. See Remark 2.6 for details and Remark 2.9 for a ¯x. Clearly k¢¸(T )k · kT k (1.2) n and thus fk¢¸(T )kgn2N is nonincreasing. Since the spectra of T and ¢¸(T ) are identical [11, 5], r(¢¸(T )) = r(T ): (1.3) 2000 Mathematics Subject Classi¯cation. Primary 47A30, 47A63 In memory of my brother-in-law, Johnny Kei-Sun Man, who passed away on January 16, 2008, at the age of ¯fty nine. 2 T.Y. Tam IEOT Antezana, Massey and Stojano® [5] proved that for any square matrix X, n lim k¢¸(X)k = r(X): (1.4) n!1 When T is invertible, the polar decomposition T = UjT j is unique [16, p.315] in which U is unitary and jT j is invertible positive so that ¢¸(T ) is also invertible. Our goal is to show that (1.4) is true for invertible operator T 2 B(H) and unitarily invariant norm under which B(H) is a Banach algebra with jjj I jjj = 1 [16, p.227-228], by extending the ideas in [17]. 2. Main results The following inequalities are known as Heinz's inequalities [10, 14]. Lemma 2.1. (Heinz) Let jjj ¢ jjj be a unitarily invariant norm on B(H). For positive A; B 2 B(H), X 2 B(H) and 0 · ® · 1, jjj A®XB1¡® jjj · jjj AX jjj ® jjj XB jjj 1¡® (2.1) jjj A®XB® jjj · jjj AXB jjj ® jjj X jjj 1¡®: (2.2) Remark 2.2. In [17] inequality (2.2) and McIntosh inequality jjj A¤XB jjj · jjj AA¤X jjj 1=2 jjj XBB¤ jjj 1=2 are used in the proof of (1.1). Yamazaki's proof [18] uses (2.2). We note that (2.2) and McIntosh's inequality follow from (2.1). n We simply write T0 := T and Tn := ¢¸(T ), n 2 N, once we ¯x 0 · ¸ · 1. Lemma 2.3. Let T 2 B(H) and 0 · ¸ · 1. Let jjj ¢ jjj be a unitarily invariant norm on B(H). Then for k; n 2 N, k k jjj (Tn+1) jjj · jjj (Tn) jjj : (2.3) k So for each k 2 N, limn!1 jjj (Tn) jjj exists. Moreover if T is invertible, then k 2¸¡1 k+1 ¸ k¡1 1¡¸ jjj (Tn+1) jTnj jjj · jjj (Tn) jjj jjj (Tn) jjj ; (2.4) 1¡2¸ k k+1 1¡¸ k¡1 ¸ jjjjTnj (Tn+1) jjj · jjj (Tn) jjj jjj (Tn) jjj : (2.5) Proof. Denote by Tn = UnTn the polar decomposition of Tn. Since Tn+1 = ¸ 1¡¸ jTnj UnjTnj , k ¸ k¡1 1¡¸ (Tn+1) = jTnj (Tn) UnjTnj : (2.6) By (2.1) k k¡1 ¸ k¡1 1¡¸ k jjj (Tn+1) jjj · jjjjTnj(Tn) Un jjj jjj (Tn) UnjTnj jjj = jjj (Tn) jjj since jjj ¢ jjj is unitarily invariant. So (2.3) is established. 2¸¡1 Suppose that T is invertible so that jTnj exists for 0 · ¸ · 1. From (2.6) k 2¸¡1 ¸ k¡1 ¸ (Tn+1) jTnj = jTnj (Tn) UnjTnj : Vol. 99 (9999) ¸-Aluthge iteration and spectral radius 3 Since jjj ¢ jjj is unitarily invariant and Un is unitary, by (2.2) k 2¸¡1 k¡1 ¸ k¡1 1¡¸ jjj (Tn+1) jTnj jjj · jjjjTnj(Tn) UnjTnj jjj jjj (Tn) Un jjj k+1 ¸ k¡1 1¡¸ = jjj (Tn) jjj jjj (Tn) jjj : Similarly (2.5) is established. ¤ 2¸¡1 When ¸ = 1=2, jTnj = I for any T 2 B(H). But the above computation does not work without assuming that T is invertible since the polar decomposition of a general T 2 B(T ) only yields T = UjT j where U is only a partial isometry [16, p.316]. The spectral norm enjoys kT k = k jT j k which is not valid for general unitarily invariant norm jjj ¢ jjj . Lemma 2.4. (Spectral radius formula) [16, p.235] Suppose that B(H) is a Banach algebra with respect to the norm jjj ¢ jjj (not necessarily unitarily invariant). For T 2 B(H), r(T ) = lim jjj T k jjj 1=k = inf jjj T k jjj 1=k: k!1 k2N In particular jjj T jjj ¸ r(T ). For B(H) to be a Banach algebra with respect to jjj ¢ jjj , the norm in Lemma 2.4 has to be submultiplicative, i.e., jjj ST jjj · jjj S jjj jjj T jjj [16, p.227]. The unitarily invariant norm jjj ¢ jjj in Lemma 2.3 need not be so. The condition jjj I jjj = 1 is k 1=k inessential for the formula r(T ) = limk!1 jjj T jjj , i.e, it is still valid even k 1=k jjj I jjj > 1. The formula r(T ) = infk2N jjj T jjj is valid for any normed algebra [6, p.236]). Lemma 2.5. Suppose that B(H) is a Banach algebra with respect to the unitarily k invariant norm jjj ¢ jjj and jjj I jjj = 1. Let 0 < ¸ < 1. Let sk := limn!1 jjj (Tn) jjj and s := s1. If T 2 B(H) is non-quasinilpotent, then s > 0. Moreover if T 2 B(H) k is invertible, then sk = s for each k 2 N. k Proof. Let T 2 B(H). By (2.3), for each k 2 N, the sequence f jjj (Tn) jjjgn2N is k nonincreasing so that sk := limn!1 jjj (Tn) jjj exists. By Lemma 2.4 and (1.3) jjj Tn jjj ¸ r(Tn) = r(T ). The spectrum σ(T ) of T is a compact nonempty set. If T is non-quasinilpotent, i.e., r(T ) > 0 so that s := s1 = lim jjj Tn jjj ¸ r(T ) > 0: (2.7) n!1 k Now assume that T is invertible. We proceed by induction to show that sk = s for all k 2 N. When k = 1, the statement is trivial. Suppose that the statement is true for 1 · k · m. Case 1: 0 < ¸ · 1=2. By (2.1)(A = jTnj, X = B = I) we have 1¡2¸ 1¡2¸ 0 < jjjjTnj jjj · jjjjTnj jjj 4 T.Y. Tam IEOT since jjj I jjj = 1 and 0 · 1 ¡ 2¸ < 1. Since T is invertible, jTnj is also invertible 2¸¡1 and thus jTnj exists. So m m jjj (Tn+1) jjj jjj (Tn+1) jjj 1¡2¸ · 1¡2¸ jjjjTnj jjj jjjjTnj jjj m 2¸¡1 · jjj (Tn+1) jTnj jjj since jjj ¢ jjj is submultiplicative m+1 ¸ m¡1 1¡¸ · jjj (Tn) jjj jjj (Tn) jjj by (2.4) m ¸ ¸ m¡1 1¡¸ · jjj (Tn) jjj jjj Tn jjj jjj (Tn) jjj : By the induction hypothesis, taking limits as n ! 1 yields sm · s¸ s(m¡1)(1¡¸) · sm¸s¸s(m¡1)(1¡¸); s1¡2¸ m+1 (m+1)¸ ¸ (m+1)¸ m+1 where s > 0 by (2.7). We have s · sm+1 · s and hence sm+1 = s . Case 2. 1=2 < ¸ < 1. Similar to Case 1, by (2.1) we have 2¸¡1 2¸¡1 jjjjTnj jjj · jjjjTnj jjj and m m jjj (Tn+1) jjj jjj (Tn+1) jjj 2¸¡1 · 2¸¡1 jjjjTnj jjj jjjjTnj jjj 1¡2¸ m · jjjjTnj Tn+1 jjj m+1 1¡¸ m¡1 ¸ · jjj (Tn) jjj jjj (Tn) jjj by (2.5) m 1¡¸ 1¡¸ m¡1 ¸ · jjj (Tn) jjj jjj Tn jjj jjj (Tn) jjj : So sm · s1¡¸ s(m¡1)¸ · sm(1¡¸)s(1¡¸)s(m¡1)¸ s2¸¡1 m+1 m+1 which leads to sm+1 = s . ¤ Remark 2.6. Suppose ¸ = 1=2. In the proof of [17, Lemma 4], the possibility that s = 0 is not considered (the spectral k ¢ k is the norm under consideration). It amounts to r(T ) = 0, that is, T is quasinilpotent [9, p.50], [13, p.381]. In the m+1 above induction proof, if ¸ = 1=2, one cannot deduce that sm+1 = s for jjj ¢ jjj , m 1=2 (m¡1)=2 m granted that s · sm+1s · s (that relies on (2.4) which is under the assumption that T is invertible) is valid.
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