PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 130, Number 4, Pages 1131–1137 S 0002-9939(01)06283-9 Article electronically published on September 19, 2001
AN EXPRESSION OF SPECTRAL RADIUS VIA ALUTHGE TRANSFORMATION
TAKEAKI YAMAZAKI
(Communicated by Joseph A. Ball)
Abstract. For an operator T ∈ B(H), the Aluthge transformation of T is e 1 1 defined by T = |T | 2 U|T | 2 . And also for a natural number n,then-th Aluthge f ^ f e transformation of T is defined by Tn = (Tn−1)andT1 = T .Inthispaper,we shall show f lim kTnk = r(T ), n→∞ where r(T ) is the spectral radius.
1. Introduction n 1 As a characterization of the spectral radius, it is well known that lim kT k n = n→∞ r(T ). This result is very famous and quite useful. On the other hand, Aluthge [1] e e 1 1 defined a transformation T of T by T = |T | 2 U|T | 2 ,whereT = U|T | is the polar decomposition of T . Te is called the Aluthge transformation of T . Many researchers have obtained their results by using Aluthge transformation, for example, [1], [2], [3], [4], [6], [7], [8]. It is easily obtained that kT k≥kTek≥r(Te)=r(T ). Recently [9], as a generalization of Aluthge transformation, for each natural f number n, we defined a transformation Tn of T by f ^] f e Tn = (Tn−1)andT1 = T. f We call Tn the n-th Aluthge transformation of T . In this paper, we shall show another characterization of the spectral radius by using n-th Aluthge transformation as follows: f Theorem 1. Let T ∈ B(H).Then lim kTnk = r(T ). n→∞
2. Proof In what follows, a capital letter means a bounded linear operator on a complex Hilbert space H.AnoperatorT is said to be positive (denoted by T ≥ 0) if (Tx,x) ≥ 0 for all x ∈ H. To prove Theorem 1, we prepare the following results.
Received by the editors October 27, 2000. 2000 Mathematics Subject Classification. Primary 47A13, 47A30. Key words and phrases. Aluthge transformation, Heinz inequality, spectral radius.
c 2001 American Mathematical Society 1131
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Theorem A ([5]). Let A and B be positive operators, and X ∈ B(H).Then kAαXBαk≤kAXBkαkXk1−α holds for all α ∈ [0, 1]. Lemma 2. For a natural number n and k =0, 1, ··· ,n+1,let n!(n − 2k +1) (2.1) D = . n k k!(n − k +1)! Then the following assertions hold:
(i) nD0 =1for all natural numbers n. (ii) nDk + nDk+1 = n+1Dk+1 for all natural numbers n and k =0, 1, ··· ,n. (iii) 2n+1Dn = 2n+2Dn+1 for all natural numbers n. [ n ] X2 n (iv) (n − 2k +1)nDk =2 , k=0 n n ≤ n where [ 2 ] is the largest integer satisfying [ 2 ] 2 . (n − 2k +1) D (v) lim n k =0for all positive integers k. n→∞ 2n Proof. (i). By (2.1), we have n!(n +1) D = =1. n 0 0!(n +1)! (ii). By (2.1), we obtain n!(n − 2k +1) n!(n − 2k − 1) D + D = + n k n k+1 k!(n − k +1)! (k +1)!(n − k)! n!{(k +1)(n − 2k +1)+(n − k +1)(n − 2k − 1)} = (k +1)!(n − k +1)! n!(n +1)(n − 2k) = (k +1)!(n − k +1)! (n +1)!(n − 2k) = = D . (k +1)!(n − k +1)! n+1 k+1
(iii). By (ii) and 2n+1Dn+1 =0,wehave
2n+2Dn+1 = 2n+1Dn + 2n+1Dn+1 = 2n+1Dn. (iv). We shall prove (iv) by induction on n. (a) The case n = 1. By (2.1), we obtain
[ 1 ] X2 (1 − 2k +1)1Dk =21D0 =2. k=0 (b) Assume that
[ n−1 ] X2 n−1 (2.2) (n − 2k)n−1Dk =2 . k=0
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··· n n−1 (c-1) The case n =2m +1form =1, 2, .Then[2 ]=[ 2 ]=m. Hence we obtain
Xm (n − 2k +1)nDk k=0 Xm =(n +1)nD0 + (n − 2k +1)nDk k=1 Xm =(n +1)n−1D0 + (n − 2k +1)(n−1Dk−1 + n−1Dk) by (i) and (ii) k=1 Xm Xm =(n +1)n−1D0 + (n − 2k +1)n−1Dk−1 + (n − 2k +1)n−1Dk k=1 k=1 mX−1 Xm = (n − 2k − 1)n−1Dk + (n − 2k +1)n−1Dk k=0 k=0 mX−1 =2 (n − 2k)n−1Dk +(n − 2m +1)n−1Dm k=0 mX−1 =2 (n − 2k)n−1Dk +2n−1Dm by n =2m +1 k=0 Xm n−1 n =2 (n − 2k)n−1Dk =2· 2 =2 by (2.2). k=0
··· n n−1 (c-2) The case n =2m +2for m =0, 1, 2, .Then[2 ]=m +1 and[ 2 ]=m. Hence we obtain
mX+1 (n − 2k +1)nDk k=0 mX+1 =(n +1)nD0 + (n − 2k +1)nDk k=1 mX+1 =(n +1)n−1D0 + (n − 2k +1)(n−1Dk−1 + n−1Dk) by (i) and (ii) k=1 mX+1 mX+1 =(n +1)n−1D0 + (n − 2k +1)n−1Dk−1 + (n − 2k +1)n−1Dk k=1 k=1 Xm mX+1 = (n − 2k − 1)n−1Dk + (n − 2k +1)n−1Dk k=0 k=0 Xm =2 (n − 2k)n−1Dk + {n − 2(m +1)+1}n−1Dm+1 k=0 n−1 n =2· 2 =2 by (2.2) and n−1Dm+1 = 2m+1Dm+1 =0.
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(v). We remark that nα (2.3) lim = 0 holds for fixed α ≥ 0. n→∞ 2n (a) The case k =0.Wehave (n +1) D (n +1) lim n 0 = 2 lim =0 by(2.3). n→∞ 2n n→∞ 2n+1 (b) The case k =1.Wehave (n − 1) D 1 (n − 1)2 lim n 1 = lim =0 by(2.3). n→∞ 2n 2 n→∞ 2n−1 (c) The case k ≥ 2. For sufficiently large n, (n − 2k +1) D n!(n − 2k +1)2 0 ≤ n k = 2n 2nk!(n − k +1)! z k}|−1 { n(n − 1) ···(n − k +2)(n − 2k +1)2 = 2nk! − − nk+11 · (1 − 1 ) ···(1 − k 2 )(1 − 2k 1 )2 nk+1 = n n n ≤ . 2nk! 2n Hence we obtain (v) by (2.3).
Lemma 3. Let T ∈ B(H).Then
en n+1 1 n−1 1 kT k≤kT k 2 kT k 2 holds for all natural numbers n. Proof. Let T = U|T | be the polar decomposition of T .Thenwehave
en 1 1 n 1 n−1 1 kT k = k(|T | 2 U|T | 2 ) k = k|T | 2 (U|T |) U|T | 2 k n−1 1 n−1 1 ≤k|T |(U|T |) U|T |k 2 k(U|T |) Uk 2 by Theorem A n+1 1 n−1 1 = kT k 2 kT k 2 .
∈ n Lemma 4. Let T B(H) and m =[2 ].Then
nD0 nD1 nDk nDm f n+1 n n−1 n n−2k+1 n n−2m+1 n kTnk≤kT k 2 kT k 2 ···kT k 2 ···kT k 2 . Proof. We shall prove Lemma 4 by induction on n. e 2 1 (a) kT k≤kT k 2 holds by Lemma 3. (b) Assume that
n−1D0 n−1D1 ] n − n−2 − kTn−1k≤kT k 2n 1 kT k 2n 1 (2.4) − D − D n−2k n 1 k n−2m n 1 m ×···×kT k 2n−1 ···kT k 2n−1 ,
n−1 where m =[ 2 ].
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··· n n−1 (c-1) The case n =2m +1form =1, 2, .Then[2 ]=[ 2 ]=m. Hence by (2.4), we have
f ^e kTnk = k(T)n−1k
n−1D0 n−1D1 en − en−2 − ≤kT k 2n 1 kT k 2n 1
n−1Dk−1 n−1Dk n−1Dm−1 n−1Dm en−2k+2 − en−2k − e3 − e − ×···×kT k 2n 1 kT k 2n 1 ···kT k 2n 1 kTk 2n 1
n−1D0 n−1D1 n+1 1 n−1 1 2n−1 n−1 1 n−3 1 2n−1 ≤ kT k 2 kT k 2 kT k 2 kT k 2
n−1Dk−1 n−2k+3 1 n−2k+1 1 2n−1 ×···× kT k 2 kT k 2
n−1Dk n−2k+1 1 n−2k−1 1 2n−1 × kT k 2 kT k 2
n−1Dm−1 − D 4 1 2 1 2n−1 2 n 1 m ×···× kT k 2 kT k 2 kT k 2n
D D + D n+1 n−1 0 n−1 n−1 0 n−1 1 = kT k 2n kT k 2n
− D + − D − D − + − D n−2k+1 n 1 k−1 n 1 k 2 n 1 m 1 n 1 m ×···×kT k 2n ···kT k 2n
n+1 nD0 n−1 nD1 n−2k+1 nDk 2 nDm = kT k 2n kT k 2n ···kT k 2n ···kT k 2n ,
by (i) and (ii) of Lemma 2, and the last inequality holds by Lemma 3. ··· n n−1 (c-2) The case n =2m +2for m =0, 1, 2, .Then[2 ]=m +1 and[ 2 ]=m. Henceby(2.4),wehave
f ^e kTnk = k(T )n−1k
n−1D0 n−1D1 en − en−2 − ≤kT k 2n 1 kT k 2n 1
n−1Dk−1 n−1Dk n−1Dm−1 n−1Dm en−2k+2 − en−2k − e4 − e2 − ×···×kT k 2n 1 kT k 2n 1 ···kT k 2n 1 kT k 2n 1
n−1D0 n−1D1 n+1 1 n−1 1 2n−1 n−1 1 n−3 1 2n−1 ≤ kT k 2 kT k 2 kT k 2 kT k 2
n−1Dk−1 n−2k+3 1 n−2k+1 1 2n−1 ×···× kT k 2 kT k 2
n−1Dk n−2k+1 1 n−2k−1 1 2n−1 × kT k 2 kT k 2
n−1Dm−1 n−1Dm 5 1 3 1 2n−1 3 1 1 2n−1 ×···× kT k 2 kT k 2 kT k 2 kT k 2
D D + D n+1 n−1 0 n−1 n−1 0 n−1 1 = kT k 2n kT k 2n
− D + − D D + D D n−2k+1 n 1 k−1 n 1 k 3 n−1 m−1 n−1 m n−1 m ×···×kT k 2n ···kT k 2n kT k 2n
D n+1 nD0 n−1 nD1 n−2k+1 nDk 3 nDm n m+1 = kT k 2n kT k 2n ···kT k 2n ···kT k 2n kT k 2n ,
by (i), (ii) and (iii) of Lemma 2, and the last inequality holds by Lemma 3.
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∞ Lemma 5. Let {an} be a sequence satisfying lim an = a, and for each natural n=1 n→∞ { }n number n,let αn,k k=1 be a positive sequence satisfying
(2.5) αn,1 + ···+ αn,k + ···+ αn,n =1 for all natural numbers n
and lim αn,k =0for fixed k =1, 2, ···.Then n→∞
lim (αn,1a1 + ···+ αn,kak + ···+ αn,nan)=a. n→∞
Proof. For any ε>0, there exists k>0 such that |an − a| <εand αn,1|a1 − a| + ··· + αn,k|ak − a| <εfor all natural numbers n>kby the as- sumptions lim an = a and lim αn,k =0.Thenwehave n→∞ n→∞
|(αn,1a1 + ···+ αn,kak + αn,k+1ak+1 + ···αn,nan) − a|
= |αn,1(a1 − a)+···+ αn,k(ak − a)
+ αn,k+1(ak+1 − a)+···+ αn,n(an − a)| by (2.5)
≤ αn,1|a1 − a| + ···+ αn,k|ak − a|
+ αn,k+1|ak+1 − a| + ···+ αn,n|an − a|
<αn,1|a1 − a| + ···+ αn,k|ak − a|
+ {1 − (αn,1 + ···+ αn,k)}ε by (2.5) < 2ε.
n Proof of Theorem 1. Let m =[2 ]. Then by Lemma 4, (iv) of Lemma 2 and Arith- metic mean-Geometric mean inequality, we have
nD0 nD1 f f n+1 n n−1 n r(T )=r(Tn) ≤kTnk≤kT k 2 kT k 2 n−2k+1 nDk n−2m+1 nDm ···kT k 2n ···kT k 2n − (n +1)nD0 n+1 1 (n 1)nD1 n−1 1 ≤ kT k n+1 + kT k n−1 2n 2n − (n 2k +1)nDk n−2k+1 1 + ···+ kT k n−2k+1 2n − (n 2m +1)nDm n−2m+1 1 + ···+ kT k n−2m+1 2n −→ r(T )asn →∞
n 1 by lim kT k n = r(T ), (iv) and (v) of Lemma 2 and Lemma 5. n→∞ References
[1] A.Aluthge, On p-hyponormal operators for 0
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[6] T.Huruya, Anoteonp-hyponormal operators, Proc. Amer. Math. Soc., 125 (1997), 3617– 3624. MR 98b:47025 [7] I.B.Jung, E.Ko and C.Pearcy, Aluthge transforms of operators,IntegralEquationsOperator Theory, 37 (2000), 437–448. CMP 2000:17 [8] K.Tanahashi, On log-hyponormal operators, Integral Equations Operator Theory, 34 (1999), 364–372. MR 2000b:47055 [9] T.Yamazaki, Characterizations of log A ≥ log B and normaloid operators via Heinz inequal- ity,preprint.
Department of Mathematics, Kanagawa University, Yokohama 221-8686, Japan E-mail address: [email protected]
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