On the power boundedness of certain Volterra operator pencils

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1 STUDIA MATHEMATICA 156 (1) (23) On he power boundedness of cerain Volerra operaor pencils by Dashdondog Tsedenbayar (Warszawa and Ulan-Baor) Absrac. Le V be he classical Volerra operaor on L 2 (, 1), and le z be a comple number. We prove ha I zv is power bounded if and only if Re z and Im z =, while I zv 2 is power bounded if and only if z =. The firs resul yields (I V ) n (I V ) n+1 = O(n 1/2 ) as n, an improvemen of [Py]. We also sudy some oher relaed operaor pencils. 1. Preliminaries. We say ha an operaor A is power-bounded if sup n A n <. We denoe by V he classical Volerra operaor (V f)() = We recall he well-known formula f(s) ds, < < 1, on L p (, 1), 1 p. (V n f)() = ( s) n 1 (n 1)! f(s) ds for n N. A generalizaion of his formula is he definiion of he Riemann Liouville inegral operaor of any fracional order α >, (J α f)() = 1 ( s) α 1 f(s) ds Γ (α) (Γ is he Euler gamma funcion) on L p (, 1), 1 p. In paricular, V = J 1. Recall ha he Ri condiion for he resolven R(λ, A) = (A λi) 1 of a bounded linear operaor A on a Banach space is R(λ, A) cons, λ > 1, λ 1 2 Mahemaics Subjec Classificaion: Primary 47A1. Key words and phrases: Volerra operaor, power-bounded operaor, resolven. [59]

2 6 D. Tsedenbayar which is equivalen o a geomeric condiion much sronger han he power boundedness of A [NaZe], [Ne2]. If he operaor A is merely power-bounded, hen he weaker Kreiss condiion R(λ, A) cons, λ > 1, λ 1 holds, bu no conversely in general. 2. Inroducion. In 1997, Allan [Al] recorded he observaion made by T. V. Pedersen ha I V is similar o (I + V ) 1, namely (1) S 1 (I V )S = (I + V ) 1 where (Sf)() = e f(), f L p (, 1), 1 p. By [Ha, Problem 15], we know ha (I + V ) 1 = 1 on L 2 (, 1). Hence I V is a power-bounded operaor on L 2 (, 1). In 1987, Pylik [Py], basing on an upper esimae for he Fejér epression for Laguerre polynomials (see [Sz, p. 198]), proved (2) (I V ) n (I V ) n+1 = O(n 1/4 ) as n + on L 2 (, 1). The same argumen gives he same resul also on L 1 (, 1), in which case i is sharp [ToZe]. By his mehod, one is unable o disinguish he delicae properies of he L p -norms. We shall show, by an algebraic argumen, he power boundedness of I V for >, on L 2 (, 1), which will improve Pylik s esimae o O(n 1/2 ). Our mehod, however, does no apply o L 1 (, 1), because I V is no power-bounded here (see [Hi, p. 247]), and (2) acually canno be improved on L 1 (, 1) as menioned above [ToZe]. We also sudy some oher relaed operaor pencils. The deails of some calculaions as well as alernaive proofs of some cases are given in [Ts]. 3. The resuls Proposiion 1. Le A and B be wo commuing power-bounded operaors on a Banach space, and 1. Then he conve combinaion A + (1 )B is a power-bounded operaor. Proof. By he binomial formula, n ( n (A + (1 )B) n k k= n ( n cons k k= ) k A k (1 ) n k B n k ) k (1 ) n k = cons( + (1 )) n = cons.

3 Power boundedness of Volerra operaor pencils 61 Theorem 1. The operaor I zv is power-bounded on L 2 (, 1) if and only if Re z and Im z =. Proof. (If ) I follows from Proposiion 1 and he power boundedness of I V (eplained above) ha I V = (1 )I + (I V ) is power-bounded for 1 on L 2 (, 1). Le m be a naural number. Noe he following eension of formula (1): (3) S 1 (I mv )S = (I (m 1)V )(I + V ) 1 where (Sf)() = e f(), f L p (, 1). We shall verify i by inducion. If m = 1, we have (1). Suppose ha (3) holds for some m. Then S 1 (I (m + 1)V )S = I S 1 (mv )S S 1 V S = (I (m 1)V )(I + V ) 1 + (I + V ) 1 I = (I (m 1)V )(I + V ) 1 + (I + V ) 1 (I + V )(I + V ) 1 = (I (m 1)V )(I + V ) 1 + (I (I + V ))(I + V ) 1 = (I mv )(I + V ) 1. This proves (3) and yields he power boundedness of I mv for all m N. Then he conve combinaion (1 )(I mv ) + (I (m + 1)V ) = I (m + )V is power-bounded for 1 and m N. (Only if ) We shall show ha he operaor I zv does no saisfy he Kreiss condiion on L 2 (, 1) for Im z. Thus I zv is no power-bounded on his space for hose z. Indeed, using he well-known formula for he resolven of V (see e.g. [Ne1, p. 27]), we obain (R(λ, I zv )f)() = f() λ 1 + z (λ 1) 2 e ( s)z/(λ 1) f(s) ds, λ 1. We have lim sup ( 1 + i/n 1) R(1 + i/n, I zv )e in = for Im z <, lim sup ( 1 i/n 1) R(1 i/n, I zv )e in = for Im z >. Of course, I zv is no power-bounded for Re z < and Im z =, because for f 1, we have lim sup (I zv ) n 1 =. Corollary 1. On L 2 (, 1), we have (I V ) n (I V ) n+1 = O(n 1/2 ) as n.

4 62 D. Tsedenbayar Proof. Se L = I µv for µ > 1, which is power-bounded by Theorem 1. Then L ω = (1 ω)i+ωl = (1 ω)i+ω(i µv ) = I ωµv is power-bounded for < ω < 1 by Proposiion 1. Now, Nevanlinna s heorem [Ne1, Theorem 4.5.3] yields ( ) lim sup n 1/2 L n ω 1/2 ω(l ω I) cons. 2π(1 ω) So, for ω = 1/µ we ge he claim. Remark 1. Corollary 1 does no follow from Nevanlinna s paper [Ne2, p. 121] because his resolven assumpion (1.35) is no saisfied for any posiive α < 1. Remark 2. Alernaively, one can also use [FoWe, Lemma 2.1] insead of [Ne1, Theorem 4.5.3]; observe ha he proof in [FoWe, Lemma 2.1] works also for power-bounded commuing pairs, or use [BoDu, Theorem 4.1]. Remark 3. I would be ineresing o know if he above esimae O(n 1/2 ) is already sharp, and if i eends, ogeher wih Theorem 1, o L p (, 1), 1 < p <. The above proof of Theorem 1 eends o hese spaces as soon as we know ha I V is power-bounded here. Perhaps he Riesz Thorin conveiy heorem [BeSh, p. 196] could be applied. Remark 4. I has been poined ou by Yuri Tomilov ha Corollary 1 also follows from [Sa] and (1), by using [FoWe] as above. However, his approach does no seem o give Theorem 1. On he oher hand, our Theorem 1 yields he corresponding informaion abou he power boundedness of he Sarason operaor pencil. Remark 5. Consider he mari ( ) 1 A =. Then I za, z C, is power-bounded if and only if z =. Theorem 2. The operaor I zv, z C, is power-bounded on L 1 (, 1) if and only if z =. Proof. We consider he following hree cases: Case <. The operaor I V is no power-bounded on L 1 (, 1) for < since as before, from he binomial formula i is clear ha lim sup (I V ) n 1 =.

5 Power boundedness of Volerra operaor pencils 63 Case >. As in [Py, p. 292] we can wrie ((I V ) n f)() ((I V ) n+1 f)() = (V (I V ) n f)() ( n ( ) ) n n ( n = ( 1) k k V k+1 f () = )( 1) k k ( s)k k k k! where k= L () n () = n k= k= ( n )( 1) k k k k!, n 1, f(s) ds = L () n (( s))f(s) ds are he Laguerre polynomials wih parameer. By summing hese formulas and using [Sz, p. 12, formula (5.1.13)], we ge ((I V ) n+1 f)() = f() L (1) n (( s))f(s) ds where L (1) n () are he Laguerre polynomials wih parameer 1. Using he classical esimaes for Laguerre polynomials [Sz, p. 177 and 198] and he formula for he norm of an inegral operaor on L 1 (, 1) given in [ToZe, Lemma 4.5], we deduce as in [ToZe, Eample 4.6] ha lim sup (I V ) n =. Case z C \ R. We show ha he operaor I zv does no saisfy he Kreiss condiion on L 1 (, 1) for Im z. Thus I zv is no power-bounded on L 1 (, 1) for hose z. Indeed, on L 1 (, 1), we have lim sup ( 1 + i/n 1) R(1 + i/n, I zv )e in = for Im z <, lim sup ( 1 i/n 1) R(1 i/n, I zv )e in = for Im z >. Remark 6. By dualiy, he same characerizaion holds on L (, 1). Proposiion 2. Le σ(q) = {}. If I Q saisfies he Ri condiion, hen so does I Q for >. Proof. We can wrie ( 1 λ R(λ, I Q) = (I Q λi) 1 = 1 = 1 [ (I Q) I + 1 λ I = 1 [ ( I Q 1 1 λ ] 1 ) I] 1. ) 1 I Q

6 64 D. Tsedenbayar Whenever 1 + λ >, i.e. λ (1 ) >, which cerainly holds for Re λ > 1, we have R(λ, I Q) 1 cons 1 = cons λ 1, 1+λ and his yields he Ri condiion by [NaZe, Lemma, p. 146] because σ(q) = {}. Remark 7. The operaor I J α saisfies he Ri condiion for < α < 1 on L p (, 1), 1 p, by [Ly2, p. 137], hence I J α saisfies he Ri condiion for all > on L p (, 1), 1 p, by Proposiion 2. Hence hese operaors are power-bounded by [Ly1, Theorem 1, p. 154] or [NaZe, Theorem, p. 147]. This observaion does no seem o follow by he mehod used above in he case α = 1, because here is no analogy of (1) and (3) for α 1. We know from Theorem 1 ha I V is power-bounded on L 2 (, 1), while I + V is no for > (for = 1 he laer also follows from he Gelfand Theorem [Ge]). This leads o he naural quesion wheher he produc (I V )(I + V ) = I 2 V 2 is power-bounded. The answer is negaive. Theorem 3. The operaor I zv 2, z C, is power-bounded on L p (, 1), 1 p, if and only if z =. Proof. We consider he following hree cases: Case <. The operaor I V 2 is no power-bounded on L p (, 1), 1 p, for < because, as before, from he binomial formula i is clear ha lim sup (I V 2 ) n 1 =. Case >. The resolven formula for V 2 is (R(λ, I V 2 )f)() = f() λ (λ 1) 3/2 sinh s f(s) ds for λ 1 (see [Hi, p. 26] or [Ne1, p. 13]). Therefore he resolven formula for I V 2 is (R(λ, I V 2 )f)() = 1 ( ( R 1 1 λ ), I V )f 2 () = f() λ 1 + 1/2 (λ 1) 3/2 sinh ( s)1/2 f(s) ds.

7 We choose f 1. Then Power boundedness of Volerra operaor pencils 65 (R(λ, I V 2 )1)() = 1 λ 1 + 1/2 (λ 1) 3/2 sinh We noe ha sinh ( s)1/2 ds = 2 λ λ 1 cosh 1/2. (λ 1)1/2 = 1/ /2 [e 1/2 /(λ 1) 1/2 Hence, for λ n = 1 + 1/n, we ge ( s)1/2 ds + e 1/2 /(λ 1) 1/2 ]. (R(λ n, I V 2 )1)() = 2n + n cosh n = n(cosh n 2), and an easy calculaion shows ha lim sup( λ n 1) (R(λ n, I V 2 )1)() =. Therefore, R(λ, I V 2 ) does no saisfy he Kreiss condiion for >. Case z C \ R. We show ha he operaor I zv 2 does no saisfy he Kreiss condiion on L p (, 1), 1 p, for Im z. Indeed, we can wrie z = (α + iβ) 2 wih α, β R, where α. In he resolven formula for I zv 2, (R(λ, I zv 2 )f)() = f() λ 1 + z 1/2 (λ 1) 3/2 sinh we se λ n = 1 + 1/n 2. Then (R(1 + 1/n 2, I z 2 V 2 )e in )() = n 2 e in + n 3 (α + iβ) We noe ha sinh[n( s)(α + iβ)]e ins ds = 1 2 We ge lim sup 1 2 ( s)z1/2 f(s) ds sinh[n( s)(α + iβ)]e ins ds. e in n(α + i(β 1)) + e n(α+iβ) 2n(α + i(β 1)) e in n(α + i(β + 1)) + e n(α+iβ) 2n(α + i(β + 1)). 1 n 2 (R(λ n, I zv 2 )e in )() =. Therefore, R(λ, I zv 2 ) does no saisfy he Kreiss condiion.

8 66 D. Tsedenbayar Acknowledgmens. I am graeful o Professors Yuri Tomilov and Jaroslav Zemánek for helpful discussions during he work on his paper. References [Al] G. R. Allan, Power-bounded elemens and radical Banach algebras, in: Linear Operaors, J. Janas, F. H. Szafraniec and J. Zemánek (eds.), Banach Cener Publ. 38, Ins. Mah., Polish Acad. Sci., 1997, [BeSh] C. Benne and R. Sharpley, Inerpolaion of Operaors, Academic Press, New York, [BoDu] F. F. Bonsall and J. Duncan, Complee Normed Algebras, Springer, Berlin, [FoWe] S. R. Foguel and B. Weiss, On conve power series of a conservaive Markov [Ge] operaor, Proc. Amer. Mah. Soc. 38 (1973), I. Gelfand, Zur Theorie der Charakere der Abelschen opologischen Gruppen, Ma. Sb. 9 (1941), [Ha] P. R. Halmos, A Hilber Space Problem Book, Van Nosrand, Princeon, [Hi] E. Hille, Remarks on ergodic heorems, Trans. Amer. Mah. Soc. 57 (1945), [Ly1] [Ly2] [NaZe] [Ne1] Yu. Lyubich, Specral localizaion, power boundedness and invarian subspaces under Ri s ype condiion, Sudia Mah. 134 (1999), , The single poin specrum operaors saisfying Ri s resolven condiion, ibid. 145 (21), B. Nagy and J. Zemánek, A resolven condiion implying power boundedness, ibid. 134 (1999), O. Nevanlinna, Convergence of Ieraions for Linear Equaions, Birkhäuser, Basel, [Ne2], Resolven condiions and powers of operaors, Sudia Mah. 145 (21), [Py] T. Pylik, Analyic semigroups in Banach algebras, Colloq. Mah. 51 (1987), [Sa] D. Sarason, A remark on he Volerra operaor, J. Mah. Anal. Appl. 12 (1965), [Sz] G. Szegő, Orhogonal Polynomials, Colloq. Publ. 23, Amer. Mah. Soc., [ToZe] [Ts] Yu. Tomilov and J. Zemánek, A new way of consrucing eamples in operaor ergodic heory, Mah. Proc. Cambridge Philos. Soc., o appear. D. Tsedenbayar, Some properies of he Volerra and Cesàro operaors, disseraion, Ins. Mah., Polish Acad. Sci., Warszawa, 22. Insiue of Mahemaics Polish Academy of Sciences Śniadeckich 8 P.O. Bo Warszawa, Poland sdnbr@impan.gov.pl Deparmen of Mahemaics Mongolian Teacher s Universiy Ulan-Baor, Mongolia Received December 31, 21 Revised version March 28, 22 (4865)