aa r X i v : . [ m a t h . F A ] M a y A short proof of the zero-two law for cosinefunctions
Jean Esterle
Abstract : Let ( C ( t )) t ∈ R be a cosine function in a unital Banach algebra. Wegive a simple proof of the fact that if lim sup t → k C ( t ) − A k <
2, then lim sup t → k C ( t ) − A k = Keywords : Cosine function, scalar cosine function, commutative local Ba-nach algebra.AMS classification : Primary 46J45, 47D09, Secondary 26A99
Recall that a cosine function taking values in a unital normed algebra A withunit element 1 A is a family C = ( C ( t )) t ∈ R of elements of A satisfying the so-calledd’Alembert equation C (0) = A , C ( s + t ) + C ( s − t ) = C ( s ) C ( t ) ( s ∈ R , t ∈ R ). (1)One can define in a similar way cosine sequences ( C ( n )) n ∈ Z . A cosine se-quence depends only on the value of C (1), since we have, for n ≥ C ( − n ) = C ( n ) = T n ( C (1)),where T n ( x ) = [ n /2] P k = C kn x n − k ( x − k is the n th -Tchebyshev polynomial.Strongly continuous operator valued cosine functions play an important rolein the study of abstract nonlinear second order differential equations, see forexample [11]. In a paper to appear in the Journal of Evolution Equations [9],Schwenninger and Zwart showed that if a strongly continuous cosine family( C ( t )) t ∈ R of bounded operators on a Banach space X satisfies lim sup t → k C ( t ) − I X k <
2, then the generator a of this cosine function is a bounded operator, sothat lim sup k C ( t ) − I X k = lim sup t → k cos ( t a ) I X − I X k =
0, and they asked whe-ther a similar zero-two law holds for general cosine functions ( C ( t )) t ∈ R . This1uestion was answered positively by Chojnacki in [5]. Using a sophisticated ar-gument based on ultrapowers, Chonajcki deduced this zero-two law from thefact that if a cosine sequence C ( t ) satisfies sup t ∈ R k C ( t ) − A k <
2, then C ( t ) = A for t ∈ R . This second result, which was obtained independently by the author in[7], is proved by Chojnacki in [5] by adapting methods used by Bobrowski, Choj-nacki and Gregosiewicz in [3] to show that if a cosine sequence ( C ( t )) t ∈ R satisfies sup t ∈ R k C ( t ) − cos ( at )1 A k < p for some a ∈ R , then C ( t ) = cos ( at )1 A for t ∈ R ,a result also obtained independently by the author in [7], which improves pre-vious results of [2], [4] and [10].The purpose of this paper is to give a short direct proof of the zero-two law.The zero-two law for complex-valued cosine functions is a folklore result, whicheasily implies that if lim sup t → ρ ( C ( t ) − A ) < t → ρ ( C ( t ) − A ) =
0, where ρ ( x ) denotes the spectral radius of an element x of a Banach algebra,see section 2. Our proof of he zero-two law is then based on the fact that if k C ( t ) − A k ≤
2, and if ρ ¡ C ¡ t ¢ − A ¢ <
1, then we have C µ t ¶ = s A − C ( t ) − A p A − u is defined by the usual series for k u k ≤
1. It follows from thisidentity and from the fact that the coefficients of the Taylor series at the originof the function t → − p − t are positive that in this situation we have °°°° C µ t ¶ − A °°°° ≤ − s − °°°° C ( t ) − A °°°° ,and the zero-two law follows.Notice that if we replace the constant 2 by a "three line argument" dueto Arendt [1] shows that if lim sup t → k C ( t ) − A k < then lim sup t → k C ( t ) − A k =
0. The proof presented here has some analogy with Arendt’s proof, andthe difficulty to estimate k ¡ A + C ¡ t ¢¢ − k is circumvented in the present paperby using the formula above.The author wishes to thank W. Chonajcki for giving information about thereference [5]. He also thanks F. Schwenninger for valuable discussions about thecontent of the paper. The zero-two law for scalar cosine functions pertains to folklore, but we couldnot find a reference in the litterature for the following certainly well-known lemma,which is a variant of proposition 3.1 of [7].2 emma 2.1.
Let c = ( c ( t )) t ∈ R be a complex-valued cosine function. Then c statis-fies one of the following conditions(i) l i msup t → | c ( t ) − | = +∞ , (ii) l i msup t → | c ( t ) − | = (iii) l i msup t → | c ( t ) − | = M : = l i msup t → | c ( t ) | < +∞ , and denote by S the set ofall complex numbers α for which there exists a sequence ( t m ) m ≥ of positivereal numbers such that l i m m →+∞ t m = l i m m →+∞ c ( t m ) = α . Then | α | ≤ M for every α ∈ S . Notice that if α ∈ S , and if a sequence ( t m ) m ∈ Z satisfies the aboveconditions with respect to α , then T n ( α ) = l i m m →+∞ T n ( C ( t m )) = l i m m →+∞ C ( nt m ),and so T n ( α ) ∈ S , and | T n ( α ) | ≤ M for n ≥
1. Now write α = cos ( z ) = +∞ P k = z k (2 k )! , andset u = Re ( z ), v = I m ( z ). We have, for n ≥ T n ( α ) = cos ( nz ) = e inu e − nv + e − inu e nv sup n ≥ | T n ( α ) | ≤ M , we have v = S ⊂ [ −
1, 1], and l i msup t → | c ( t ) − | ≤ S {1}, and let α ∈ S \ {1}. We have α = cos ( u ) for some u ∈ R . We see as above that cos ( nu ) ∈ S for every n ≥
1. If u / π is irrationnal, thenthe set { e inu } n ≥ T , and so S = [ −
1, 1] since S is closed,and in this situation l i msup t → | c ( t ) − | = u / π is rational, and let s ≥ e ius =
1. Then e i π s = e i pu sor some positive integer p , and so cos ¡ π s ¢ ∈ S . Let ( t m ) m ≥ be a sequence of positive reals such that l i m m →+∞ t m = l i m m →+∞ c ( t m ) = cos ¡ π s ¢ , let q ≥
2, and let β be a limit point of the sequence c ¡ t m s q − ¢ n ≥ . There exists y ∈ R such that cos ( y ) = β and such that s q − y = π s + k π , with k ∈ Z . Then y = (1 + k s ) π s q . Since g cd (1 + k s , s q ) =
1, there exists apositive integer r such that r y − π s q ∈ π Z , so that cos ¡ π s q ¢ ∈ S . This implies that cos ³ p π s q ´ ∈ S for p ≥ q ≥
1, and S is dense in [ −
1, 1]. Since S is closed, we obtainagain S = [ −
1, 1], which implies that l i msup t → | c ( t ) − | =
2. So if neither (i) nor(ii) holds, we have S = {1}, which implies (iii). ä Notice that if a cosine function C = ( C ( t )) t ∈ R in a Banach algebra A satisfies sup | t |≤ η k C ( t ) k ≤ M < +∞ for some η >
0, then sup | t |≤ L k C ( t ) k < +∞ for every L >
0, since sup | t |≤ n η k C ( t ) k ≤ sup k y k≤ M k T n ( y ) k for every n ≥
1, where T n de-notes the n -th Tchebyshev polynomial. In particular if a complex-valued cosinefunction c = ( c ( t )) t ∈ R satisfies (iii), then the identity(1 − c ( s − t ))(1 − c ( s + t )) = ( c ( s ) − c ( t )) c is continuous on R , whichimplies that c ( t ) = cos ( t a ) for some a ∈ C .If A is commutative and unital, we will denote 1 A the unit element of A , andwe will denote by b A the space of all characters on A , equipped with the Gelfandtopology, i.e. the compact topology induced by the weak ∗ topology on the unitball of the dual space of A . Proposition 2.2.
Let C = ( C ( t )) be a cosine function in a unital Banach algebraA . If l i msup t → ρ ( C ( t ) − A ) < then l i msup t → ρ ( C ( t ) − A ) = A is genarated by ( C ( t )) t ∈ R .Let χ ∈ b A . Then the cosine complex-valued function ( χ ( C ( t ))) t ∈ R satisfies condi-tion (iii) of the lemma, and so there exists a χ ∈ C such that we have χ ( C ( t )) = cos ( t a χ ) ( t ∈ R ).Set u χ = Re ( a χ ), v χ = I m ( a χ ). We have ρ ( C ( t ) − ≥ | − cos ( t u χ ) ch ( t v χ ) | .If the family ( u χ ) χ ∈ b A were unbounded, there would exist a sequence ( t n ) n ≥ of real numbers converging to zero and a sequence ( χ n ) n ≥ of characters of A such that cos ( t n u χ n ) = −
1, and we would have ρ ( C ( t n ) − ≥ n ≥
1. So thefamily ( u χ ) χ ∈ b A is bounded. If the family ( v χ ) χ ∈ b A were unbounded, there wouldexist a sequence ( t ′ n ) n ≥ of real numbers converging to zero and a sequence( χ ′ n ) n ≥ of characters of A such that l i m n →+∞ ch ( t ′ n v χ ′ n ) = +∞ . But this wouldimply that l i msup t → ρ ( C ( t )) = +∞ . Hence the family ( a χ ) χ ∈ b A is bounded, andwe have l i msup t → ρ ( C ( t ) − A ) = l i m t → sup χ ∈ b A | cos ( t a χ ) − | = ä Set α n = n ! 12 ¡ − ¢ . . . ¡ − n + ¢ for n ≥
1, with the convention α =
0, andfor | z | <
1, set p − z = +∞ X n = ( − n α n z n ,so that ¡ p − z ¢ = − z , and p − t is the positive square root of 1 − t for t ∈ ( −
1, 1). Also Re ¡ p − z ¢ > | z | < t ∈ [0, 1), +∞ X n = ( − n − α n t n = − p − t .Since ( − n − α n ≥ n ≥
1, the series P +∞ n = | α n | = P +∞ n = ( − n − α n is convergent,and we have +∞ X n = | α n | t n = − p − t (0 ≤ t ≤ A be a commutative unital Banach algebra, and let x ∈ A such that k x k ≤
1. Set p A − x = +∞ X n = ( − n α n x n .Then ¡ p A − x ¢ = A − x , and we have °°° A − p A − x °°° = °°°° +∞ X n = ( − n α n x n °°°° ≤ +∞ X n = | α n |k x k n = − p − k x k (2)Notice also that if A is commutative, then we have Re ³ χ ³p A − x ´´ = Re ³p − χ ( x ) ´ ≥ χ ∈ b A ). (3)We obtain the following formula Lemma 3.1.
Let ( C ( t )) t ∈ R be a cosine function in a unital Banach algebra A . As-sume that k C ( t ) − A k ≤ and that ρ ( ¡ C ¡ t ¢ − ¢ < Then we haveC µ t ¶ = s A − A − C ( t )2 .Proof : The abstract version of the formula si n ¡ u ¢ = − cos ( u )2 gives1 A − C µ t ¶ = A − C ( t )2 , C µ t ¶ = A − A − C ( t )2 = s A − A − C ( t )2 , C µ t ¶ − s A − A − C ( t )2 C µ t ¶ + s A − A − C ( t )2 = A is commutative. Let χ ∈ b A . Since ρ ¡ C ¡ t ¢ − A ¢ <
1, wehave Re ¡ χ ¡ C ¡ t ¢¢¢ >
0. Since Re µ χ µq A − A − C ( t )2 ¶¶ ≥ C ¡ t ¢ + q A − A − C ( t )2 isinvertible in A , and C ¡ t ¢ − q A − A − C ( t )2 = ä Theorem 3.2.
Let ( C ( t )) t ∈ R be a cosine sequence in a Banach algebra. If l i msup t → k C ( t ) − A k < then l i msup t → k C ( t ) − A k = η > | t | ≤ η , k C ( t ) − A k < C µ t ¶ = s A − A − C ( t )2 .Using (1), we obtain, for | t | ≤ η , °°°° C µ t ¶ − A °°°° ≤ − s − °°°° C ( t ) − A °°°° .Set l = l i msup t → k C ( t ) − A k . We obtain l ≤ − s − l ≤ l = ä Notice that the proof above gives a little bit more than the zero-two law : if k − C ( t ) k ≤ ρ ¡ − C ¡ t ¢¢ < | t | ≤ η , then we have, for n ≥ sup | t |≤ − n η k C ( t ) − A k ≤ u n ,where the sequence u n satisfies u = u n + = − q − u n for n ≥
1, and l i m n →+∞ u n =
0, which gives an explicit control on the convergence to 0 of k C ( t ) − A k as t → t → −p − t are positive was used in [6] to show that k x − x k ≥ x of a Banach algebra such that | x | ≥ T ( t )) t > ina Banach algebra A satisfies lim sup t → + k T ( t ) − T (( n + t ) k < n ( n + + n for some n ≥
1, then there exists an idempotent J of A such that lim t → k T ( t ) − J k =
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