Approximation by Polynomials in a Weighted Space of Infinitely Differentiable Functions with an Application to Hypercyclicity

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1 E extracta mathematicae Vol 27, Núm 1, ) Approximation by Polynomials in a Weighted Space of Infinitely Differentiable Functions with an Application to Hypercyclicity I Kh Musin Institute of Mathematics with Computer Centre Chernyshevskii str, 112, Ufa, , Russia musin@matemanrbru Presented by Alfonso Montes Received July 10, 2011 Abstract: A space of infinitely differentiable functions defined on an open cone of R n and of prescribed growth near the boundary of the cone and at infinity is considered The problem of polynomial approximation in this space is studied It is shown that every linear continuous operator on this space that commutes with each partial derivative operator and is not a scalar multiple of the identity is hypercyclic Key words: Hypercyclic operators, polynomial approximation AMS Subject Class 2010): 41A10, 47A16 1 Introduction Let Ω be an open connected cone in R n with apex at the origin, Ω be the closure of Ω in R n Let h m ) m=1 be a sequence of positive functions h m CΩ) such that for all m N there exists a m 0 such that for all x Ω h m x) h m+1 x) ln 1 dx) ) + a m, where dx) is the distance from x Ω to the boundary Ω of Ω, t + = t for t 0, and t + = 0 for t < 0 Let ψ m ) m=1 be a sequence of positive functions ψ m CΩ) such that for each m N: ψ m x) a) lim x Ω, x x = +, where is the Euclidean norm in Rn, b) there exists b m 0 such that for all x Ω ψ m x) ψ m+1 x) ln1 + x ) b m 75

2 76 i kh musin Let φ m x) = h m x) + ψ m x), x Ω, m N For each m N let { E m = f C m Ω) : p m f) = Obviously, E m+1 E m m N) Let φ = {φ m } m=1 and E φω) = m=1 x Ω, α m } D α f)x) expφ m x)) < E m Under usual operations of addition and multiplication by complex numbers E φ Ω) is a linear space Endow E φ Ω) with the topology of projective limit of the spaces E m Obviously, E φ Ω) is a Fréchet space In view of condition b) on the family ψ m ) m=1 the space E φ Ω) is invariant under multiplication by polynomials In Section 3 it is shown that linear differential operators with constant coefficients are continuous on E φ Ω) In this paper the following two problems are considered: 1 approximation by polynomials in E φ Ω); 2 hypercyclicity of linear continuous operators on E φ Ω) commuting with each partial derivative operator The motivation to study the first problem is the following In [10] MM Mannanov has considered the problem of description of a dual space to a weighted space of holomorphic functions on an unbounded convex domain of C n with given majorants of growth near the boundary and at infinity in terms of the Laplace transform of linear continuous functionals on this space To solve the problem he used the known scheme of the proof of the Polya- Martineau-Ehrenpreis theorem see for details [7, Theorem 453]) and developed methods of the article [12] Therefore he had to consider some space of infinitely differentiable functions on an unbounded convex domain of R 2n with given majorants of growth near the boundary of a domain and at infinity But for this space approximation problems for example, approximation by polynomials or by a system of exponentials) have not been studied There was no need to study them to obtain the main result of [10] Note that problems of approximation by polynomials in E φ Ω) are studied here under minimal conditions on weight functions h m and ψ m In Section 2 the following theorem is proved Theorem 1 The polynomials are dense in E φ Ω)

3 approximation by polynomials in a weighted space 77 From this theorem it easily follows that E φ Ω) is a separable space Theorem 1 helps us to study the second problem Recall that a linear continuous operator T on a separable locally convex space X is called hypercyclic if there is an element x X such that its orbit Orb{x, T } = {x, T x, T 2 x, } is dense in X Hypercyclic operators have been extensively studied since the 1980 s For background, development, the most known results of theory of hypercyclic operators we refer the reader to the surveys of K-G Grosse- Erdmann [5] where, in particular, results on hypercyclicity of operators in the real analysis setting are described), [6] and to the articles by R Gethner and JH Shapiro [2], G Godefroy and JH Shapiro [3], A Montes-Rodríguez and N Salas [11] Note that C Kitai [8] and R Gethner and JH Shapiro [2] provided a useful sufficient condition the so-called Hypercyclicity Criterion) for an operator to be hypercyclic It was refined afterwards by many authors see [1], [5], [6]) In some cases it is convenient to use the following result established by G Godefroy and JH Shapiro [3] see also [4, Corollary 110]) with the help of the Hypercyclicity Criterion Theorem A Godefroy-Shapiro Criterion) Let X be a separable Fréchet space and T : X X be a linear continuous operator Suppose that λ <1 kert λ) and λ >1 kert λ) both span a dense subspace of X Then T is hypercyclic In Section 3, Theorem 1 and Theorem A are used to prove the following statement For notation, see below Theorem 2 Every linear continuous operator on E φ Ω) that commutes with each partial derivative operator and is not a scalar multiple of the identity is hypercyclic Let us fix some notation For u = u 1,, u n ), v = v 1,, v n ) R n C n ), u, v = u 1 v u n v n and u denotes the Euclidean norm in R n C n ) For α = α 1,, α n ) Z n +, x = x 1,, x n ) R n, z = z 1,, z n ) C n, α = α α n, x α = x α 1 1 xαn n, z α = z α 1 1 zαn n, D α = α x α 1 1 xαn n, D α z = α z α 1 1 zαn n For multi-indices α = α 1,, α n ), β = β 1,, β n ) Z n + the notation β α indicates that β j α j for j = 1, 2,, n

4 78 i kh musin For a positive function Φ CΩ) such that For the sake of simplicity, we put ) Φx) := inf x, y + Φy), x R n y Ω If S n 1 = {x R n : x = 1} we set For each σ prω), let For a function u : 0, ) R, let θ m x) := expφ m x)), x R n prω) := Ω S n 1 ψ m,σ t) := ψ m σt), t > 0 u[e]x) := ue x ), x 0 For Ω R n and ε > 0, let Ω ε) be the ε-enlargement of Ω let Φx) lim = + we set x Ω, x x 2 On approximation by polynomials in E φ Ω) For a lower semi-continuous function u : [0, ) R such that Note that u x) < on [0, ) and ux) lim = + 1) x + x u ) x) = xy uy), x 0 y 0 u x) lim = + x + x Lemma 1 Let a lower semi-continuous function u : [0, ) R satisfies the condition 1) Then u[e]) x) + u [e]) x) x ln x x, x > 0

5 approximation by polynomials in a weighted space 79 Thus, Proof Let x > 0 There exist numbers t 0 and ξ 0 such that u[e]) x) = xt ue t ), u [e]) x) = xξ u e ξ ) u[e]) x) + u [e]) x) = xt ue t ) + xξ e ξ η uη) ) η 0 Hence, for each η 0 u[e]) x) + u [e]) x) xt ue t ) + xξ e ξ η + uη) Putting here η = e t we have u[e]) x) + u [e]) x) xt + xξ e ξ+t Consequently, u[e]) x) + u [e]) x) xy e y ) y 0 y R xy e y ) = x ln x x Proof of Theorem 1 Let ω be the function on R n defined as follows: ωt) = c ω exp 1 for t < 1, 1 t 2 ) ωt) = 0 for t 1, where c ω > 0 is chosen so that ωt) dt = 1 R n For ε > 0 let ω ε t) = ε n ω t ε ), t Rn For each ν N let { K ν = x Ω : x ν, distx, Ω) 1 } ν Obviously, the closed sets K ν are non-empty for ν ν 0 ν 0 is some positive integer) and K ν int K ν+1, ν=ν 0 K ν = Ω For ν ν 0 let r ν = ν 1 η ν x) = K 2rν ) ν ν+1) and ω rν x y) dy, x R n

6 80 i kh musin Obviously, η ν C0 Rn ), η ν x) = 1 for x K r ν) ν, η ν x) = 0 for x / K 3r ν) ν, 0 η ν x) 1 for x R n Since for each α Z n + ) x y D α η ν )x) = 1 we have for all x R n r n+ α ν D α η ν )x) 1 K 2r ν ) ν r n+ α ν m α rν n+ α µ D α ω) r ν dy, x R n, ) D α ω)t) µ K ν 2rν) t R n ) M α1 + ν) n, K 2rν) ν r n+ α ν where m α = D α ω)t), M α = m n απ 2 t R n Γ n +1), Γ is the Gamma-function and 2 µ denotes n-dimensional Lebesgue measure Thus, for each α Z n + D α η ν )x) M α 4 n+ α ν + 1) 3n+2 α, x R n, 2) where M α > 0 does not depend on ν ν 0 Now let f E φ Ω) This means that f C Ω) and for each m N there exists c m > 0 such that D α f)x) c m θ m x), x Ω, α m 3) Let us approximate f by polynomials in E φ Ω) There are three steps in the proof 1 For every positive integer ν ν 0 let f ν x) = fx)η ν x), x Ω Obviously, f ν E φ Ω) Note that pf ν ) K ν+1 Let us show that f ν f in E φ Ω) as ν First note that for each m N x Ω f ν x) fx) θ m x) fx) 1 η ν x)) = x Ω θ m x) p m+1 f) exp x Ω\K ν fx) θ m x) φm+1 x) φ m x) ) x Ω\K ν Let T ν = Ω {x R n : x > ν}, S ν = Ω \ K ν T ν ), ν ν 0 Since for each m N one has see the properties of the families h m ) m=1 and ψ m) m=1 ) that φ m+1 x) φ m x) ln1 + ν) + a m + b m, x T ν, φ m+1 x) φ m x) ln ν + a m + b m, x S ν, )

7 approximation by polynomials in a weighted space 81 then Hence, for each m N exp φm+1 x) φ m x) ) ) x Ω\K ν 0, ν + 4) x Ω f ν x) fx) θ m x) 0 as ν 5) Further, for x Ω and α Z n + with α > 0 we have D α f ν x) fx)) = CαD β β f)x)d α β η ν )x) 6) x Ω 1 α m β α, β < α + β α, β < α D α f)x)η ν x) 1), where Cα β = n j=1 Cβ j α j and C β j α j are the combinatorial numbers Denote by F ν x) the first term of the right-hand side in 6) We have for each m N Cα D β β f)x)d α β η ν )x) F ν x) β α, β < α θ m x) θ m x) x K ν+1 \ K ν 1 α m With the help of the inequalities 2) and 3) we have for each s N x Ω 1 α m F ν x) θ m x) x K ν+1 \ K ν 1 α m 2 mn 4 n+m p m+s f)ν + 1) 3n+2m e φmx) φ m+sx) 7) Let R ν = K ν+1 \ K ν ) {ν x ν + 1}, P ν = K ν+1 \ K ν ) { x ν} Note that for each k N φ k x) φ k+1 x) ln1 + ν) a k b k, x R ν, φ k x) φ k+1 x) ln ν a k b k, x P ν Thus, for each s N one obtains x R ν, 1 α m F ν x) θ m x) 2mn 4 n+m p m+s f)ν + 1) 3n+2m 1 + ν) s e am a m+s b m b m+s,

8 82 i kh musin x P ν, 1 α m F ν x) θ m x) x Ω, 1 α m 2mn 4 n+m p m+s f)ν + 1) 3n+2m ν s e a m a m+s b m b m+s Letting s = 3n + 2m + 1 we get from these two estimates and 7) that for each m N F ν x) 0 as ν 8) θ m x) Now let us consider the second term in 6) For arbitrary m N we have x Ω 1 α m D α f)x)η ν x) 1) θ m x) p m+1 f) exp x Ω \ K ν 1 α m D α f)x) θ m x) φm+1 x) φ m x) ) x Ω\K ν ) Using 4) we get x Ω, 1 α m D α f)x)η ν x) 1) θ m x) 0, ν From this, 8) and 5) it follows that p m f ν f) 0 as ν for each m N Thus, the sequence f ν ) ν=1 converges to f in E φω) as ν 2 Fix a positive integer ν ν 0 Let h 0 be an entire function of exponential type at most 1 such that h L 1 R) and hx) 0 for x R For example, we can take hz) = sin2 z 2 z 2, z C By the Paley-Wiener theorem [9] there exists a function g CR) with port in [ 1, 1] such that hz) = 1 1 gt)e izt dt, z C From this representation it follows that for each k Z + h k) x) 2 max gt), x R 9) t 1 Let Hz 1, z 2,, z n ) = hz 1 )hz 2 ) hz n ) It is an entire function in C n From 9) it follows that there exists a positive constant C H > 0 such that for each α Z n + D α H)x) C H, x R n 10)

9 approximation by polynomials in a weighted space 83 Let R n Hx) dx = A Define a function f ν on R n as follows: fν x) = f ν x), x Ω; fν x) = 0, x R n \ Ω Obviously, fν C R n ) For λ > 1 let f ν,λ x) = λn fν y)hλx y)) dy, x R n A R n It is clear that f ν,λ C R n ) Moreover, fν,λ admits holomorphic continuation in C n Note that for all α Z n + and x R n D α fν,λ )x) λ n D α fν )y) Hλx y)) dy A R n max D α f ν )y) λn Hλx y)) dy y K ν+1 A R n = max y K ν+1 D α f ν )y) Let f ν,λ be the restriction of f ν,λ on Ω Obviously, f ν,λ E φ Ω) Let us show that f ν,λ f ν in E φ Ω) as λ + Take an arbitrary m N and let rλ) = λ 2n 2n+1 λ > 1) Note that for all α Z n +, x Ω D α f ν,λ )x) D α f ν )x) = λn D α fν )y) D α fν )x) ) Hλx y)) dy A R n = λn D α fν )y) D α fν )x) ) Hλx y)) dy A {y R n : y x rλ)} + λn D α fν )y) D α fν )x) ) Hλx y)) dy A {y R n : y x >rλ)} Denote the terms on the right-hand side of this equality by I 1,α x) and I 2,α x), respectively Let C ν,m = D β fν )t) Then x Ω, α m t R n, β m+1 I 1,α x) π n 2 nch C ν,m AΓ n 2 + 1) λ n I 2,α x) 2C ν,m x Ω, α m A u >λ 1 2n+1 2n+1, Hu) du

10 84 i kh musin From these two estimates it follows that D α f ν,λ )x) D α f ν )x) 0 x Ω, α m as λ + Hence, p m f ν,λ f ν ) 0 as λ + Since m N was arbitrary then f ν,λ f ν in E φ Ω) as λ + 3 For fixed λ > 0 and positive integer ν ν 0 let us approximate f ν,λ by polynomials in E φ Ω) For N N let U N x) = H0) + For x R n we have Hx) U N x) N k=1 1 i 1 n 1 i 1 n Using the inequality 10) we get 1 i k n 1 i N+1 n ξ [0,x] k H x i1 x ik 0)x i1 x ik k! N+1 H ξ)x i1 x in+1 x i1 x in+1 N + 1)! Hx) U N x) C Hn N+1 x N+1 N + 1)!, x R n 11) Let V N x) = λn fν y)u N λx y)) dy, x R n A R n It is clear that V N is a polynomial of degree at most N We claim that the sequence V N ) N=1 converges to f ν,λ in E φ Ω) as N Let m N be arbitrary For α Z n + and x Ω we have D α f ν,λ )x) D α V N )x) = λn D α fν )y) Hλx y)) U N λx y)) ) dy A R n Using the inequality 11) and taking into account that f ν has a bounded port we can find positive constants C 1 and C 2 depending on n, λ, ν and m) such that for all N N, α Z n + with α m, x Ω D α f ν,λ )x) D α V N )x)) C 1 C N x )N+1 N + 1)!

11 approximation by polynomials in a weighted space 85 Thus, for each N N p m f ν,λ V N ) C 1C N 2 N + 1)! x Ω 1 + x ) N+1 θ m x) Furthermore, x Ω 1 + x ) N+1 θ m x) x Ω = exp 1 + x ) N+1 2 N+1 exp = 2 N+1 exp = 2 N+1 exp = 2 N+1 exp e ψ mx) r>0, σ prω) N + 1) lnr + 1) ψm rσ) )) r>1, σ prω) t>0, σ prω) σ pr Ω) N + 1) ln r ψm rσ) )) +) N + 1)t ψm e t σ) )) +) N + 1)t ψm,σ e t ) ))) +) t>0 ) + ) ψ m,σ [e]) N + 1) σ prω) Now applying Lemma 1 we obtain x Ω 1 + x ) N+1 θ m x) Thus, for each N N 2 N+1 max p m f ν,λ V N ) C 1C N 2 2N+1 N + 1)! max 1, 1, ) N + 1) N+1 e N+1 e inf σ prω) ψ m,σ[e]) N+1) ) N + 1) N+1 e N+1 e inf 12) σ prω) ψ m,σ[e]) N+1) Since N! N N e N for all N N, we get C 1 C N 2 2N+1 N + 1)! N + 1) N+1 e N+1 e inf σ prω) ψ m,σ [e]) N+1) C 1 CN 2 2N+1 e inf σ prω) ψ m,σ [e]) N+1) 13)

12 86 i kh musin Note that uniformly for σ prω) one has lim ξ + This is so because for each σ prω) ψ m,σ[e]) ξ) ξ = + 14) ψ m,σ[e]) ξ) ξt ψ m,σe t ), ξ > 0, t > 0, and ψm,σe t ) = e t r ψ m,σ r) ) r 0 r 0, σ prω) e t r ψ m rσ) ) = e t x ψ m x) ) x Ω Using 13) and 14) we have from 12) that p m f ν,λ V N ) 0 as N From the conclusions of all three above steps, one derives that each function f E φ Ω) can be approximated by polynomials in E φ Ω) 3 Application of Theorem 1 to hypercyclicity The following auxiliary results will be used in the proof of Theorem 2 Lemma 2 Partial derivative operators are continuous on E φ Ω) Proof Let m N be arbitrary Since φ m x) φ m+1 x) a m b m for all x Ω, we get for each f E φ Ω), all α Z n + with α m and j = 1,, n that Thus, D α x j f )) x) p m+1f)e φ m+1x) p m+1 f)e φmx)+am+bm, x Ω ) p m f e a m+b m p m+1 f), f E φ Ω) x j This means that the operators x j are continuous on E φ Ω) Corollary 1 Let P x) = a α x α be a polynomial in R n N N) α N Then the operator a α D α is continuous on E φ Ω) α N

13 approximation by polynomials in a weighted space 87 Note that for each z C n the function f z ξ) := expi ξ, z ) belongs to E φ Ω) since for each m N p m f z ) 1 + z ) m exp φ m Im z) ) So for each functional Φ E φ Ω)) its Fourier-Laplace transform ˆΦz) = Φe i ξ,z ) is well defined everywhere on C n Lemma 3 Let Φ E φ Ω)) Then ˆΦ is an entire function on C n Moreover, for each α Z n + Dz α ˆΦ)z) = Φ iξ) α e i ξ,z ), z C n 15) Proof Fix Φ as in the hypothesis as well as an arbitrary point ζ C n For each z C n such that z ζ < 1 let Using the inequality g z,ζ ξ) := e i ξ,z e i ξ,ζ i ξ, z ζ e i ξ,ζ, ξ R n 16) D α g z,ζ )ξ) 1 + ζ ) α 1 + ξ + ξ 2 )e ξ z ζ 2 e ξ, Im ζ, α Z n +, positivity of functions h m and condition a) on the system ψ m ) m=1 for each m N we can find a constant C > 0 depending on ζ and m such that p m g z,ζ ) C z ζ 2 17) Since Φ is a continuous functional then Φg z,ζ ) = o z ζ ), z ζ And now since Φ is linear we get ˆΦz) ˆΦζ) = n j=1 Φ iξ j e i ξ,ζ ) z j ζ j ) + o z ζ ), z ζ Therefore, ˆΦ is holomorphic at the point ζ Since ζ C n was arbitrary, then ˆΦ is an entire function The second part of the statement is evident The lemma is proved Lemma 4 Let O be a non-empty open set in C n {expi ξ, z )} z O is complete in E φ Ω) Then the system

14 88 i kh musin Proof Let S be an arbitrary linear continuous functional on E φ Ω) such that S e i ξ,z ) = 0 for each z O By using the Hahn-Banach theorem, our task is to show that S is a zero functional By Lemma 3 the Fourier-Laplace transform Ŝ of S is an entire function on Cn So by the uniqueness theorem Ŝz) = 0 for each z C n Now using 15) we get Sξ α ) = 0 for all α Z n + Thus, Sp) = 0 for each polynomial p By Theorem 1 polynomials are dense in E φ Ω) Therefore, S = 0 and the proof is complete Denote by LE φ Ω)) the set of linear continuous operators on E φ Ω) Let T LE φ Ω)) Define the function F T on Ω C n by the rule F T ξ, z) = T f z )ξ) For each fixed z C n let f j,z ξ) := iξ j expi ξ, z ) Lemma 5 Let T LE φ Ω)) Then the function F T is an entire function in the second variable Proof Let T be a linear continuous operator on E φ Ω) Then for each k N there exist numbers c k > 0 and m N such that for all g E φ Ω) p k T g)) c k p m g) 18) Let ξ Ω, ζ C n be arbitrary points For each z C n such that z ζ < 1 consider the function g z,ζ see 16)) From 18) it follows that T g z,ζ )ξ) c k p m g z,ζ )e φ kξ), ξ Ω From this, the inequality 17) and linearity of T we get for each ξ Ω F T ξ, z) F T ξ, ζ) = n T f j,ζ )ξ)z j ζ j ) + o z ζ ), z ζ j=1 Therefore, for each fixed ξ Ω, F T ξ, z) is holomorphic at the point ζ as a function of z Since ζ C n was arbitrary, then the assertion of lemma is proved Proof of Theorem 2 Since T commutes with each partial derivative operator, then for each z C n and j = 1,, n D j T f z ) = T D j f z ) = T iz j f z ) = iz j T f z ) From this it follows that for each z C n there is a complex number a T z) such that T f z ) = a T z)f z 19)

15 approximation by polynomials in a weighted space 89 Thus, for all z C n, ξ Ω we have F T ξ, z) = a T z)e i ξ,z Using Lemma 5 we get that a T is an entire function on C n Taking into account that T is not a scalar multiple of the identity and Lemma 4 we conclude that a T is not a constant function Note that, if Ω is a non-empty open set in C n, then the system {T f z )} z Ω is complete in E φ Ω) It is easy to show using the representation 19) and Lemma 4 Consider the sets W 1 = {z C n : a T z) < 1} and W 2 = {z C n : a T z) > 1} They are open in C n Let X 0 be the linear span of the system {T f z )} z W1, Y 0 be the linear span of the system {T f z )} z W2 The sets X 0 and Y 0 are dense in E φ Ω) Therefore, linear spans of the sets kert λ) and kert λ) are dense in E φ Ω) λ <1 λ >1 Thus, all the conditions of Theorem A are fulfilled Hence, the operator T is hypercyclic Corollary 2 Let P x) = R n N N) Then the operator α m α N a α x α be a non-constant polynomial in a α D α is hypercyclic in E φ Ω) Acknowledgements The author is very grateful to the referees for their careful reading, valuable comments and suggestions This work was ported by the grants RFBR , References [1] J Bès, A Peris, Hereditarily hypercyclic operators, J Funct Anal ), [2] RM Gethner, JH Shapiro, Universal vectors for operators on spaces of holomorphic functions, Proc Amer Math Soc ), [3] G Godefroy, JH Shapiro, Operators with dense, invariant, cyclic vector manifolds, J Funct Anal ), [4] F Bayart, E Matheron, Dynamics of Linear Operators, Cambridge Tracts in Mathematics, 179, Cambridge University Press, Cambridge, 2009 [5] K-G Grosse-Erdmann, Universal families and hypercyclic operators, Bull Amer Math Soc ), [6] K-G Grosse-Erdmann, Recent developments in hypercyclicity, RACSAM Rev R Acad Cienc Exactas Fís Nat Ser A Mat ), [7] L Hörmander, An Introduction to Complex Analysis in Several Variables, D Van Nostrand Co, Inc, Princeton, NJ-Toronto, Ont-London 1966

16 90 i kh musin [8] C Kitai, Invariant Closed Sets for Linear Operators, Ph D Thesis, University of Toronto, Toronto, 1982 [9] P Koosis, Introduction to H p Spaces, Cambridge Tracts in Mathematics, 115 Cambridge University Press, Cambridge, 1998 [10] MM Mannanov, Description of a certain class of analytic functionals Russian), Sibirsk Mat Zh 31 3) 1990), 62 72; translation in Siberian Math J 31 3) 1990), [11] A Montes-Rodríguez, NH Salas, Supercyclic subspaces, Bull London Math Soc ), [12] VV Napalkov, Spaces of analytic functions of given growth near the boundary Russian), Izv Akad Nauk SSSR Ser Mat ), ; translation in Math USSR-Izv ),