Rate of growth of Dfrequently hypercyclic functions


 Gladys Bridges
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1 Rate of growth of Dfrequently hypercyclic functions A. Bonilla Departamento de Análisis Matemático Universidad de La Laguna
2 Hypercyclic Definition A (linear and continuous) operator T in a topological vector space X is said to be hypercyclic if there exists a vector x X, also called hypercyclic, whose orbit {T n x : n N} is dense in X.
3 Hypercyclic Definition A (linear and continuous) operator T in a topological vector space X is said to be hypercyclic if there exists a vector x X, also called hypercyclic, whose orbit {T n x : n N} is dense in X. Examples: T a Birkhoff (1929) D MacLane (1952/53)
4 Frequently hypercyclic Definition (Bayart and Grivaux, 2006) Let X be a topological vector space and T : X X a operator. Then a vector x X is called frequently hypercyclic for T if, for every nonempty open subset U of X, dens{n N : T n x U} > 0. The operator T is called frequently hypercyclic if it possesses a frequently hypercyclic vector.
5 Frequently hypercyclic Definition (Bayart and Grivaux, 2006) Let X be a topological vector space and T : X X a operator. Then a vector x X is called frequently hypercyclic for T if, for every nonempty open subset U of X, dens{n N : T n x U} > 0. The operator T is called frequently hypercyclic if it possesses a frequently hypercyclic vector. The lower density of a subset A of N is defined by #{n A : n N} dens(a) = liminf N, N where # denotes the cardinality of a set.
6 Remark A vector x X is frequently hypercyclic for an operator T in X if and only if for every nonempty open subset U of X, there is a strictly increasing sequence (n k ) of positive integers and some C > 0 such that n k Ck and T n k x U para todo k N.
7 Remark A vector x X is frequently hypercyclic for an operator T in X if and only if for every nonempty open subset U of X, there is a strictly increasing sequence (n k ) of positive integers and some C > 0 such that n k Ck and T n k x U para todo k N. Theorem (Bayart and Grivaux, 2006) D is frequently hypercyclic.
8 Differences between hypercyclic and frequently hypercyclic vectors
9 Differences between hypercyclic and frequently hypercyclic vectors Let T : X X be, X a Frechet space HC(T ) = { set of hypercyclic vectors } FHC(T ) = { set of frequently hypercyclic vectors }
10 Differences between hypercyclic and frequently hypercyclic vectors Let T : X X be, X a Frechet space HC(T ) = { set of hypercyclic vectors } FHC(T ) = { set of frequently hypercyclic vectors } Then HC(T ) = k 1 n 0 {x X;T n x B k } where {B k } k 1 is a numerable basis of open sets.
11 Differences between hypercyclic and frequently hypercyclic vectors Let T : X X be, X a Frechet space Then HC(T ) = { set of hypercyclic vectors } FHC(T ) = { set of frequently hypercyclic vectors } HC(T ) = k 1 n 0 {x X;T n x B k } where {B k } k 1 is a numerable basis of open sets. Hence if T is hypercyclic, HC(T ) is a dense G δ set in X, thus it is residual.
12 Differences between hypercyclic and frequently hypercyclic vectors Let T : X X be, X a Frechet space Then HC(T ) = { set of hypercyclic vectors } FHC(T ) = { set of frequently hypercyclic vectors } HC(T ) = k 1 n 0 {x X;T n x B k } where {B k } k 1 is a numerable basis of open sets. Hence if T is hypercyclic, HC(T ) is a dense G δ set in X, thus it is residual. In the known examples of FHC operators, in particular, D and T a FHC(T ) is a dense set of first category.
13 Order of growth and Dfrequently hypercyclic function Given a entire function f is defined for 1 p <, ( 1 2π 1/p,r M p (f,r) = f (re it ) dt) p > 0. 2π 0
14 Order of growth and Dfrequently hypercyclic function Given a entire function f is defined for 1 p <, ( 1 2π 1/p,r M p (f,r) = f (re it ) dt) p > 0. 2π Theorem (K.G. GrosseErdmann, 1990) 0 Let 1 p. (a) For any function ϕ : R + R + with ϕ(r) as r there is a Dhypercyclic entire function f with M p (f,r) ϕ(r) er r for r > 0 sufficiently large. (b) There is no Dhypercyclic entire function f that satisfies M p (f,r) C er r for r > 0, where C > 0.
15 [Bayart and Grivaux, 2006] D is frequently hypercyclic.
16 [Bayart and Grivaux, 2006] D is frequently hypercyclic. [ B. and GrosseErdmann, 2006] ε > 0 there exist entire functions Dfrequently hypercyclic with f (z) Ce (1+ε)r for z = r.
17 Does there exist an entire function Dfrequently hypercyclic f with M (f,r) er r a for z = r sufficiently large?
18 Does there exist an entire function Dfrequently hypercyclic f with M (f,r) er r a for z = r sufficiently large? a 1 2 : No
19 Does there exist an entire function Dfrequently hypercyclic f with M (f,r) er r a for z = r sufficiently large? a 1 2 : No a < 0: Yes
20 Does there exist an entire function Dfrequently hypercyclic f with M (f,r) er r a for z = r sufficiently large? a 1 2 : No a < 0: Yes 0 a < 1 2 :?
21 Does there exist an entire function Dfrequently hypercyclic f with M p (f,r) er r a for z = r sufficiently large?
22 Does there exist an entire function Dfrequently hypercyclic f with M p (f,r) er r a for z = r sufficiently large?
23 Theorem (BlascoBGrosseErdmann, PEMS 2009) Let ψ : R + R + be a function with ψ(r) 0 as r. a)if 1 p 2, there is no Dfrequently hypercyclic entire function f that satisfies M p (f,r) ψ(r) er r 1/2p for r > 0 sufficiently large. b)if 2 < p, there is no Dfrequently hypercyclic entire function f that satisfies M p (f,r) ψ(r) er r 1/4 for r > 0 sufficiently large.
24 Lemma Let ψ : R + R + be a function with ψ(r) 0 as r, 1 < p 2 and f is an entire function that satisfies M p (f,r) ψ(r) er r 1/2p for r > 0 sufficiently large. Then m 1 m f (n) (0) q 0. n=0 where q is the conjugate exponent of p.
25 Proof of theorem For p=1, is consequence of GrosseErdmann s theorem. (There is no Dhypercyclic entire function f that satisfies where C > 0.) M p (f,r) C er r for r > 0,
26 Proof of theorem For p=1, is consequence of GrosseErdmann s theorem. (There is no Dhypercyclic entire function f that satisfies where C > 0.) M p (f,r) C er r for r > 0, For 1 < p 2 is consequence of Lemma. Because given the open set U = {g H(C) : g(0) > 1}
27 Proof of theorem For p=1, is consequence of GrosseErdmann s theorem. (There is no Dhypercyclic entire function f that satisfies where C > 0.) M p (f,r) C er r for r > 0, For 1 < p 2 is consequence of Lemma. Because given the open set U = {g H(C) : g(0) > 1} dens{n N : f (n) 1 U} = liminf m m #{n m : f (n) (0) > 1} m 1 liminf m m f (n) (0) q = 0, n=0
28 Proof of theorem For p=1, is consequence of GrosseErdmann s theorem. (There is no Dhypercyclic entire function f that satisfies where C > 0.) M p (f,r) C er r for r > 0, For 1 < p 2 is consequence of Lemma. Because given the open set U = {g H(C) : g(0) > 1} dens{n N : f (n) 1 U} = liminf m m #{n m : f (n) (0) > 1} m 1 liminf m m f (n) (0) q = 0, n=0 which shows that f is not frequently hypercyclic for the differentiation operator.
29 Proof of the lemma a)let p=2 be. Assume that ψ is decreasing.
30 Proof of the lemma a)let p=2 be. Assume that ψ is decreasing. Then ( M 2 (f,r) = n n=0 a 2 r 2n) 1/2 e r ψ(r) r 1/4 for r > 0 sufficiently large.
31 Proof of the lemma a)let p=2 be. Assume that ψ is decreasing. Then ( M 2 (f,r) = n n=0 a 2 r 2n) 1/2 e r ψ(r) r 1/4 for r > 0 sufficiently large. Hence, for big values of r, f (n) (0) 2 r 2n+1/2 e 2r n=0 ψ(r) 2 (n!) 2 C
32 Fix m 1. Using Stirling formula, we see that the function r r 2n+1/2 e 2r (n!) 2 has its maximum at n + 1/4 of order 1/ n and an inflection point at n n + 1/4 ± Hence, if m n < 2m then
33 Fix m 1. Using Stirling formula, we see that the function r r 2n+1/2 e 2r (n!) 2 has its maximum at n + 1/4 of order 1/ n and an inflection point at n n + 1/4 ± Hence, if m n < 2m then
34 Fix m 1. Using Stirling formula, we see that the function r r 2n+1/2 e 2r (n!) 2 has its maximum at n + 1/4 of order 1/ n and an inflection point at n n + 1/4 ± Hence, if m n < 2m then 3m m r 2n+1/2 e 2r ψ(r) 2 (n!) 2 C 1 ψ(m) 2 1 n n C 1 ψ(m) 2.
35 Integrating f (n) (0) 2 r 2n+1/2 e 2r n=0 ψ(r) 2 (n!) 2 on [m,3m] we obtain for m sufficiently large that C 2m 1 m f (n) (0) 2 Cψ(m) 2, n=m
36 Integrating f (n) (0) 2 r 2n+1/2 e 2r n=0 ψ(r) 2 (n!) 2 on [m,3m] we obtain for m sufficiently large that C 2m 1 m f (n) (0) 2 Cψ(m) 2, n=m hence 2m 1 m f (n) (0) 2 0. n=m
37 Integrating f (n) (0) 2 r 2n+1/2 e 2r n=0 ψ(r) 2 (n!) 2 on [m,3m] we obtain for m sufficiently large that C 2m 1 m f (n) (0) 2 Cψ(m) 2, n=m hence This implies 2m 1 m f (n) (0) 2 0. n=m m 1 m f (n) (0) 2 0. n=0
38 If 1 p < 2, using HausdorffYoung inequality ( a n q r qn) 1 ( ) q M p n z n=0 n=0a n,r and the same ideas for p = 2.
39
40
41 Theorem (BonetB., CAOT 2011) For any function ϕ : R + R + with ϕ(r) as r. If 1 p, there is an entire function Dfrequently hypercyclic f with M p (f,r) ϕ(r) er for z = r sufficiently large. r 1/2p
42 For 1 p and a weight function v, and B p, = B p, (C,v) := {f H(C) : supv(r)m p (f,r) < } r>0 B p,0 = B p,0 (C,v) := {f H(C) : lim r v(r)m p (f,r) = 0}. These spaces are Banach spaces with the norm f p,v := supv(r)m p (f,r). r>0
43 Remarks 1.The polynomials are contained and dense in B p,0 for all 1 p.
44 Remarks 1.The polynomials are contained and dense in B p,0 for all 1 p. 2.B p,0 is separable.
45 Remarks 1.The polynomials are contained and dense in B p,0 for all 1 p. 2.B p,0 is separable. 3.The inclusion B p, H(C) is continuous.
46 Lemma v(r) Let v be a weight function such that sup <. Then the r>0 v(r + 1) differentiation operators D : B p, B p, and D : B p,0 B p,0 are continuous. Example: v(r) = e ar,r > 0,a > 0.
47 Theorem (Grivaux, 2011) Let T be a bounded operator on a complex separable Banach space X. Suppose that, for any countable set A T, span{ker(t λi) : λ T \ A} is dense in X. Then T is frequently hypercyclic.
48 Lemma The following conditions are equivalent for a weight v and 1 p < : (i) {e θz : θ = 1} B p,0. (ii) There is θ C, θ = 1, such that e θz B p,0. (iii) lim r v(r) er r 1 2p = 0
49 Proof of the Lemma Consider f (z) = e z,z C, and write z = r(cost + isint). Now, we can apply the Laplace methods for integrals, for r > 0, 2π ( π ) 1/2e ( 1 ) 2πM p (e z,r) p = e rp cost dt = rp + e rp O. 0 2rp rp
50 Proof of the Lemma Consider f (z) = e z,z C, and write z = r(cost + isint). Now, we can apply the Laplace methods for integrals, for r > 0, 2π ( π ) 1/2e ( 1 ) 2πM p (e z,r) p = e rp cost dt = rp + e rp O. 0 2rp rp This yields, for a certain constant c p > 0 depending only on p, M p (f,r) = c p e r r 1 2p ( 1 ) + e r O. r p 1
51 Proof of the Lemma Consider f (z) = e z,z C, and write z = r(cost + isint). Now, we can apply the Laplace methods for integrals, for r > 0, 2π ( π ) 1/2e ( 1 ) 2πM p (e z,r) p = e rp cost dt = rp + e rp O. 0 2rp rp This yields, for a certain constant c p > 0 depending only on p, M p (f,r) = c p e r r 1 2p ( 1 ) + e r O. r p 1 This implies that for each 1 p < there are d p,d p > 0 and r 0 > 0 such that for each θ = 1 and each r > r 0 d p e r r 1 2p M p (e θz,r) D p e r r 1 2p (1)
52 Theorem Let v be a weight function such that lim r v(r) er r 1 2p = 0 for some 1 p. If the differentiation operator D : B p,0 B p,0 is continuous, then D is frequently hypercyclic.
53 Idea of the proof We must to prove for any countable set A T, E T\A = span({e θz : θ = 1,θ T \ A}) is contained and dense in B p,0.
54 Idea of the proof We must to prove for any countable set A T, E T\A = span({e θz : θ = 1,θ T \ A}) is contained and dense in B p,0. To prove the density, we define the following vector valued functions on the closed unit disc D: H : D B p,0, H(ζ )(z) := e ζ z, ζ D. The function H is well defined, H is holomorphic on D and H : D B p,0 is continuous.
55 We proceed with the proof that E T\A is dense in B p,0 and apply the HahnBanach theorem.
56 We proceed with the proof that E T\A is dense in B p,0 and apply the HahnBanach theorem. Assume that u (B p,0 ) vanishes on E T\A. We must show u = 0.
57 We proceed with the proof that E T\A is dense in B p,0 and apply the HahnBanach theorem. Assume that u (B p,0 ) vanishes on E T\A. We must show u = 0. Since the function u H is holomorphic in D, continuous at the boundary and vanishes at the points ζ T \ A, it is zero in D. In particular (u H) (n) (0) = u(h (n) (0)) = u(z n ) = 0, hence u vanishes on the polynomials. As the polynomials are dense in B p,0, we conclude u = 0.
58 Corollary Let ϕ(r) be a positive function with lim r ϕ(r) =. For each 1 p there is an entire function f such that M p (f,r) ϕ(r) er that is frequently hypercyclic for the differentiation operator D on H(C). r 1 2p
59 Proof It is possible to find a positive increasing continuous function ψ(r + 1) ψ(r) ϕ(r) with lim r ψ(r) = and sup <. r>0 ψ(r)
60 Proof It is possible to find a positive increasing continuous function ψ(r + 1) ψ(r) ϕ(r) with lim r ψ(r) = and sup <. r>0 ψ(r) Define v(r) = r 1 2p ψ(r)e r for r r 0, with r 0 large enough to ensure that v(r) is non increasing on [r 0, [, and v(r) = v(r 0 ) on [0,r 0 ]. One can take r 0 = 1 2p.
61 The following diagram represents our knowledge of possible or impossible growth rates e r /r a for frequent hypercyclicity with respect to the differentiation operator D two weeks ago.
62 The following diagram represents our knowledge of possible or impossible growth rates e r /r a for frequent hypercyclicity with respect to the differentiation operator D two weeks ago.
63 Theorem (DrasinSaksman, in a couple weeks in ArXiv) For any C > 0 there is an entire function Dfrequently hypercyclic f with M p (f,r) C er r a(p) where a(p) = for p [2, ] and a(p) = 2p for p (1,2]
64 Theorem (DrasinSaksman, in a couple weeks in ArXiv) For any C > 0 there is an entire function Dfrequently hypercyclic f with M p (f,r) C er r a(p) where a(p) = for p [2, ] and a(p) = 2p for p (1,2] The construction is direct with no functional analysis, but uses remarkable polynomials of RudinShapiro and de la Vallee Poussin.(Conference in honour of P. Gauthier and K. Gowrisankaran, june 2023, 2011)
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