cyclic vectors in dirichlet type spaces
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Cyclic vectors in Dirichlet-type spaces Constanze Liaw (Baylor - PowerPoint PPT Presentation

Cyclic vectors in Dirichlet-type spaces Constanze Liaw (Baylor University) at TeXAMP 2013 This presentation is based on joint work with C. B en eteau, A. Condori, D. Seco, A. Sola. Thanks to NSF for their support. Broader Impacts of the


  1. Cyclic vectors in Dirichlet-type spaces Constanze Liaw (Baylor University) at TeXAMP 2013

  2. This presentation is based on joint work with C. B´ en´ eteau, A. Condori, D. Seco, A. Sola. Thanks to NSF for their support.

  3. Broader Impacts of the problem of cyclicity Invariant subspace problem and cyclic vectors: Does every bounded operator T on a Hilbert space H have a non-trivial closed invariant subspace (i.e. T ( W ) ⊂ W )? NO, IF one can find an operator T such that every 0 � = ϕ ∈ H is cyclic (i.e. H = clos span { T n ϕ : n ∈ N } ). Structure (basic building blocks) of a function space determined by its cyclic vectors Brown–Shields conjecture For physicists, the cyclicity of an operator means that the spectrum has multiplicity one

  4. One complex variable

  5. Dirichlet-type spaces and cyclic vectors Consider the Dirichlet-type spaces D α , i.e. bounded analytic functions on the unit disk D ⊂ C with norm k =0 ( k + 1) α | a k | 2 < ∞ , where f ( z ) = � ∞ D α = � ∞ k =0 a k z k � f � 2 Bergman A 2 = D − 1 ; Hardy H 2 = D 0 ; and Dirichlet D = D 1 A vector f is cyclic (under the forward shift) for D α if D α = span { z k f ( z ) : k ∈ N ∪ { 0 }} The constant function 1 is cyclic for D α f ∈ D α cyclic, implies f ( z ) � = 0 for z ∈ D “The fewer zeros the easier is cyclicity.”

  6. Optimality Note f is cyclic in D α iff p n � p n f − 1 � 2 N n ( f, α ) := inf D α → 0 as n → ∞ If f ( z ) = 1 − z , then p n = (order n Taylor poly. of 1 /f ) yields � p n f − 1 � 2 D α = n + 2 Two types of results: Optimal sequence of polynomials p n The optimal rate of decay of these norms N n ( f, α ) as n → ∞

  7. Example of explicit optimal approximants For f ( z ) = 1 − z, optimal for n � k � H 2 : � z k , C n ( z ) = 1 − n + 1 k =0 n n � 1 − H k +1 � 1 � z k , � D : R n ( z ) = H n = k , H n +2 k =0 k =2 n � k ( k + 3) � A 2 : � z k . S n ( z ) = 1 − ( n + 1)( n + 4) k =0

  8. Rate of decay Let H n = � n 1 k and note that H n ≈ log n for large n . k =2 Definition For α < 1 , we set ϕ α ( n ) = n α − 1 , n ∈ N . For α = 1 , we use ϕ 1 ( n ) = 1 /H n , n ∈ N . Theorem (B´ en´ eteau–Condori–L.–Seco–Sola, J. d’A. accepted) Suppose f ∈ D α , α ≤ 1 , can be extended analytically to some strictly bigger disk. Suppose also that f does not vanish in D . Then there exists a constant C 0 so that the optimal norm satisfies N n ( f, α ) ≤ C 0 ϕ α ( n + 1) . Moreover, for polynomial f with zero on T , and α = 1 , 0 , − 1 , there is a constant C 1 so that C 1 ϕ α ( n + 1) ≤ N n ( f, α ) . Polynomials that have no zeros in D are cyclic in D α for α ≤ 1 .

  9. Partial result on the Brown–Shields conjecture

  10. Outer Vectors in H 2 are cyclic iff they are outer For α ≥ 0 : If f cyclic in D α , then f outer Logarithmic capacity Non-tangentially f ∗ ( ζ ) = lim z → ζ ∈ T f ( z ) For f ∈ D , f ∗ exists outside a set of logarithmic capacity zero Zero set Z ( f ) = { ζ ∈ T : f ∗ ( ζ ) = 0 } Brown–Shields: If f ∈ D is cyclic, then Z ( f ) has capacity zero Brown–Shields Conjecture (1984) A vector f ∈ D is cyclic iff it is outer and has Z ( f ) capacity zero. Brown–Cohn: For any closed set of logarithmic capacity zero E ⊂ T , there exists a cyclic function f in D with Z ( f ) = E .

  11. Two weak versions of the Brown–Shields conjecture: Theorem (Hedenmalm–Shields 1990, Richter–Sundberg 1994) A vector f ∈ D is cyclic, if it is outer and Z ( f ) is countable. Theorem (El-Fallah–Kellay–Ransford 2006) The condition ‘countable’ can be replaced by one which is closer to ‘capacity zero’, but VERY complicated.

  12. Theorem (B´ en´ eteau–Condori–L.–Seco–Sola, J. d’A. accepted) Suppose f ∈ D and log f ∈ D . Then f is cyclic in D . Theorem (B´ en´ eteau–Condori–L.–Seco–Sola, J. d’A. accepted) Let f ∈ H ∞ and q = log f ∈ D α , α ≤ 1 . Suppose there exist polynomials q n of degree ≤ n that approach q in D α norm with sup Re( q ( z ) − q n ( z )) + log � q n − q � ≤ C z ∈ D for some constant C > 0 . Then f is cyclic in D α . Brown–Cohn’s examples satisfy above assumptions.

  13. Two complex variables

  14. Dirichlet-type space on the bidisk Bidisk D 2 = { ( z 1 , z 2 ) ∈ C 2 : | z 1 | < 1 , | z 2 | < 1 } Holomorphic f : D 2 → C belongs to the Dirichlet-type space D α if its power series f ( z 1 , z 2 ) = � ∞ � ∞ l =0 a k,l z k 1 z l 2 satisfies k =0 ∞ ∞ ( k + 1) α ( l + 1) α | a k,l | 2 < ∞ � � � f � 2 α = k =0 l =0 Function f ∈ D α is cyclic , if D α := span { z k 1 z l 2 f : k = 0 , 1 , . . . ; l = 0 , 1 , . . . } Let P n , n ∈ N , be the polynomials of the form n n � � c k,l z k 1 z l p n = 2 k =0 l =0 n →∞ f is cyclic iff N n ( f, α ) := inf p n ∈ P n � p n f − 1 � 2 → 0 D α

  15. Reductions to functions of one variable

  16. Reduction to functions of one variable Consider � ∞ � � a k z Mk z Nk J α,M,N := f ∈ D α : f = , 1 2 k =0 e.g. f ( z 1 , z 2 ) = 1 − z 1 z 2 ∈ J α, 1 , 1 Consider the mappings L M,N ( F )( z 1 , z 2 ) = F ( z M 1 · z N L M,N : D 2 α → D α via 2 ) , R M,N ( f )( z ) = f ( z 1 /M , 1) R M,N : J α,M,N → D 2 α via If f ∈ J α,M,N , there exist constants such that c 2 � R ( f ) � D 2 α ≤ � f � α ≤ c 1 � R ( f ) � D 2 α Note the change from D α for bidisk to D 2 α for disk!

  17. Theorem (B´ en´ eteau–Condori–L.–Seco–Sola, submitted 2013) Let f ∈ J α,M,N have the property that R ( f ) = f ( z 1 /M , 1) is a function that admits an analytic continuation to the closed unit disk, whose zeros lie in C \ D . Then f is cyclic in D α , and there exists a constant C = C ( α, f, M, N ) such that N n ( f, α ) ≤ Cϕ 2 α ( n + 1) . This result is sharp in the sense that, if R ( f ) has at least one zero on T , then there exists c = c ( α, f, M, N ) such that for large n : cϕ 2 α ( n + 1) ≤ N n ( f, α ) . � n 2 α − 1 � for 2 α < 1 Here ϕ 2 α ( n ) = increases if α > 1 / 2 . 1 / � n 1 for 2 α = 1 k =2 k

  18. Examples Functions like f ( z 1 , z 2 ) = 1 − z 1 , f ( z 1 , z 2 ) = (1 − z 1 z 2 ) N , N ∈ N , and f ( z 1 , z 2 ) = z 2 1 z 2 2 − 2(cos θ ) z 1 z 2 + 1 , θ ∈ R , satisfy the assumptions of the theorem Polynomial g ( z 1 , z 2 ) = 1 − z 1 z 2 is not cyclic in D α for α > 1 / 2 , although it is only zero for z 1 = z 2 = 1 Notice that g is outer, but its zero set { z 1 = z 2 = 1 } has non-zero logarithmic capacity

  19. Open problems The Brown-Shields conjecture for functions on the bidisk: Is the condition that f ∈ D is outer and the zero set of f (on the boundary) has logarithmic capacity 0 sufficient for f to be cyclic? Sub-problem: Characterize the cyclic polynomials f ∈ D α for each α ≤ 1 .

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