Orthogonal polynomials and zeros of optimal approximants Daniel Seco (with Bénéteau, Khavinson, Liaw, and Simanek) Universitat de Barcelona Optimal Point Configurations and Orthogonal Polynomials, CIEM, Castro Urdiales, 20/04/2017 Seco (UB) Approximants vs. OP CIEM 1 / 11
Our basics Definition ω k Let ω 0 = 1, ω k > 0, and lim ω k + 1 = 1. Then ∞ H 2 � a k z k , || f || 2 � | a k | 2 ω k < ∞} . ω = { f ∈ Hol ( D ) : f ( z ) = ω = k ∈ N k = 0 Seco (UB) Approximants vs. OP CIEM 2 / 11
Our basics Definition ω k Let ω 0 = 1, ω k > 0, and lim ω k + 1 = 1. Then ∞ H 2 � a k z k , || f || 2 � | a k | 2 ω k < ∞} . ω = { f ∈ Hol ( D ) : f ( z ) = ω = k ∈ N k = 0 Definition f is cyclic (in H 2 ω ) if P f is dense in H 2 ω . Seco (UB) Approximants vs. OP CIEM 2 / 11
Our basics Definition ω k Let ω 0 = 1, ω k > 0, and lim ω k + 1 = 1. Then ∞ H 2 � a k z k , || f || 2 � | a k | 2 ω k < ∞} . ω = { f ∈ Hol ( D ) : f ( z ) = ω = k ∈ N k = 0 Definition f is cyclic (in H 2 ω ) if P f is dense in H 2 ω . n →∞ f cyclic ⇔ ∃{ p n } n ∈ N ∈ P : || p n f − 1 || ω → 0. Seco (UB) Approximants vs. OP CIEM 2 / 11
Our basics Definition ω k Let ω 0 = 1, ω k > 0, and lim ω k + 1 = 1. Then ∞ H 2 � a k z k , || f || 2 � | a k | 2 ω k < ∞} . ω = { f ∈ Hol ( D ) : f ( z ) = ω = k ∈ N k = 0 Definition f is cyclic (in H 2 ω ) if P f is dense in H 2 ω . n →∞ f cyclic ⇔ ∃{ p n } n ∈ N ∈ P : || p n f − 1 || ω → 0. f regular + Z ( f ) ∩ D = ∅ ⇒ cyclic ⇒ Z ( f ) ∩ D = ∅ . Seco (UB) Approximants vs. OP CIEM 2 / 11
Our approach Our approach: find p ∗ n that minimizes � p n f − 1 � among P n , that is, the orthogonal projection of 1 onto P n f . Seco (UB) Approximants vs. OP CIEM 3 / 11
Our approach Our approach: find p ∗ n that minimizes � p n f − 1 � among P n , that is, the orthogonal projection of 1 onto P n f . Theorem (BCLSS, ’15; FMS, ’14) j = 0 c j z j only solution to Mc = b where n ( z ) = � n p ∗ c = ( c j ) n M j , k = < z j f , z k f > ω , b k = < 1 , z k f > ω . j = 0 , Seco (UB) Approximants vs. OP CIEM 3 / 11
Our approach Our approach: find p ∗ n that minimizes � p n f − 1 � among P n , that is, the orthogonal projection of 1 onto P n f . Theorem (BCLSS, ’15; FMS, ’14) j = 0 c j z j only solution to Mc = b where n ( z ) = � n p ∗ c = ( c j ) n M j , k = < z j f , z k f > ω , b k = < 1 , z k f > ω . j = 0 , Useful techniques for studying cyclicity of a fixed function. Cyclic ⇔ � p ∗ n f − 1 � → 0 ⇔ p ∗ n ( 0 ) → 1 / f ( 0 ) . Seco (UB) Approximants vs. OP CIEM 3 / 11
Our approach Our approach: find p ∗ n that minimizes � p n f − 1 � among P n , that is, the orthogonal projection of 1 onto P n f . Theorem (BCLSS, ’15; FMS, ’14) j = 0 c j z j only solution to Mc = b where n ( z ) = � n p ∗ c = ( c j ) n M j , k = < z j f , z k f > ω , b k = < 1 , z k f > ω . j = 0 , Useful techniques for studying cyclicity of a fixed function. Cyclic ⇔ � p ∗ n f − 1 � → 0 ⇔ p ∗ n ( 0 ) → 1 / f ( 0 ) . So we want to know about these polynomials! Today: where are their zeros? Which points of the plane may be zeros of such polynomials? (for a fixed space) Seco (UB) Approximants vs. OP CIEM 3 / 11
The connection between OP and OA n p ∗ � n ( z ) f ( z ) = Π n ( 1 )( z ) = � 1 , ϕ k f � ϕ k ( z ) f ( z ) k = 0 Seco (UB) Approximants vs. OP CIEM 4 / 11
The connection between OP and OA n p ∗ � n ( z ) f ( z ) = Π n ( 1 )( z ) = � 1 , ϕ k f � ϕ k ( z ) f ( z ) k = 0 Theorem (BKLSS ’16) n p ∗ � n ( z ) = ϕ k ( z ) ϕ k ( 0 ) f ( 0 ) . k = 0 Seco (UB) Approximants vs. OP CIEM 4 / 11
The connection between OP and OA n p ∗ � n ( z ) f ( z ) = Π n ( 1 )( z ) = � 1 , ϕ k f � ϕ k ( z ) f ( z ) k = 0 Theorem (BKLSS ’16) n p ∗ � n ( z ) = ϕ k ( z ) ϕ k ( 0 ) f ( 0 ) . k = 0 We can also (in general) recover OPs from opt. apprs. Seco (UB) Approximants vs. OP CIEM 4 / 11
The connection between OP and OA n p ∗ � n ( z ) f ( z ) = Π n ( 1 )( z ) = � 1 , ϕ k f � ϕ k ( z ) f ( z ) k = 0 Theorem (BKLSS ’16) n p ∗ � n ( z ) = ϕ k ( z ) ϕ k ( 0 ) f ( 0 ) . k = 0 We can also (in general) recover OPs from opt. apprs. In H 2 , − 1 : z j ∈ Z ( ϕ n ) } . Z ( p ∗ n ) = { z j In fact, these zero sets characterize cyclic functions. Seco (UB) Approximants vs. OP CIEM 4 / 11
So where are the zeros? Easy trick: it is enough to study n = 1, Seco (UB) Approximants vs. OP CIEM 5 / 11
So where are the zeros? Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f : � zf � 2 If ( z − z 0 ) optimal for f ⇒ z 0 = < f , zf > . Seco (UB) Approximants vs. OP CIEM 5 / 11
So where are the zeros? Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f : � zf � 2 If ( z − z 0 ) optimal for f ⇒ z 0 = < f , zf > . If z 0 / ∈ D , z 0 is achieved by f ( z ) = 1 / ( z − z 0 ) in any space. Seco (UB) Approximants vs. OP CIEM 5 / 11
So where are the zeros? Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f : � zf � 2 If ( z − z 0 ) optimal for f ⇒ z 0 = < f , zf > . If z 0 / ∈ D , z 0 is achieved by f ( z ) = 1 / ( z − z 0 ) in any space. If we rotate the variable, the result also rotates, and the set of points that are achieved is connected so the only unknown is: which points in [ 0 , 1 ] are possible? Seco (UB) Approximants vs. OP CIEM 5 / 11
So where are the zeros? Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f : � zf � 2 If ( z − z 0 ) optimal for f ⇒ z 0 = < f , zf > . If z 0 / ∈ D , z 0 is achieved by f ( z ) = 1 / ( z − z 0 ) in any space. If we rotate the variable, the result also rotates, and the set of points that are achieved is connected so the only unknown is: which points in [ 0 , 1 ] are possible? nondec CSI � zf � 2 � zf � If ω non decreasing, then > ≥ 1 ⇒ z 0 / ∈ D . | < f , zf > | � f � Seco (UB) Approximants vs. OP CIEM 5 / 11
So where are the zeros? Easy trick: it is enough to study n = 1, ...and then we can solve the problem for each fixed function f : � zf � 2 If ( z − z 0 ) optimal for f ⇒ z 0 = < f , zf > . If z 0 / ∈ D , z 0 is achieved by f ( z ) = 1 / ( z − z 0 ) in any space. If we rotate the variable, the result also rotates, and the set of points that are achieved is connected so the only unknown is: which points in [ 0 , 1 ] are possible? nondec CSI � zf � 2 � zf � If ω non decreasing, then > ≥ 1 ⇒ z 0 / ∈ D . | < f , zf > | � f � If ∃ k , n : ω k > 4 ω k + n + 1 then f k , n ( z ) = z k T n ( 1 + z 1 − z ) makes z 0 ∈ D . Seco (UB) Approximants vs. OP CIEM 5 / 11
The general problem Problem Let ω fixed (decreasing). � � | < f , zf > | : f ∈ H 2 What is the value of U ω := sup ? ω � zf � 2 Seco (UB) Approximants vs. OP CIEM 6 / 11
The general problem Problem Let ω fixed (decreasing). � � | < f , zf > | : f ∈ H 2 What is the value of U ω := sup ? ω � zf � 2 Is the supremum a maximum? Uniqueness? Seco (UB) Approximants vs. OP CIEM 6 / 11
The general problem Problem Let ω fixed (decreasing). � � | < f , zf > | : f ∈ H 2 What is the value of U ω := sup ? ω � zf � 2 Is the supremum a maximum? Uniqueness? If so, what is the extremal function? Seco (UB) Approximants vs. OP CIEM 6 / 11
The general problem Problem Let ω fixed (decreasing). � � | < f , zf > | : f ∈ H 2 What is the value of U ω := sup ? ω � zf � 2 Is the supremum a maximum? Uniqueness? If so, what is the extremal function? Theorem U ω = �J ω � 2 Seco (UB) Approximants vs. OP CIEM 6 / 11
The general problem Problem Let ω fixed (decreasing). � � | < f , zf > | : f ∈ H 2 What is the value of U ω := sup ? ω � zf � 2 Is the supremum a maximum? Uniqueness? If so, what is the extremal function? Theorem U ω = �J ω � 2 If U ω > 1 then ∃ ! f ∗ (up to f ∗ ( z ) → Cf ∗ ( ze i θ ) ). Seco (UB) Approximants vs. OP CIEM 6 / 11
The general problem Problem Let ω fixed (decreasing). � � | < f , zf > | : f ∈ H 2 What is the value of U ω := sup ? ω � zf � 2 Is the supremum a maximum? Uniqueness? If so, what is the extremal function? Theorem U ω = �J ω � 2 If U ω > 1 then ∃ ! f ∗ (up to f ∗ ( z ) → Cf ∗ ( ze i θ ) ). f ∗ ( z ) = � ∞ n = 0 P n ( �J ω � ) z n Seco (UB) Approximants vs. OP CIEM 6 / 11
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