Partial Classification of Polynomials and an Orthonormal Basis Construction on the Associated Basin of Attraction
PPartial Classification of Polynomials and an Orthonormal BasisConstruction on the Associated Basin of Attraction
James TiptonDecember 27, 2018
Abstract
In the paper
Infinite product representations for kernels and iterations of functions , the authorsassociate certain Fatou subsets with reproducing kernel Hilbert spaces. They also present a method forconstructing an orthonormal basis for said Hilbert space, but the method depends on the polynomialof the given Fatou set. We provide a partial classification of those polynomials the method applies to.
Complex Dynamics
Recall that R : C → C has an attracting fixed point at z ∈ C if | R (cid:48) ( c ) | <
1. The point z is calledan attracting fixed point because all points within a certain neighborhood of z are “attracted” to z under repeated iteration of R . The n th iterate of R is denoted by R ◦ n ( z ) = R ◦ R ◦ · · · ◦ R ( z ) (cid:124) (cid:123)(cid:122) (cid:125) n times The basin of attraction of R at the attracting fixed point z is the following subset of C : B R,z = { z ∈ C : lim n →∞ R ◦ n ( z ) = z } For many polynomials with an attracting fixed point, the basin of attraction is a fractal.
Reproducing Kernel Hilbert Spaces
A reproducing kernel Hilbert space (RKHS) on C is a Hilbert space of functions on C in whichevery linear evaluation functional is bounded. Uniquely associated to each RKHS is a kernel function K : C × C → C with the reproducing property: (cid:104) f ( z ) , K ( z, w ) (cid:105) H = f ( w )Since a RKHS is, in particular, a Hilbert space, it must have an orthonormal basis (ONB). AlthoughONBs are guaranteed to exist, explicitly constructing an ONB is a much harder task. a r X i v : . [ m a t h . F A ] D ec INTRODUCTION
Kernel Functions on Basins of Attraction If R satisfies sufficient conditions, then one may construct a kernel function, represented as an infiniteproduct, on a subset of B R,z . See [1] for the general result. In particular, if R is a polynomial and z = 0, then the map K : C × C → C defined by K ( z, w ) = ∞ (cid:89) n =0 (cid:16) R ◦ n ( z ) R ◦ n ( w ) (cid:17) (1)is a kernel function on all of B R, [3]. The infinite product involves iterates of the map R and themap 1 + zw , which is a kernel function on C . The kernel function 1 + zw can be used to construct anONB under certain circumstances. The ONB Construction
First we take a moment to recall multi-index notation. Suppose J is an index set, then J ∞ = { v : v ∈ J N for some N = 1 , , . . . } Denote the RKHS associated to the previous kernel function (1) by H . The constant function ( z ) = 1plays a crucial role in the construction, and in fact belongs to H . Consider a family of operators on H , { S i : H → H} . For each v = ( v , v , . . . , v N ) ∈ J N , define b v : C → C b v ( z ) = ( S v S v · · · S v N )( z )The next theorem, due to the authors of [1], gives sufficient conditions for the functions b v to form anONB. Theorem 1.
If a family of operators { S i : H → H} Ni =1 satisfies the Cuntz relations: S ∗ i S j = δ ij I, N (cid:88) i =1 S i S ∗ i = I then B = { b v : v ∈ J ∞ } is an ONB for H . In our particular set-up, the family we wish to consider is F = { S , S } where S f ( z ) = f ( R ( z ))and S f ( z ) = zf ( R ( z )). This family of operators can be shown to satisfy the Cuntz relations whencertain conditions are met, which we discuss now. The Dagger Conditions
The family F of interest depends on the map R that is chosen. It can be shown that if R satisfies forall z ∈ B R, , M ( z ) < ∞ where M ( z ) is the number of solutions to R ( ζ ) = z , counting multiplicity, and either1 M ( z ) (cid:88) R ( ζ )= z e i ( ζ ) e j ( ζ ) = δ ij , ∀ i, j ∈ J ( † )or1 M ( z ) (cid:88) R ( ζ )= z e i ( ζ ) e j ( ζ ) = δ ij , ∀ i, j ∈ J, ( ‡ ) PARTIAL CLASSIFICATION OF THE DAGGER CONDITIONS then F satisfies the Cuntz relations [1]. The functions, e i ( z ), are taken from any ONB for the RKHSassociated to the underlying kernel function of the infinite product kernel function. For the family F that we are interested in, we have that e ( z ) = 1 and e ( z ) = z ; this comes from the underlying kernelfunction 1 + zw mentioned earlier. For ease of exposition we will refer to the above conditions as thedagger conditions. A natural question is when does the map R satisfy either of the above conditions?In the context of the underlying kernel function 1 + zw , the † condition becomes (cid:88) R ( ζ )= z M ( z ) ( † ) (cid:88) R ( ζ )= z ζ = 0 = (cid:88) R ( ζ )= z ζ ( † ) (cid:88) R ( ζ )= z | ζ | = M ( z ) , ( † )and the ‡ condition becomes (cid:88) R ( ζ )= z M ( z ) ( ‡ ) (cid:88) R ( ζ )= z ζ = 0 ( ‡ ) (cid:88) R ( ζ )= z ζ = M ( z ) ( ‡ )We examine which polynomials R satisfy the dagger conditions and offer a classification for R tosatisfy the ‡ condition. The purpose of the dagger conditions is to construct an ONB for the RKHS corresponding to thekernel function on B R, . Thus our interest lies only with those polynomials with an attracting fixedpoint at 0, even though the dagger conditions do not require R to have such a property. The first twocases of either dagger condition is quite easily characterized. Proposition 2. If P ( z ) is a degree n polynomial with an attracting fixed point at then the followinghold:a) P ( z ) satisfies † and ‡ .b) P ( z ) satisfies † and ‡ if and only if a n − = 0 .Proof. a) Since B P, is completely invariant with respect to P , we know that if P ( ζ ) = z for some z ∈ B P, ,then we must have that ζ ∈ Ω. By the Fundamental Theorem of Algebra there are n solutionsto P ( ζ ) = z , counting multiplicity. Thus (cid:80) P ( ζ )= z n = M ( z ).b) We have that P ( z ) satisfies † and ‡ if and only if (cid:80) P ( ζ )= z ζ = 0 = (cid:80) P ( ζ )= z ζ which by Vieta’sformulas is equivalent to a n − = 0. PARTIAL CLASSIFICATION OF THE DAGGER CONDITIONS
Note:
The conditions † and ‡ are equivalent to each other since (cid:80) P ( ζ )= z ζ = (cid:80) P ( ζ )= z ζ The next proposition will complete our characterization of ‡ . Proposition 3.
A polynomial P of degree n ≥ , with an attracting fixed point at , satisfies ‡ ifand only if a n − = − na n Proof.
The polynomial P satisfies ‡ if and only if (cid:88) R ( ζ )= z ζ = n which is equivalent to a n − = − na n , since Vieta’s formulas and the Newton-Girard formulas give n = (cid:88) R ( ζ )= z ζ = − e = − a n − a n Combining the last two propositions, we may characterize those polynomials satisfying the ‡ con-dition: Theorem 4.
Suppose P is a polynomial of degree n ≥ with an attracting fixed point at . Thepolynomial P satisfies ‡ if and only if P ( z ) = n (cid:80) k =1 a k z k where a n − = 0 , and a n − = − na n . Note:
Since P has an attracting fixed point at 0, we have also that | a | < † condition seems to be more challenging. It is easy to find polynomials whichsatisfy † at a particular point. But the dagger conditions are required to hold for all z ∈ B P, . As suchwe introduce “partial” conditions, † c and ‡ c , each meaning that the corresponding set of equationshold precisely at the point c ∈ C . The next proposition will help us determine which polynomialsmight satisfy † . Proposition 5.
Suppose c ∈ B P, .a) If P ( z ) satisfies ‡ c , then P ( z ) satisfies ‡ .b) If P ( z ) satisfies both † c and ‡ c then the equation P ( z ) = c has only real solutions.Proof. a) If P ( z ) satisfies ‡ c , then we must have that (cid:88) P ( ζ )= c n, (cid:88) P ( ζ )= c ζ = 0 , (cid:88) P ( ζ )= c ζ = n By Vieta’s formulas, we have that a n − = 0. Thus we have that ‡ and ‡ hold by an applicationof Proposition 1. By an application of the Newton-Girard formulas we have that n = (cid:88) P ( ζ )= c ζ = − e = − a n − a n Applying the same formula again we find that (cid:88) P ( ζ )= z ζ = − a n − a n = n So P actually satisfies ‡ . PARTIAL CLASSIFICATION OF THE DAGGER CONDITIONS b) Suppose that P ( z ) satisfies both † c and ‡ c , so in particular, we have that (cid:88) R ( ζ )= c ζ = (cid:88) R ( ζ )= c | ζ | which requires that (cid:80) R ( ζ )= c Im ζ = 0. Thus we have that (cid:88) R ( ζ )= c Re ζ = (cid:88) R ( ζ )= c | ζ | If ζ is not real, then | ζ | > Re ζ , so the above equality holds only if all solutions to P ( ζ ) = c are real.So if P ( z ) has a non-real zero, then P ( z ) cannot satisfy both † and ‡ . The only interesting polyno-mials which might satisfy † are those P ( z ) which have the property that ‡ c is not satisfied for any c ∈ Ω.Such a polynomial requires the property that P ( z ) + c has at least one non-real zero for all c ∈ Ω.An example of a polynomial with this property is any cubic of the form P ( z ) = az + bz + c with b ascalar multiple of a . We show now that no polynomial can satisfy both † and ‡ . This fact can be usedin turn to show that the aforementioned property is a necessary condition for a polynomial to satisfy † . Proposition 6. If P is a polynomial with an attracting fixed point at , then P cannot satisfy both † and ‡ .Proof. Suppose to the contrary that P satisfies both † and ‡ . By Proposition 5, P ( z ) = c has only realsolutions, for any c ∈ B P, . Thus the inverse image of B P, under P must be a subset of R . However, B P, is an open set, and P is a continuous map, so that the inverse image of B P, under P must bean open set. But no subset of R is open as a subset of C . Thus it cannot be that P satisfies both † and ‡ .We can now state a necessary condition for a polynomial to satisfy † . Proposition 7. If P is a polynomial with an attracting fixed point at that satisfies † , then for all c ∈ B R, , the equation P ( z ) = c has at least one non-real solution.Proof. Suppose that P satisfies † , and to the contrary, that P ( z ) = c has only real solutions. If ζ issuch a solution, then | ζ | = ζ . Since P satisfies † , we have also that, by the previous observation,that P satisfies ‡ . By Proposition 2, P also satisfies ‡ and ‡ , so that P must satisfy ‡ . But P cannot satisfy both † and ‡ , so that P ( z ) = c must have at least one non-real solution.Another approach to showing that a polynomial satisfies one of the dagger conditions is to write itas a product of two polynomials, each of which satisfy the same dagger condition. This is equivalentto determining whether the product of two polynomials, both satisfying the same dagger condition,will satisfy a dagger condition. This works quite well for the ‡ condition. Proposition 8. If R ( z ) and Q ( z ) satisfy ‡ , then R ( z ) Q ( z ) satisfies ‡ .Proof. Suppose that R and Q both satisfy ‡ and let S ( z ) = R ( z ) Q ( z ). Let a i denote the coefficients of R , b i denote the coefficients of Q , and suppose deg R = r , deg Q = q so that deg S = r + q . Theorem 4tells us that a r − = − ra r b q − = − rb q a r − = 0 = b q − . If c i denotes the coefficients of S , thenwe have that c r + q = a r b q , c r + q − = 0, and c r + q − = − ( r + q ) c r + q S satisfies the ‡ condition. EXAMPLES
Figure 1: The domain of K(z,w): B R, Here we present an example of a polynomial satisfying ‡ , and an example of a polynomial satisfying † . Example 9.
Consider the polynomial R ( z ) = iz − iz − i z . We see that is an attracting fixedpoint of R since R (0) = 0 and | R (cid:48) (0) | = √ < . So we have that the map K ( z, w ) = ∞ (cid:89) n =0 (cid:16) R ◦ n ( z ) R ◦ n ( w ) (cid:17) is a kernel function on B R, . The polynomial R has coefficients: a = i , a = − i , a = − i , and a = a = 0 . Since a = 0 and a = − a , Theorem 4 tells us that R satisfies ‡ . This in turn showsthat the operators S and S , defined by: S f ( z ) = f ( R ( z )) and S f ( z ) = zf ( R ( z )) satisfy the Cuntz relations. So we may apply Theorem 1 to conclude that the functions b v ( z ) form anONB for the Hilbert space associated to K . Recall that b v ( z ) = ( S v S v · · · S v N )( z ) where v ∈ J ∞ . The first few basis elements are: , z, R ( z ) , zR ( z ) , R ◦ ( z ) , zR ◦ ( z ) , R ( z ) R ◦ ( z ) , zR ( z ) R ◦ ( z ) , . . . So the basis elements may be calculated recursively, but obtaining a general formula appears to requirea general formula for R ◦ n . Example 10.
Consider the polynomial Q ( z ) = z + z which also has an attracting fixed point at . So the map K ( z, w ) = ∞ (cid:89) n =0 (cid:16) Q ◦ n ( z ) Q ◦ n ( w ) (cid:17) is a kernel function on B Q, . There is a RKHS associated to K , however, we can’t use the daggerconditions to construct an ONB. By Theorem 4, Q doesn’t satisfy ‡ , in particular, the condition ‡ .It turns out that Q does satisfy † ; this follows from Proposition 2 and the following observation: (cid:88) Q ( ζ )=0 | ζ | = (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) i (cid:112) / (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) − i (cid:112) / (cid:12)(cid:12)(cid:12) = 3 APPENDIX: NEWTON-GIRARD IDENTITIES AND THE VIETA FORMULA
Figure 2: The domain of K(z,w): B Q, It can be shown in a similar fashion that Q satisfies † c for c = i √ . However Q doesn’t satisfy † c forall c ∈ B Q, : for c = i ∈ B Q, and with the aid of WolframAlpha, we have that (cid:88) Q ( ζ )= i | ζ | > Here we discuss some open questions pertaining to this paper.1.
Classify the † condition . The main issue lies with † . Since the sum involves the modulus ofthe roots, Vieta’s formula may not be applied.2. Find a polynomial that satisfies † or show that no polynomial satifies † . Just havingone example would be a nice starting point; but so would knowing that no examples exist.3. What if we don’t count multiplicity?
Much of the theory presented in [1] should still workif we don’t count the multiplicity of the solutions to P ( z ) = c . The biggest issue with this changewould be in the application of Vieta’s formula, since it does use multiplicity.4. Classify polynomials satisfying † . This could be another starting point for classifying the † condition. Understanding when † is satisfied could help to understand when † c is satisfied.5. Generalize the results presented here to other “underlying” kernel functions . Thereare other underlying kernel functions one could use other than 1 + zw . However, changing theunderlying kernel function will change the dagger conditions. There are some kernel functionsfor which the approach presented here might still work, in particular, kernel functions of theform 1 + ( zw ) n , where n is a positive integer. Here we take a brief look at the Newton-Girard identities and Vieta’s formula; both quintessentialtools in this paper. See [2] for a more in depth historical introduction.
Theorem 11 (Vieta’s formula) . Suppose P ( z ) = n (cid:80) j =0 a j z j satisfies a n (cid:54) = 0 . If z , . . . , z n are the roots(counting multiplicity) of P , then (cid:88) ≤ j < ··· Proof. By assumption we have that n (cid:88) j =0 a j z j = a n n (cid:89) j =1 ( z − z j ) = a n z n + a n n (cid:88) k =1 (cid:16) ( − k (cid:88) ≤ j < ··· From the proof of Vieta’s formula we have n (cid:89) j =1 ( z − z j ) = z n + n (cid:88) l =1 (cid:16) ( − l (cid:88) ≤ j < ··· References [1] Alpay, D., Jorgensen, P., Lewkowicz, I., & Martziano, I. (2015). Infinite product representationsfor kernels and iterations of functions. In Recent advances in inverse scattering, Schur analysisand stochastic processes (pp. 67-87). Birkhuser, Cham.[2] Funkhouser, H. G. (1930). A short account of the history of symmetric functions of roots ofequations. The American mathematical monthly, 37(7), 357-365.[3] Tipton, J. E. (2016). Reproducing kernel Hilbert spaces and complex dynamics. The University ofIowa.(pp. 67-87). Birkhuser, Cham.[2] Funkhouser, H. G. (1930). A short account of the history of symmetric functions of roots ofequations. The American mathematical monthly, 37(7), 357-365.[3] Tipton, J. E. (2016). Reproducing kernel Hilbert spaces and complex dynamics. The University ofIowa.