Zacytuj

Introduction

Weighted shift operators have been studied by many authors in different contexts, for instance the work by N. K. Nikol’skiĭ in the spaces spaces p, [1518], R. Gellar in Banach spaces, [24] and Grabiner in Banach algebras of power series spaces, [57].

Forward weighted operators (multiplication and integration operators) play a remarkable role in the study of bases in spaces of analytic functions and have been considered by many Russian mathematicians, [11]. The Gončarov polynomials, that under certain conditions are a basis in analytic spaces [1, 9], are related to the backward weighted operator (derivation operator).

We work with Köthe spaces and weighted shifts on them (generalized integration and derivation operators). We characterize the forward shift-invariant isomorphisms and then determine some some quasi-power bases. Our results include, as particular cases, those of Nagnibida for the multiplication and integration operators on the space of analytic functions on a disc, [11] and Prada for the multiplication operator on infinite power series spaces, [19, 20]. Using the backward shift operator we get conditions for the Gončarov polynomials to be a basis.

Basic results

Denote by λp(A), 1 ≤ p < ∞, the Köthe (echelon) space given by the matrix A=(ank)n=0$\begin{array}{} \displaystyle A=(a_{n}^{k})_{n=0}^{\infty } \end{array}$, 0<ankank+1$\begin{array}{} \displaystyle 0<a_{n}^{k}\leq a_{n}^{k+1} \end{array}$ for all n, k, that is

λp(A)={x=(xn)n=0,xn:n=0(|xn|ank)p<,k=0,1,2,}.$$\begin{array}{} \displaystyle \lambda^{p}(A)=\left\{ x=(x_{n})_{n=0}^{\infty },\, x_{n}\in \mathbb{C}:\sum_{n=0}^\infty \left( \left| x_{n} \right| a_{n}^{k}\right)^{p} <\infty,\, \forall k=0,1,2,\dots\right\}. \end{array}$$

λ p(A) is a Frèchet space, [14], with the norms

xk=(n=0(|xn|ank)p)1/p,k=0,1,2,$$\begin{array}{} \displaystyle \|x\|_{k}=\left(\sum_{n=0}^\infty \left(\left| x_{n}\right| a_{n}^{k}\right)^{p}\right)^{1/p}, \quad k=0,1,2,\dots \end{array}$$

When p = 0, ∞, we have

λ0(A) ={x=(xn),xn:xk=sup(|xn|ank)<,k=0,1,2,}.λ(A) ={x=(xn),xn:xk=lim(|xn|ank) =0,k=0,1,2,}.$$\begin{array}{} \displaystyle \begin{aligned} \lambda^{0}(A) &=\left\{x= (x_{n}), x_n\in\mathbb{C}:\left\Vert x\right\Vert _{k}=\sup \left( \left\vert x_{n}\right\vert a_{n}^{k}\right) <\infty, \; \forall k=0,1,2,\dots\right\}. \\ \lambda^{\infty }(A) &=\left\{x= (x_{n}), x_n\in\mathbb{C}:\left\Vert x\right\Vert _{k}=\lim \left( \left\vert x_{n}\right\vert a_{n}^{k}\right) \text{ }=0,\; \forall k=0,1,2,\dots\right\}. \end{aligned} \end{array}$$

The canonical basis in the spaces λp(A), p = 0, 1 ≤ p < ∞, is denoted by δn=(δn,k)k=0$\begin{array}{} \displaystyle \delta _{n}=(\delta_{n,k})_{k=0}^{\infty } \end{array}$, where dn,k is the Kronecker delta.

The dual space of λp(A), 1 ≤ p < ∞, p = 0, 1p+1q=1$\begin{array}{} \displaystyle \frac{1}{p}+\frac{1}{q}=1 \end{array}$ is given by

(λp(A))×={(xn)n=0,xn:(n=0|xn|q(ank)q)1q<, for a suitable k},1<p<.(λ1(A))×={(xn)n=0,xn:supn0{|xn|ank}<, for a suitable k},p=1.(λ0(A))×={(xn)n=0,xn:n=0|xn|ank<, for a suitable k},p=0.$$\begin{array}{} \displaystyle \begin{aligned} \left( \lambda^{p}(A)\right) ^{\times} & =\left\{(x_n)_{n=0}^\infty,x_n\in\mathbb{C}: \left(\sum_{n=0}^\infty \frac{|x_n|^q}{(a_n^k)^q}\right)^{\frac{1}{q}}<\infty, \text{ for a suitable }k \right\}, \ 1<p<\infty.\\\\ \left( \lambda^{1}(A)\right) ^{\times}& =\left\{(x_n)_{n=0}^\infty,x_n\in\mathbb{C}: \sup_{n\ge0}\left\{ \frac{|x_n|}{a_n^k}\right\}<\infty, \text{ for a suitable }k \right\}, \ p=1.\\\\ \left( \lambda^{0}(A)\right) ^{\times} & =\left\{(x_n)_{n=0}^\infty,x_n\in\mathbb{C}: \sum_{n=0}^\infty \frac{|x_n|}{a_n^k}<\infty, \text{ for a suitable } k\right\},\ p=0. \end{aligned} \end{array}$$

Recall that the coordinate operators are continuous, [14].

λp(A), p ∈ [1, ∞), p = 0 is the projective limit of the Banach spaces p(ak), c0(ak), diagonal transformations of p, c0, with the usual topology:

p(ak)={x=(xn)n=0:(xnank)n=0p},1p<c0(ak)={x=(xn)n=0:(xnank)n=0c0}.$$\begin{array}{} \displaystyle \begin{aligned} \ell^p(a^k)&=\left\{x=(x_n)_{n=0}^\infty: \, (x_n \, a_{n}^{k})_{n=0}^\infty\in\ell^p\right\},\quad 1\le p<\infty \\ c_0(a^k)&=\left\{x=(x_n)_{n=0}^\infty: \, (x_n \, a_{n}^{k})_{n=0}^\infty\in c_0\right\}. \end{aligned} \end{array}$$

The space 1(ak) is a Banach algebra if and only if the following condition holds

C(k)>0:am+nkC(k)amkank,n,m=0,1,2,$$\begin{array}{} \displaystyle \exists\,C(k)>0:\,a_{m+n}^{k}\leq C(k)a_{m}^{k}a_{n}^{k},\quad\forall n,m=0,1,2,\dots \end{array}$$

The space λ1(A) is nuclear if and only if

k,r(k):ankanr(k)1$$\begin{array}{} \displaystyle \forall k,\,\exists r(k):\,\frac{a_{n}^{k}}{a_{n}^{r(k)}}\in \ell^{1} \end{array}$$

and therefore λp(A) = λ1(A) = λ0(A), 1 ≤ p < ∞ [14].

If λ = (λn) is a sequence of nonzero complex numbers with λ0 = 1 to simplify computations, the operator Jλ defined by

Jλ(δn)=λn+1λnδn+1$$\begin{array}{} \displaystyle J_{\lambda}(\delta _{n})=\frac{\lambda_{n+1}}{\lambda_{n}}\delta _{n+1} \end{array}$$

is called the generalized integration operator. If λn=1n!$\begin{array}{} \displaystyle \lambda_{n}=\frac{1}{n!} \end{array}$, Jλ = J is the integration operator and if λn = 1, Jλ = U, is the multiplication one (shift operator), see [11].

We assume that the operator Jλ, where the sequence (λn) are positive real numbers without lost of generality, is continuous on λp(A), that is the following condition is fulfilled

k,r=r(k):supn0(λn+1λnank+1ank)<$$\begin{array}{} \displaystyle \forall k,\,\exists r=r(k)\,:\,\sup\limits_{n\ge 0}\left( \frac{\lambda _{n+1}}{\lambda_{n}}\frac{a_{n}^{k+1}}{a_{n}^{k}}\right) <\infty \end{array}$$

If (dn) is a sequence of positive real numbers, the operator D given by

D(δn)=dn1dnδn1$$\begin{array}{} \displaystyle D(\delta _{n})=\frac{d_{n-1}}{d_{n}}\delta _{n-1} \end{array}$$

is called the generalized derivation operator, being the usual derivation when dn=1n!$\begin{array}{} \displaystyle d_{n}=\frac{1}{n!} \end{array}$.

Isomorphisms commuting with Jλ. Bases in Köthe spaces

We characterize the isomorphisms between Köthe spaces that commute with the generalized integration operator Jλ determining some bases, related with it, on λ1(A).

Theorem 1

Let T : λ1(A) → λ1(A) be a continuous linear operator.{1λnTnx}n=0$\begin{array}{} \displaystyle \left\{ \frac{1}{\lambda_{n}}T^{n}x\right\}_{n=0}^{\infty} \end{array}$, xλ1(A) is a basis in λ1(A) if and only if there exists an isomorphism S : λ1(A) → λ1(A) such that T ∘ S = S ∘ Jλ and x = 0.

Proof. If {1λnTnx}$\begin{array}{} \displaystyle \left\{ \frac{1}{\lambda_{n}}T^{n}x\right\} \end{array}$, n ≥ 0, is a basis in λ1(A), then there exists an isomorphism S such that Sδn=1λnTnx$\begin{array}{} \displaystyle S\delta_{n}=\frac{1}{\lambda_{n}}T^{n}x \end{array}$, n = 0, 1, 2,... It follows that 0 = x and for n ∈ ℕ

(SJλ)δn=λn+1λnSδn+1=1λn(TTnx)=(TS)δn.$$\begin{array}{} \displaystyle (S \circ {J_\lambda }){\delta _n} = \frac{{{\lambda _{n + 1}}}}{{{\lambda _n}}}S{\delta _{n + 1}} = \frac{1}{{{\lambda _n}}}(T \circ {T^n}x) = (T \circ S){\delta _n}. \end{array}$$

Conversely

Tnx=(Tn1TS)δ0=(Tn1SJλ)δ0=(SJλn)δ0=λnSδn.$$\begin{array}{} \displaystyle T^{n}x=(T^{n-1}TS)\delta _{0}=(T^{n-1}SJ_{\lambda})\delta _{0}=(SJ_{\lambda}^{n})\delta _{0}=\lambda_{n}S\delta _{n}. \end{array}$$

Corollary 2

{1λnJλnx}$\begin{array}{} \displaystyle \left\{ \frac{1}{\lambda_{n}}J_{\lambda}^{n} x\right\} \end{array}$, xλ1(A) is a basis in λ1(A) if and only if there exists an isomorphism T : λ1(A) → λ1(A) that commutes with Jλ and x = T δ0.

Proposition 3

[13] A linear operator T : λ1(A) → λ1(A) is continuous and commutes with Jλ if and only if

T=m=0bmλmJλm,b=(bm)m=0=Tδ0$$\begin{array}{} \displaystyle T=\sum_{m=0}^{\infty}\frac{b_{m}}{\lambda_{m}} J_{\lambda}^{m}, \quad b=(b_{m})_{m=0}^{\infty}=T\delta _{0} \end{array}$$

and the condition

k,r=r(k):supn{m=0|bm|λm+nλmλnam+nkanr}<$$\begin{array}{} \displaystyle \forall k,\,\exists r=r(k):\,\sup_{n }\left\{\sum_{m=0}^{\infty} | b_{m}| \frac{\lambda_{m+n}}{\lambda_{m}\lambda_{n}} \frac{a_{m+n}^{k}}{a_{n}^{r}}\right\} <\infty \end{array}$$

is fulfilled.

Proposition 4

(c.f. [13]). Let T be a linear operator from λ1(A) onto itself commuting with Jλ and b = (bn) = T (δ0) (b0 ≠ 0). If T−1is the formal operator given by the inverse matrix of T , c = (cn) = T−1(δ0) and k,

k,r=r(k):supm0,n0{λm+nλmλnam+nkamkanr}<,$$\begin{array}{} \displaystyle \forall k,\,\exists r= r(k):\, \sup_{m\ge0,n\ge0} \left\{ \frac{\lambda_{m+n}}{\lambda_{m}\lambda_{n}} \frac{a_{m+n}^{k}}{a_{m}^{k}a_{n}^{r}}\right\} <\infty, \end{array}$$

then T is an isomorphism if and only if b, cλ1(A).

Remark 1

Recall that the matrix (ti,j) of a continuous linear operator T commuting with Jλ is lower triangular so, formally, (ti,j) has an inverse of the same type if T δ0 = (bn) with b0 ≠ 0. The operator T−1 given by this inverse matrix is always linear and commutes with Jλ . Then a continuous operator T is an isomorphism if and only if T−1 is continuous and T−1 can be written

T1=n=0cnλnJλn,c=(cn)=T1(δ0).$$\begin{array}{} \displaystyle T^{-1}=\sum_{n=0}^{\infty}\frac{c_{n}}{\lambda_{n}}J_{\lambda}^{n}, \quad c=(c_{n})=T^{-1}(\delta _{0}). \end{array}$$

Theorem 5

(c.f. [13]). Assume the following conditions:

am+nkCkamkank$\begin{array}{} \displaystyle a_{m+n}^{k}\leq C_{k}a_{m}^{k}a_{n}^{k} \end{array}$, ∀k, that is, the spaces ℓ1(ak) are Banach algebras.

λm+nmλn, ∀m, n.

LetT=n=0bnλnJλn$\begin{array}{} \displaystyle T=\sum\limits_{n=0}^{\infty}\frac{b_{n}}{\lambda_{n}}J_{\lambda}^{n} \end{array}$be a linear operator on λ1(A) commuting with Jλ.

Then T is an isomorphism if and only if any of the following equivalent conditions are satisfied:

The sequence(bnλn)$\begin{array}{} \displaystyle (\frac{b_{n}}{\lambda_{n}}) \end{array}$is an exponential (invertible) element of all the Banach algebras ℓ1(bk), bnk=λnank$\begin{array}{} \displaystyle b_{n}^{k}=\lambda_{n}a_{n}^{k} \end{array}$, for all k.

The sequence (bn) ∈ λ1(A) and

n=0bnλnzn0,|z|rk,rk=limn(λnank)1nfor allk.$$\begin{array}{} \displaystyle \mathop {\sum }\limits^\infty_{n = 0} \frac{{{b_n}}}{{{\lambda _n}}}{z^n} \ne 0,\quad |z| \le {r_k},\,\,{r_k} = \mathop {\lim }\limits_n {({\lambda _n}a_n^k)^{\frac{1}{n}}}\,\,forall\,k. \end{array}$$

Corollary 6

(c.f. [13]). Let T be a linear operator commuting with Jλ

T=n=0bnλnJλn,b=(bn)=Tδ0,b00.$$\begin{array}{} \displaystyle T=\sum_{n=0}^\infty\frac{b_{n}}{\lambda_{n}} J_{\lambda}^{n},\,b=(b_{n})=T\delta _{0},\,b_{0}\neq 0. \end{array}$$

Suppose the following conditions are satisfied:

k,Ck>0:am+nkCkamkank,λm+n Cλmλn,m,n.$$\begin{array}{} \displaystyle \begin{aligned} \forall k,\,\exists C_{k}>0:\,\ a_{m+n}^{k}&\leq C_{k}a_{m}^{k}a_{n}^{k}, \\ \lambda_{m+n} &\le C\lambda _{m}\lambda_{n,}\, \forall m,n. \end{aligned} \end{array}$$

If(bnλn)$\begin{array}{} \displaystyle \left( \frac{b_{n}}{\lambda_{n}}\right) \end{array}$is an exponential (invertible) element of λ1(B), B=(bnk)=(λnank)$\begin{array}{} \displaystyle B=\left( b_{n}^{k}\right) =(\lambda_{n}a_{n}^{k}) \end{array}$, then the system

{λnTn(bjλj)j=0}n=0$$\begin{array}{} \displaystyle \left\{ \lambda_{n}T^{n}\left( \frac{b_{j}}{\lambda_{j}}\right) _{j=0}^{\infty }\right\} _{n=0}^{\infty } \end{array}$$

is a basis in λ1(A).

Proposition 7

[13] Let T be a linear operator on λ1(A) commuting with Jλ, T=n=0bnλnJλn$\begin{array}{} \displaystyle T=\sum\limits_{n=0}^{\infty}\,\frac{b_{n}}{\lambda_{n}} J_{\lambda}^{n} \end{array}$, b0 ≠ 0.

If there existsMk=limnλn+1λnan+1kank$\begin{array}{} \displaystyle M_{k}=\lim\limits_{n\to\infty}\frac{\lambda_{n+1}}{\lambda_{n}} \frac{a_{n+1}^{k}}{a_{n}^{k}} \end{array}$, Mk ≠ 0, for a suitable k, then the functionϕ(z)=n=0bnλnzn$\begin{array}{} \displaystyle \phi (z)=\sum\limits_{n=0}^{\infty} \frac{b_{n}}{\lambda_{n}}z^{n} \end{array}$is an holomorphic one with no zeros in a disc D(0, ρ), with ρ ≥ Mk.

Iflimnλn+1λnan+1kank=$\begin{array}{} \displaystyle \lim\limits_{n\to\infty}\frac{\lambda_{n+1}}{\lambda_{n}}\frac{a_{n+1}^{k}}{a_{n}^{k}}=\infty \end{array}$for a suitable k, then the functionϕ(z)=n=0bnλnzn$\begin{array}{} \displaystyle \phi (z)=\sum\limits_{n=0}^{\infty}\frac{b_{n}}{\lambda_{n}}z^{n} \end{array}$is an entire function without zeros.

Proposition 8

(c.f. [13]). Let T be a linear operator on λ1(A) commuting with Jλ

T=n=0bnλnJλn, b00.$$\begin{array}{} \displaystyle T=\sum_{n=0}^{\infty}\,\frac{b_{n}}{\lambda_{n}} J_{\lambda}^{n},\text{ }b_{0}\neq 0. \end{array}$$

Suppose that

k,Mk=supn{λn+1λnan+1kank}<.$$\begin{array}{} \displaystyle \forall k,\,\exists M_{k}=\sup_{n}\left\{ \frac{\lambda_{n+1}}{\lambda_{n}} \frac{a_{n+1}^{k}}{a_{n}^{k}}\right\} <\infty. \end{array}$$

If the functionϕ(z)=n=0bnλnzn$\begin{array}{} \displaystyle \phi (z)=\sum\limits_{n=0}^{\infty}\frac{b_{n}}{\lambda_{n}}z^{n} \end{array}$is holomorphic without zeros in a discDρ$\begin{array}{} \displaystyle \mathbb{D}_{\rho} \end{array}$, ρ > supk {Mk} or ρ = ∞, then T is an isomorphism from λ1(A) onto itself.

Proposition 9

(c.f. [13]). If for a suitable k,

limnλnank=,limnλn+1λnan+1kank=,limnsuplog(n+1)log(λn+1λnan+1kank)=0,$$\begin{array}{} \displaystyle \lim\limits_{n\to\infty} \lambda_{n}a_{n}^{k}=\infty,\quad \lim\limits_{n\to\infty} \frac{\lambda_{n+1}}{\lambda_{n}} \frac{a_{n+1}^{k}}{a_{n}^{k}}=\infty,\quad \lim\limits_{n\to\infty} \sup \frac{\log (n+1)}{\log (\frac{\lambda_{n+1}}{\lambda_{n}} \frac{a_{n+1}^{k}}{a_{n}^{k}})}=0, \end{array}$$

then the only entire functions without zeros that give continuous linear operators on λ1(A) are the constants.

Example 10

The space of holomorphic functions, H(DR)$\begin{array}{} \displaystyle \mathscr{H}(\mathbb{D}_R) \end{array}$, on the discDR=D(0,R)$\begin{array}{} \displaystyle \mathbb{D}_R=\mathbb{D}(0,R) \end{array}$, 0 < R ≤ ∞ is a Köthe space λ1(A), withA=(ank)=(tkn)$\begin{array}{} \displaystyle A=(a_{n}^{k})=(t_{k}^{n}) \end{array}$, where (tk) is an increasing sequence of real positive numbers converging to R.

If λn = 1, ∀n then a continuous linear operatorT=n=0bnUn$\begin{array}{} \displaystyle T=\sum \limits_{n=0}^{\infty}{b_n}U^{n} \end{array}$onH(DR)$\begin{array}{} \displaystyle \mathscr{H}(\mathbb{D}_{R}) \end{array}$, commuting with the multiplication operator U, is an isomorphism if and only if the functionϕ(z)=n=0bnznH(DR)$\begin{array}{} \displaystyle \phi (z)=\sum\limits_{n=0}^{\infty}{b_{n}}z^{n}\in\mathscr{H}(\mathbb{D}_{R}) \end{array}$and has no zeros in the discDR$\begin{array}{} \displaystyle \mathbb{D}_{R} \end{array}$, see [11].

Ifλn=1n!$\begin{array}{} \displaystyle \lambda_{n}=\frac{1}{n!} \end{array}$, ∀n then Jλ = J and a linear continuous operator T onH(DR)$\begin{array}{} \displaystyle \mathscr{H}(\mathbb{D}_{R}) \end{array}$, commuting with J, is an isomorphism if and only if the functionϕ(z)=n=0bnznH(DR)$\begin{array}{} \displaystyle \phi(z)=\sum\limits_{n=0}^\infty b_{n}z^{n} \in \mathscr{H}(\mathbb{D}_{R}) \end{array}$and b0 ≠ 0, see [11].

Example 11

The space λ1(A) = Λ(α), A = (en) with (αn) an increasing sequence of positive numbers going to infinity, is an infinite power series space.

If λn = 1, ∀n, and αm+nC + αn + αm, ∀m, n, then a continuous linear operator T on Λ(α), commuting with U, is an isomorphismif and only if the sequence T δ0 = (bn) ∈ Λ(α) and the functionϕ(z)=n=0bnzn$\begin{array}{} \displaystyle \phi (z)=\sum\limits_{n=0}^{\infty}b_nz^{n} \end{array}$has no zeros in the closed disk D(0,1) (iflimnαnn=0$\begin{array}{} \displaystyle \lim\limits_{n\to\infty} \frac{\alpha_{n}}{n}=0 \end{array}$or has no zeros in the complex plane (iflimnαnn>0$\begin{array}{} \displaystyle \lim\limits_{n\to\infty} \frac{\alpha _{n}}{n}>0 \end{array}$[20].

Ifλn=1n!$\begin{array}{} \displaystyle \lambda_{n}=\frac{1}{n!} \end{array}$, ∀n, αm+nC + αn + αm, ∀m,n, andlimnαnn<$\begin{array}{} \displaystyle \lim\limits_{n\to\infty}\frac{\alpha _{n}}{n}<\infty \end{array}$, then a continuous linear operator T is an isomorphism on Λ(α) commuting with J if and only if T δ0 = (bn) ∈ Λ(α) and b0 ≠ 0.

Example 12

The conditions of the proposition 9 are fulfilled, for instance, if λln = 1 orλn=1n!$\begin{array}{} \displaystyle \lambda_{n}=\frac{1}{n!} \end{array}$andank=enαk$\begin{array}{} \displaystyle a_{n}^{k}=e^{n^{\alpha}k} \end{array}$, α > 0.

Two continuous operators commuting with Jλ commute with each other [2] but the converse is not true. For example, take an operator given by an infinite two-block matrix

(a0,0a0,1a1,0a0,1),  a0,1,a1,00,a0,0a1,1$$\begin{array}{} \displaystyle \begin{pmatrix} a_{0,0} & a_{0,1} \\ a_{1,0} & a_{0,1} \end{pmatrix}, \qquad a_{0,1},a_{1,0}\neq 0,\quad a_{0,0}\neq a_{1,1} \end{array}$$

and the operator Jλ2$\begin{array}{} \displaystyle J_{\lambda}^{2} \end{array}$. We show that for certain spaces the result is true.

Theorem 13

Let T be a linear operator from λp(A) to λp(A), p = 0, p ∈ [1, +∞) commuting with Jλ , T=n=0bnλnJλn$\begin{array}{} \displaystyle T=\sum\limits_{n=0}^\infty \frac{b_{n}}{\lambda_{n}}J_{\lambda}^{n} \end{array}$and{λnUn(bj+1λj+1)j=0}n=0$\begin{array}{} \displaystyle \left\{ \lambda_{n}U^{n}\left( \frac{b_{j+1}}{\lambda_{j+1}}\right) _{j=0}^{\infty }\right\} _{n=0}^{\infty } \end{array}$is a basis of λ1(A). Then any continuous linear operator S on λp(A),commuting with T , commutes with Jλ.

Proof. It is similar to the proof of theorem 3.5 in [20].

Gončarov polynomials in a nuclear Köthe space

Conditions for the generalized Gončarov polynomials to be a basis in the nuclear space spaces λ1(A) are given.

Given a sequence of complex numbers (zn)n=0$\begin{array}{} \displaystyle (z_n)_{n=0}^{\infty} \end{array}$, the Gončarov polynomials Gn(z;z0,...,zn−1) are recursively defined by

G0(z)=1G1(z;z0)=zz0Gn(z;z0,,zn1)=znn!k=0n1zknk(nk)!Gk(z;z1,,zk1).$$\begin{array}{} \displaystyle \begin{aligned} G_{0}(z) &=1 \\ G_{1}(z;z_{0}) &=z-z_{0} \\ &\dots \\ G_{n}(z;z_{0},\dots,z_{n-1}) &=\frac{z^{n}}{n!}-\sum_{k=0}^{n-1}\frac{z_{k}^{n-k}}{(n-k)!} G_{k}(z;z_{1},\dots,z_{k-1}). \end{aligned} \end{array}$$

The generalized Gončarov polynomials Qn(z;z0,...,zn−1) are given by

Q0(z)=1G1(z;z0)=d1(zz0)Qn(z;z0,,zn1)=dnznk=0n1dnkzknkQk(z;z1,,zk1)$$\begin{array}{} \displaystyle \begin{aligned} Q_{0}(z) &=1 \\ G_{1}(z;z_{0}) &=d_{1}(z-z_{0}) \\ &\dots\\ Q_{n}(z;z_{0},\dots,z_{n-1}) &=d_{n}z^{n}- \sum_{k=0}^{n-1}d_{n-k}z_{k}^{n-k}Q_{k}(z;z_{1},\dots,z_{k-1}) \end{aligned} \end{array}$$

where (dn) is a sequence of positive real numbers.

Recall that if X is a locally convex space, a biorthogonal system {ei, fi}, eiX, fiX, fi(ej) = δij, is complete, if the finite linear combinations of (ei) are dense in X, see [14].

If we define the functionals Dm, Lm, m ≥ 0 on H(DR)$\begin{array}{} \displaystyle \mathscr{H}(\mathbb{D}_{R}) \end{array}$ by

Dm(f(z)) n=mxnn!(nm)!zmnmLm(f(z)) =n=mxndnmdnzmnm,f(z)=n=0xnznH(DR),$$\begin{array}{} \displaystyle \begin{aligned} D_{m}(f(z)) &\sum_{n=m}^{\infty}x_{n}\frac{n!}{(n-m)!} z_{m}^{n-m} \\ L_{m}(f(z)) &=\sum_{n=m}^{\infty}x_{n}\frac{d_{n-m}}{ d_{n}}z_{m}^{n-m} , \end{aligned}\qquad f(z) =\sum\limits_{n=0}^{\infty}x_{n}z^{n} \in \mathscr{H}(\mathbb{D}_{R}), \end{array}$$

then {Gm(z;z0,z1,,zm1);Dm}m=0$\begin{array}{} \displaystyle \left\{ G_{m}(z;z_{0},z_{1},\dots,z_{m-1});D_{m}\right\} _{m=0}^{\infty } \end{array}$ and {Qm(z;z0,z1,,zm1);Lm}m=0$\begin{array}{} \displaystyle \left\{ Q_{m}(z;z_{0},z_{1},\dots,z_{m-1});L_{m}\right\} _{m=0}^{\infty } \end{array}$ are biorthogonal systems for H(DR)$\begin{array}{} \displaystyle \mathscr{H(}\mathbb{D}_{R}) \end{array}$.

Theorem 14

(c.f. [8]). If λ1(A) is nuclear, a complete biorthogonal system, (ei, fi), fi = ( fi,j), is a Schauder basis for λ1(A) if and only ifk ∈ ℕ there exists r = r(k) ∈ ℕ such that:

supi,j(|fi,j|ajreik)<.$$\begin{array}{} \displaystyle \underset{i,j}{\sup }\left( \frac{\left\vert f_{i,j}\right\vert }{a_{j}^{r}} \left\Vert e_{i}\right\Vert _{k}\right) <\infty. \end{array}$$

Theorem 15

(c.f. [9]). Let (tk) be a sequence such that tk < tk+1andlimktk=R$\begin{array}{} \displaystyle \lim\limits_{k\to\infty}t_{k}=R \end{array}$, 0 < R ≤ ∞. The Gončarov polynomials Gn(z; z0,...,zn−1) are a Schauder basis inH(DR)$\begin{array}{} \displaystyle \mathscr{H}(\mathbb{D}_{R}) \end{array}$, if and only ifk ∈ ℕ, there exists r = r(k) such that

supn0supmn{m!|zn|mn(mn)!(tr)mj=0n(tk)jj!|Gnj(0;zj,,zn1)|}<.$$\begin{array}{} \displaystyle \sup_{n\ge0}\sup_{m\ge n}\left\{ \frac{m!| z_{n}|^{m-n}}{(m-n)!(t_{r})^{m}}\sum_{j=0}^{n}\frac{(t_{k})^{j}}{j!}\left\vert G_{n-j}(0;z_{j},\dots,z_{n-1})\right\vert \right\}<\infty . \end{array}$$

Theorem 16

(c.f. [10]). The generalized Gončarov polynomials Qn(z; z0,...,zn−1) are a basis inH(DR)$\begin{array}{} \displaystyle \mathscr{H}(\mathbb{D}_{R}) \end{array}$, 0 < R ≤ ∞, if and only ifk ∈ ℕ, ∃r = r(k) such that

supn0supmn{dmndm(tr)m|zn|mnj=0ndj(tk)j|Qnj(0;zj,,zn1)|}<.$$\begin{array}{} \displaystyle \sup_{n\ge0}\sup_{m\ge n}\left\{ \frac{d_{m-n}}{d_{m}(t_{r})^{m}}\left\vert z_{n}\right\vert ^{m-n}\sum_{j=0}^{n}d_{j}(t_{k})^{j}\,\left\vert Q_{n-j}(0;z_{j},\dots,z_{n-1})\right\vert \right\} <\infty. \end{array}$$

The generalized Gončarov polynomials {Qn(z;z0,,zn1)}n=0$\begin{array}{} \displaystyle \left\{Q_{n}(z;z_{0},\dots,z_{n-1})\right\}_{n=0}^{\infty } \end{array}$ are a complete system in a nuclear space λ1(A) and Ln ∈ (λ1(A)) if and only if

supmn(dmndmamr|zn|mn)<.$$\begin{array}{} \displaystyle \sup_{m\ge n}\left( \frac{d_{m-n}}{d_{m}a_{m}^{r}}\left\vert z_{n}\right\vert ^{m-n} \right)<\infty. \end{array}$$

Proposition 17

If λ1(A) is a nuclear space, the generalized Gončarov polynomials{Qn(z;z0,,zn1)}n=0$\begin{array}{} \displaystyle \left\{ Q_{n}(z;z_{0},\dots,z_{n-1})\right\}_{n=0}^{\infty } \end{array}$are a basis in λ1(A) if and only ifk ∈ ℕ, ∃r = r(k) ∈ ℕ such that:

supn0{supmn(dmndmarm|znmn|)j=0|Qnj(0;zj,,zn1)djajk|}<.$$\begin{array}{} \displaystyle \sup_{n\ge 0}\left\{ \sup_{m\ge n}\left( \frac{d_{m-n}}{d_{m}a_{r}^{m}}\left\vert z_{n}^{m-n}\right\vert \right) \sum_{j=0}^{\infty}\left\vert Q_{n-j}(0;z_{j},\dots,z_{n-1})d_{j}a_{j}^{k}\right\vert \right\} <\infty. \end{array}$$

Proof. Follows easily from Theorem 14.

eISSN:
2444-8656
Język:
Angielski
Częstotliwość wydawania:
Volume Open
Dziedziny czasopisma:
Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics