平移算子的一致收敛性

IF 0.5 Q3 MATHEMATICS Advances in Pure and Applied Mathematics Pub Date : 2022-02-26 DOI:10.4236/apm.2022.1212054
N. Tsirivas
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引用次数: 0

摘要

设θ为固定正数,θ∈(0,1),设λ = (λn)n∈n为固定的非零复数序列,使λn→∞。我们将对每个(t, z)∈[0,θ]×C应用函数gn: [0, θ]×C→C,定义为gn((t, z)) = z + λne。我们将考虑在[0,θ]×C上连续函数的空间C([0, θ]×C)具有紧网络上一致收敛的拓扑结构,并设ρ为C([0, θ]×C)中的通常度规。对于一个完整的函数f∈H(C),我们将表示f′:[0,θ]× C→C, f′(t, z)) = f(z)对于每一个(t, z)∈[0,θ]× C。我们将证明方程:lim n→+∞ρ((x◦gyn, f′))= 0不存在任何解(x, yn),其中x∈H(C)和yn是一个严格递增的自然数子序列,f∈H(C)是一个给定的非常数完整函数。当f是一个常数整函数时,根据G. Costakis给出的结果,上式有无穷多个解。
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Uniform Convergence of Translation Operators
Let θ be a fixed positive number, θ ∈ (0, 1) and let λ = (λn)n∈N be a fixed sequence of non-zero complex numbers, so that λn→∞. We shall apply the functions gn : [0, θ]× C→C, defined as gn((t, z)) = z + λne for each (t, z) ∈ [0, θ]× C. We shall consider the space C([0, θ]× C) of continuous functions on [0, θ]× C, as endowed with the topology of uniform convergence on compacta and let ρ be the usual metric in C([0, θ]×C). For an entire function f ∈ H(C) we shall denote that f̄ : [0, θ]× C→C, f̄((t, z)) = f(z) for every (t, z) ∈ [0, θ]× C. We will prove that the equation: lim n→+∞ ρ((x ◦ gyn , f̄)) = 0 does not have any solution (x, yn) where x ∈ H(C) and yn is an strictly increasing subsequence of natural numbers and f ∈ H(C) is a given non-constant entire function. When f is a constant entire function, then the above equation has infinitely several solutions, according to a result provided by G. Costakis.
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来源期刊
CiteScore
0.70
自引率
0.00%
发文量
12
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