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Let $r,x\in \mathbb{R}$ , $0<r<1$ and $A_{n,r}(f)=\sum_{k=-n}^{n}r^{|k|}e^{ikx}f(r^k)$, for $f\in C(0,\infty)$, $n\in \mathbb{N}\cup \{\infty\}$. Еxamine whether the $A_{n,r}$ bounded linear functional. If there is (uniformly) $\lim_{n \to \infty}A_{n,r}$ and $s-\lim_{n \to \infty}A_{n,r}$ for $r\in (0,1)$?

I can prove that $A_{n.r}$ are bounded linear functionals, but I can't do anything else. (Sorry for my bad English.)

Laki888
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1 Answers1

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The functionals converge in the norm, because of the estimate $$ |r^{|k|}e^{ikx}f(r^k)| \le r^{|k|} \|f\|_{C(0,\infty)}$$ where the right hand side is a convergent geometric series.

Post No Bulls
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  • $|A_{n,r}(f)|=|\sum_{k=-n}^{n}r^{|k|}e^{ikx}f(r^k)|\leq \sum_{k=-n}^{n}r^{|k|}|e^{ikx}||f(r^k)|\leq \sum_{k=-n}^{n}r^{|k|}||f||C \leq ||f||_C \cdot(\frac{1+r}{1-r})$Can it be concluded from this uniform boundedness of $A{n,r}$? – Laki888 Dec 25 '13 at 07:48
  • @Laki888 Yes, because the number $r$ is fixed: only $n$ changes. But uniform boundedness is not enough. You need $|A_{n,r}-A_{m,r}|$ to be small. This is where the convergent series helps: the norm of the difference is bounded by some tail of the convergent series. – Post No Bulls Dec 25 '13 at 07:52