Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\{f_{n}(z)\}$ be a sequence of analytic functions in the upper half plane (in a Hilbert space $H$) and continuous on the real axis, such that

(1) $0<|f_{n}(x)|\leq 1$ for all $x\in \mathbb R$, for all $n$, and

(2) $\sup_{x\in\mathbb R}|f_{n}(x)|\to 0$ as $n\to \infty$. and

(3) $f_{n}\in L^{2}(\mathbb R)$ for all $n$, such that the sequence of $L^{2}$-norms is uniformly bounded.

If $\{a_{k}\}_{k\geq 1}$ is a sequence of real numbers with no limit point, how I can construct a sequence of analytic functions in the upper half plane and continuous on the real axis, say $\{g_{n}(z)\}$ in $H$, in terms of the sequence $\{f_{n}\}$, with the following properties:

(1') $0<\lim\limits_{n\to\infty}\big(\sup_{x\in\mathbb R}|g_{n}(x)|\big)\leq 1$, and

(2') $|g_{n}(a_{k})|\to 0$ as $n\to \infty$, for all $k$.

(3') $g_{n}\in L^{2}(\mathbb R)$ for all $n$.

i.e, the only difference in the new sequence is that the sup is not going to 0.

share|improve this question
    
Have you tried $g_n(z)=\sum_{i=1}^n c_{k n} f_k(z)$, where $c_{k n}$ are chosen so that (1) and (2) hold? Are there obstructions to the existence of such $c_{k n}$? –  user31373 Jun 30 '12 at 17:17
    
@Leonid: I'm trying to use your sequence and find conditions on the $c_{kn}$'s to see if there are any obstructions, but it is very hard to set up such conditions! –  Miley Jun 30 '12 at 17:28
1  
They will probably be in terms of $g_n$ and $a_n$... you may want to begin with a concrete example, maybe $f_n(z)=e^{-z^2}/n$ and some thing like $a_k=ik$? By the way, it's a bad idea to use $i$ for an index in complex analysis. –  user31373 Jun 30 '12 at 17:34
add comment

1 Answer

If by "in terms of the sequence $\{f_n\}$" you mean that the definition of $g_n(x)$ has no dependence on $x$ except in the arguments of functions $f_j$, that is impossible. Consider the case where the $f_n$ are all constants.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.