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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.

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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
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

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.

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