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Let $f : \mathbb{C} \to \mathbb{C}$ be a continuous function with $f(0)=0$.

Let $\{a_i\}_{i\in \mathbb{N}}$ be a set of scalars in $\mathbb{C}$ such that $$\exists C > 0 : \forall i\in \mathbb{N} : |a_i| \leq C $$ Can we always find a holomorphic function $g$ on $B(0,C+1)$ (the open disk of radius $C+1$) such that $g(0)=0$ and $\sum_{i \in \mathbb{N}} |(f-g)(a_i)|^2 < +\infty$ ?

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Choose $a_i$ dense in $B(0,C)$, then the condition implies that $f=g$ on $B(0,C)$. Indeed, for such an $x$, take $a_{\sigma(i)} \longrightarrow x$ (by density). As $\sum |f-g|(a_i)^2<+\infty$, $f-g(a_{\sigma(i)})\longrightarrow 0$. Thus by continuity $f(x)=g(x)$.

Answer is no in general.

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  • $\begingroup$ He/she said "can we always find", so answer is just "no". $\endgroup$ – mathworker21 May 25 at 23:57
  • $\begingroup$ Logically speaking, no is the answer and showing a counter example is the proof. But there is no harm in being a little more specific. $\endgroup$ – FXV May 26 at 5:18
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No. Take $\{a_i\} = \{e^{2\pi i \theta} : \theta \in [0,1] \cap \mathbb{Q}\}$ and $f$ any continuous function with $f(0) = 0$ and $f(z) = 1$ for $|z|=1$. Now suppose $g$ is a holomorphic function on $B(0,2)$ with $\sum_i |g(a_i)-1|^2 < \infty$. The mean value property, $g(0) = \frac{1}{2\pi}\int_0^{2\pi} g(e^{i\theta})d\theta$, together with the continuity of $g$, clearly implies we can't have $g(0) = 0$.

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