# Can we find a holomorphic function $g$ on an open disk such that $\sum_{i \in \mathbb{N}} |(f-g)(a_i)|^2 < +\infty$?

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 $$\sum_{i \in \mathbb{N}} |(f-g)(a_i)|^2 < +\infty$$ ?

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 $$f(x)=g(x)$$.
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 $$g(0) = 0$$.