Let $G := \{z : -1 < \Im z < 1\}$ and let $f \in H(G)$ be such that
(1) $|f(z)| \leq \frac{1}{1-| \Im z |}$ for all $z \in G$, and
(2) $\lim_{x \rightarrow \infty} f(x) = 0$.
Set $f_{n}(x) := f(z + n)$ and show that $f_{n} \rightarrow 0$ locally uniformly on $G$.
Attempt: Let $0 < r < 1$ and let $G_{r} := \{z : |\Im z| < r\}$. Then $1-|\Im z| > 1 - r \implies \frac{1}{1-|\Im z|} < \frac{1}{1-r}$. Take a subsequence $(f_{n_{\ell}})_{\ell}$ of $(f_{n})_{n}$ so that the family of $g_{n_{k}} := f_{n_{k}} \mid_{G_{r}}$ is uniformly bounded. Montel's theorem implies this family is normal. So there is a subsequence $(n_{k}) \subset \mathbb{N}$ of $(n_{\ell})$ such that $g_{n_{k}} \rightarrow g$ locally uniformly for some $g \in H(G_{r})$ (it is holomorphic by local uniform convergence). Now, for positive real $x$ we have $g_{n_{k}}(x) \rightarrow 0 = g(x)$ as $k \rightarrow \infty$. This comes from condition (b). Therefore $g(x) = 0$ for all positive real $x$, whence $g \equiv 0$ on all of $G_{r}$ by the Identity Theorem. That is, $g_{n_{k}}(z) \rightarrow g(z) = 0$ for any $z \in G_{r}$. We conclude that any subsequence of $(g_{n})$ has a further subsequence converging to $0$ locally uniformly. Therefore $g_{n} \rightarrow 0$ locally uniformly. Since $0 < r < 1$ was arbitrary, $f_{n} \rightarrow 0$ locally uniformly on $G$.
What's wrong? I feel odd about the conclusion. I know I showed (hopefully) the result holds for any $0 < r < 1$, but can I necessarily conclude what I want? Is there more to show here?
Let $0 < r < 1$ and let $G_{r}$ be the closure of $\{z : |\Im z| < r\}$ (so $G_{r} $is compact). Then $1-|\Im z| > 1 - r \implies \frac{1}{1-|\Im z|} < \frac{1}{1-r}$. Take a subsequence $(f_{n_{\ell}})_{\ell}$ of $(f_{n})_{n}$ so that the family of $g_{n_{k}} := f_{n_{k}} \mid_{G_{r}}$ is uniformly bounded. Montel's theorem implies this family is normal. So there is a subsequence $(n_{k}) \subset \mathbb{N}$ of $(n_{\ell})$ such that $g_{n_{k}} \rightarrow g$ locally uniformly for some $g \in H(G_{r}^{\circ})$ (it is holomorphic by local uniform convergence). Now, for positive real $x$ we have $g_{n_{k}}(x) \rightarrow 0 = g(x)$ as $k \rightarrow \infty$. This comes from condition (b). Therefore $g(x) = 0$ for all positive real $x$, whence $g \equiv 0$ on all of $G_{r}^{\int}$ by the Identity Theorem. That is, $g_{n_{k}}(z) \rightarrow g(z) = 0$ for any $z \in G_{r}^{\int}$. We conclude that any subsequence of $(g_{n})$ has a further subsequence converging to $0$ locally uniformly. Therefore $g_{n} \rightarrow 0$ locally uniformly. Since $0 < r < 1$ was arbitrary, $f_{n} \rightarrow 0$ locally uniformly on $G$.