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Let $f_{n}(z), g(z)$ be entire functions, for all $n\geq 1$. Suppose that $g(x)$ doesn't vanish on $\mathbb H\cup\mathbb R$ (so we have $\frac{f_{n}(z)}{g(z)}$ analytic on $\mathbb H\cup\mathbb R$). Let $M_{n}=\sup_{x\in \mathbb R}\big|\frac{f_{n}(x)}{g(x)}\big|\leq 1$, where the sup attained at the points $a_{n}\in \mathbb R$, (i.e. $M_{n}=\big|\frac{f_{n}(a_{n})}{g(a_{n})}\big|$) . Also, $M_{n}\to 0$, and the sequence $\frac{f_{n}(z+a_{n})}{M_{n}g(z+a_{n})}$ converges uniformly on compact subsets of $\mathbb H$ to an analytic function $F(z)$. How to prove that $F(z)=\frac{f(z)}{g(z)}$ for some entire function $f$.

Edit: $\mathbb H$= open upper half plane.

Note: The sequence $\frac{f_{n}(z+a_{n})}{M_{n}g(z+a_{n})}$ is bounded by 1 in $\mathbb H$, and the functions $f_{n},g$ are not necessary real on the real line.

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    @George are there any more hypothesis missing from this question? Such as $f_n(z)$ and $g(z)$ mapping the real line to itself?2012-06-14

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