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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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Since $F$ and $g$ are known to be analytic, you can let $f=Fg$. Am I missing something? –  user31373 Jun 13 '12 at 20:41
    
@Leonid: I think it's slightly more delicate than that. $F$ is the limit of the functions $\frac{f_n(z+a_n)}{M_n g(z+a_n)}$, and the issue is that the sequence $a_n$ may not converge. –  J. Loreaux Jun 13 '12 at 20:56
    
@Leonid: I think if we have something like $\frac{f_{n}(z)}{g(z)}\to F(z)$ then the choice $f=Fg$ would work! However, with my editing ($f_{n}, g$ are entire functions not just analytic) make the problem harder! –  George Jun 13 '12 at 21:17
    
Okay, is it even true! –  George Jun 14 '12 at 0:08
    
@George are there any more hypothesis missing from this question? Such as $f_n(z)$ and $g(z)$ mapping the real line to itself? –  JSchlather Jun 14 '12 at 3:29

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