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Say we have a sequence of entire functions $g_{n}$, and $g_{n} \rightarrow g$ uniformly on compact sets in $\mathbb{C}$. I want to show that if each $g_{n}$ has only real zeros, then all the zeros of $g$ must also be real.

My best guess was to assume $g$ has a non real zero $w$ and take a compact set $K$ containing no real numbers and containing $w$. Then $g_{n}$ never vanishes on $K$ yet $g$ does. What if we look at the the sequence of functions $\frac{1}{g_{n}}$ analytic on $K$. Is it true $\frac{1}{g_{n}} \rightarrow \frac{1}{g}$ uniformly on compact sets? Because then we would get $\frac{1}{g}$ is analytic in $K$, a contradiction. Thanks for the help!

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As pointed out by Tau, you must assume that the limit $f$ is non zero.

Then one can use Rouché's theorem:

Theorem. Let $f,g$ be two holomorphic functions in a neighborhood of $\overline {D(a,r)}$. We suppose that for $|z-a|=r$, we have $|f(z)-g(z)| < |f(z)|$. Then $f$ and $g$ have the same number of zeroes in $D(a,r)$.

Now, in your situation, you choose $a$ a non-real zero of $f$, and $r$ such that $f(z)$ is never zero on the circle $|z-a|=r$, therefore $|f(z)| > \delta >0$ there. Then for $n$ sufficiently big, you will have $|f(z)-f_n(z)| < \delta < |f(z)|$ on $|z-a|=r$. Up to choosing a smaller $r$ so that $\overline D(a,r)$ does not intersect the real axis, you're done.

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    Hmm. Somehow the final ) wasn't made part of the link. It should be. I don't know how to control that here.2011-05-15
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That is not true. Take for example $f_n(z)= \frac{z}{n}$. Then $f_n$ vanishes only at 0 but $f=0$ for all $z$.