Suppose $\Omega$ is a region, $f_n\in H(\Omega)$ for $n=1,2,3,\ldots,$ none of the functions $f_n$ has a zero in $\Omega$, and $\{f_n\}$ converges to $f$ uniformly on compact subsets of $\Omega$. Prove that either $f$ has no zero in $\Omega$ or $f(z) =0$ for all $z\in\Omega$. If $\Omega'$ is a region that contains every $f_n(\Omega)$, and if $f$ is not constant, prove that $f(\Omega)\subset\Omega'$.
For the first statement, suppose $f\not\equiv 0$ on $\Omega$. Write $f(z) = (f(z)-f_n(z))+f_n(z)$. Let $z_0\in\Omega$ and consider a small disc $D(z_0)$ centered at $z_0$ whose closure is contained in $\Omega$. My plan is to use Rouche's theorem on $f(z)-f_n(z)$ and $f_n(z)$ using uniform convergence so that $|f(z)-f_n(z)|<|f_n(z)|$ on $\partial D(z_0)$. Now $f_n$ has no root, I can conclude $f$ has no root which proves the first statement. To do this, I first choose the minimum value of $f_n$ on $\partial D(z_0)$ and choose $n$ large so that $\sup_{z\in\partial D(z_0)}|f(z)-f_n(z)|<|f_n(z)|$. But this $n$ depends on $f_n$ which is clearly problematic. How can I resolve this problem?
For the second statement, I can use the argument principle: Suppose $f(\Omega)\not\subset \Omega'$. Then there is $w_0\notin\Omega'$ such that $f(z_0) = w_0$ for some $z_0\in\Omega$. Let $g(z) = f(z) – w_0$ and $g_n(z) = f_n(z)-w_0$ for $z\in\Omega$. Then $g_n$ has no root on $\Omega$ by assumption and $g_n\to g$ uniformly on each compact subset. Hence
$$\int_{\gamma}{g'_n(\zeta)\over g_n(\zeta)}\ d\zeta\to\int_{\gamma}{g'(\zeta)\over g(\zeta)}\ d\zeta,$$
as $n\to\infty$ where $\gamma$ is a small circle contained in $\Omega$ centered at $z_0$. LHS is $0$ and RHS is $\geq 1$ which is a contradiction.
Best Answer
Using Rouche's theorem: Following @Conrad's hint, since $f$ is not identically zero, zeros of $f$, if exist, are discrete so for given $w\in\Omega$, we can find a small disc $D(w)$ centered at $w$ contained in $\Omega$ such that $f$ has no zero on $\partial D(w)$. Let $0<\delta_w = \min_{z\in\partial D(w)}|f(z)|$. Then by the uniform convergence, we can find $n$ large so that $|f_n(z) - f(z)|<\delta_{w}/2$ on $\partial D(w)$. Hence, $|f_n(z)-f(z)|<|f(z)|$ on $\partial D(w)$ so by Rouche's theorem, the number of roots of $f$ and $f_n$ are the same on $D(w)$ and $f_n$ has no root on $D(w)$. Since $w$ was arbitrary, this proves the statement.
Using argument principle: This is from @TedShifrin. Just as the proof of part 2, using uniform convergence, $$\int_{\partial D(w)}{f_n'(\zeta)\over f_n(\zeta)}\ d\zeta\to\int_{\partial D(w)}{f'(\zeta)\over f(\zeta)}\ d\zeta$$ as $n\to\infty$ and LHS is zero so RHS is zero.