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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.

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    $\begingroup$ for the first part you use that $f$ not identically zero means that for any $w$ there are circles centered at $w$ and as close as you want on which $f$ has no zeroes, so picking one such $C_w$ and taking $\min_{z \in C_w}|f(z)|=\delta_w>0$ then by uniform convergence you can find $n$ st $|f_n(z)-f(z)|<\delta/2, z \in C_w$ then Rouche etc $\endgroup$
    – Conrad
    Oct 8, 2022 at 2:53
  • $\begingroup$ @Conrad I think you're showing $|f_n(z) -f(z)|<|f(z)|$ on $C_{w}$. $\endgroup$ Oct 8, 2022 at 3:13
  • $\begingroup$ yes that's the point so $f_n,f$ have same number of zeroes inside $\endgroup$
    – Conrad
    Oct 8, 2022 at 3:15
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    $\begingroup$ You can use the first result to show the second part: Suppose $f$ is not constant and $f(z_0) = w_0 \notin \Omega'$ for some $z_0 \in \Omega$. Let $g_n(z) = f_n(z)-w_0$ and $g(z) = f(z)-w_0$. Note that the $g_n$ have no zeros in $\Omega$ and $g_n \to g$ uniformly on compact sets. Since $g(z_0) = 0$, we see by the first part that it must be constant and hence so is $f$ which is a contradiction. $\endgroup$
    – copper.hat
    Oct 8, 2022 at 6:31

1 Answer 1

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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.

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