# Rudin Real and Complex analysis exercise 10.20

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.

• 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 Oct 8, 2022 at 2:53
• @Conrad I think you're showing $|f_n(z) -f(z)|<|f(z)|$ on $C_{w}$. Oct 8, 2022 at 3:13
• yes that's the point so $f_n,f$ have same number of zeroes inside Oct 8, 2022 at 3:15
• 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. Oct 8, 2022 at 6:31

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.