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I could not solve the following problem:

Let $\{f_n\}_n$ be a sequence holomorphic functions in the unit disk $\mathbb{D}$ such that all $f_n$ are zero-free in $\mathbb{D}$, $|f_n|<1$ for all $n\geq1$ and $lim_{n\rightarrow\infty}f_n(0)=0$. Prove that the sequence converges uniformly on compacts of $\mathbb{D}$ to 0.

This is what I tried:

Since $|f_n|<1$, the sequence $\{f_n\}_n$ is uniformly bounded in any compact of $\mathbb{D}$ and hence by Montel theorem, $\{f_n\}_n$ is a normal family. Then, there is a subsequence $\{f_{n_k}\}_k$ that converges to a function $f$ uniformly over compact subsets of $\mathbb{D}$. Now, the sequence $\{f_{n_k}\}_k$ has no zeros and has limit $0$ in $z=0$ by the hypothesis, so $f(0)=0$. By Hurwitz theorem, $f$ is identically zero. But I cannot prove that $\{f_n\}_n$ converges to $f$ uniformly over compact sets. In fact, if I prove the uniform convergence for $\{f_n\}_n$ instead of a subsequence, then we are done.

Can someone help me?

PD: I looked for the same question but I did not find it. If it is duplicated, I apologize.

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  • $\begingroup$ I'd say a family satisfying the Montel condition that doesn't converge compactly to $f$ has a subsequence that converges compactly to some $g \ne f$ $\endgroup$ – reuns Feb 26 '17 at 0:03
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$\scriptstyle \text{It's hard to state it clearly but there is what I tried :}$

Assume $f_n$ doesn't converge compactly to the zero function then we can take $C >0$ and a subsequence $g_k$ such that $\sup_{|z| \le 1/2} |g_k(z)| > C$. Also $g_k$ satisfies the Montel condition and has a subsequence that converges uniformly to $g$ on $|z| \le 1/2$, and hence it can't be the zero function, a contradiction :

Assume $f_n$ has a convergent subsequence $h_m$ which converges compactly to a non-identically zero holomorphic function $h$, since $h(0) = \lim_m h_m(0) = 0$ then $z=0$ is an isolated zero of order $M \ge 1$ and we can choose $r$ small enough such that $\int_{|z| = r} \frac{h'(z)}{h(z)}dz = 2i \pi M$. But this is impossible since $\frac{h'_m}{h_m}$ converges uniformly to $\frac{h'}{h}$ on $|z| = r$ and hence $\lim_m \int_{|z| = r} \frac{h_m'(z)}{h_m(z)} = \int_{|z| = r} \frac{h'(z)}{h(z)}dz = 2i \pi M$ i.e. for $m$ large enough $h_m$ has $M$ zeros on $|z| < r$, a contradiction.

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  • $\begingroup$ Your answer helped me to solve the problem, thank you so much. Just a couple of comments: First, if we assume that $\{f_n\}_n$ does not converge compactly to zero, then there exist a compact $K$ such that $\{f_n\}_n$ does not converge to $0$ over $K$, but I think we cannot assume that $K=\overline{D(0,\frac{1}{2})}$. Second, when you obtain the sequence ${h_m}$, I think it converges uniformly to $h$ over all compact sets (not only $K$) because it is the definition of normal family. Since $h$ has a zero in $0$, Hurwitz theorem states that $h$ is identically zero and we are done. $\endgroup$ – Daniel Gil Feb 26 '17 at 10:37

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