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Let $V=\{f\in\mathcal{O}(\mathbb{D}):f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\text{ with }|a_{n}|\leq n^{2}\text{ for all }n\}$. Prove that there exists $h\in V$, such that $|f'(\frac{1}{2})|\leq|h'(\frac{1}{2})|$ for all $f\in V$.

I want to use Montel's Theorem:

Let $\mathcal{F}$ be a family of holomorphic functions on Ω. If $\mathcal{F}$ is uniformly bounded on every compact subset of Ω, then $\mathcal{F}$ is equicontinuous on every compact subset of Ω, and hence $\mathcal{F}$ is a normal family.

My initial thought is to first prove that the set $V$ is uniformly bounded on every compact subset, but I'm not quite sure how to show that. Also, How do I use the Montel's theorem to prove above? I guess my question is how is showing the existence of $h$ relate to $V$ is a normal family?

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In fact, the problem does not require Montel's theorem necessarily: simply let $$ h(z) = \sum_{n=0}^\infty n^2 z^n, $$ and observe that $|f'(\frac{1}{2})|\leq |h'(\frac{1}{2})|$ for all $f\in V$.

However, I'll present how Montel's theorem implies that such supremum is attained in $V$ as a maximum. For any compact set $K\subset\mathbb{D}$, observe that there exists $r\in (0,1)$ such that $$ K\subset D(0,r)=\{z\in\mathbb{D}:|z|<r\}. $$ Therefore, this leads to a simple estimate that $$ |f(z)|=|\sum_{j=0}^\infty a_j z^j|\leq \sum_{j=0}^\infty j^2 r^j <\infty,\quad\forall z\in K. $$ This shows that $V$ is a normal family. Now, let $f_n\in V$ be a sequence such that $$|f'_n(\frac{1}{2})| \to \sup_{f\in V}|f'(\frac{1}{2})|. $$ Since $V$ is normal, by passing to a subsequence, we may assume that $f_n \to h$ locally uniformly on $\mathbb{D}$. By Cauchy's integral formula, this of course implies that each $k$-th derivative $f_n^{(k)}$ converges locally uniformly to $h^{(k)}$. Since $V$ is closed, we have $h\in V$ and that $$ \sup_{f\in V}|f'(\frac{1}{2})| = \lim_{n\to\infty}\sup_{f\in V}|f'(\frac{1}{2})| = |h'(\frac{1}{2})|, $$as desired.

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  • $\begingroup$ Would you please explain what it means by normal family? $\endgroup$ – Ya G Dec 11 '18 at 20:21
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    $\begingroup$ The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $\sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $h\in V$ of approaching sequence $f_n\in V$ such that $|f'_n(c)| \to \sup|f'(c)|$. $\endgroup$ – Song Dec 11 '18 at 20:25
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    $\begingroup$ This explanation is very rough, so if you are interested, see e.g. Rudin, Real and Complex Analysis. $\endgroup$ – Song Dec 11 '18 at 20:33

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