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Let $\left\{f_{n}:n=1,2,3\ldots\right\}$ a uniformly bounded sequence of holomorphic functions on a unit disk $D$.Suppose there exists a point $a\in D$,so that $\lim_{n\rightarrow\infty}f^{\left(k\right)}_{n}\left(a\right)=0$ for each $k=1,2,3...$ ($f^{\left(k\right)}_{n}$ is the $kth$ derivative of $f_{n}$). Show that $f_{n}\rightarrow 0$ uniformaly on each compact subset of $D$.

My thought: since $\left\{f_{n}\right\}$ is uniformaly bounded,from montel theory it's a normal family which means that $\left\{f_{n}\right\}$ has a subsequence that convergents on each compact subsets of $D$. Additionally the limit of above subsequence is holomorphic on each compact subset. And also the limit function on compact subsets containing $a$ is constant because of identity principle and the condition $\lim_{n\rightarrow\infty}f^{\left(k\right)}_{n}\left(a\right)=0$. Then I don't know how to approach to the result.

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  • $\begingroup$ Please tell us what $D$ is. $\endgroup$ – zhw. Jul 11 '16 at 6:57
  • $\begingroup$ @Hmm That's a different problem $\endgroup$ – zhw. Jul 11 '16 at 6:58
  • $\begingroup$ Uniformly, not uniformaly. $\endgroup$ – Mariano Suárez-Álvarez Jul 11 '16 at 7:23
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I suppose $D=\{|z|<1\}$. Being our sequence $\{f_n\}_n$ of holomorphic functions on $D$, uniformly bounded, then by Montel theorem, admits some converging subsequence $\{f_{n_\nu}\}_\nu$: let's call $f$ the limit function.

Since $f_n^{(k)}(a)\stackrel{n\to+\infty}{\longrightarrow}0$ for all $k\ge1$, then this condition holds even for the subsequence $f_{n_\nu}^{(k)}(a)\stackrel{\nu\to+\infty}{\longrightarrow}0$ for all $k\ge1$; but it's clear that $f_{n_\nu}\to f$ wrt compact subsets topology implies that $f_{n_\nu}^{(k)}\to f^{(k)}$ for every $k\ge0$, thus $$ f_{n_\nu}^{(k)}(a)\stackrel{\nu\to+\infty}{\longrightarrow} f^{(k)}(a)\;\;\;\;\forall k\ge1 $$

from which we get that $f^{(k)}(a)=0$ for all $k\ge1$ and thus (expanding $f$ near $a$ in Taylor series first, and then applying identity principle) we get $f\equiv f(a)$ on the whole $D$.

Now, this argument can be done for every converging subsequence of $\{f_n\}_n$, thus every converging subsequence of it, converges to the same function (the constant function $f(a)$), thus by the Montel converging criterion, we get that $\{f_n\}_n$ converges, and it converges necessarely to $f(a)$ (wrt the topology above).

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  • $\begingroup$ Firstly,thank you for your answer.But I still have problem about your answer. I didn't assume $f(x)=0$ I only assume that $k=1,2,3...$. Additionally when we use the identity principle, we can only use it on the compact subsets containing $a$ since the subsequence is convergent uniformly on compact subsets. How can we transition from compact subset to the whole disk. $\endgroup$ – mike Jul 11 '16 at 9:11
  • $\begingroup$ @mike: I edited. This is not a complete answer, it's rather a hint. I don't have much time to dedicate to it... however try by yourself! :-) $\endgroup$ – Joe Jul 11 '16 at 12:23

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