Let $(g_n)_n$ be a sequence of holomorphic functions on $U$, where $U$ is the open unit disk. Suppose the first $k$ derivatives of $g_k$ at zero all vanish, $g_k(0) = 0$, and finally that $g_n$ converges pointwise to a holomorphic function $g$ on $U$.

Must $g$ be the zero function? (I'm guessing yes. It would for example suffice to show that $g_k'$ converges pointwise to $g'$.)

  • $\begingroup$ Why not a nonzero constant function? $\endgroup$ – Joey Zou Jul 8 '16 at 18:46
  • $\begingroup$ Hmm I was implicitly thinking of $0$th derivative as the value of the function, but I'll edit that in explicitly. $\endgroup$ – Evan Chen Jul 8 '16 at 18:47

No. I'll construct a sequence $g_k$ such that $g_k(z) \to z$ pointwise on $U$.

Let $A_k = \{z = r e^{i\theta}\; : \; 2/k \le r \le 1, 0 \le \theta \le 2\pi - 1/k\}$, and $B_k$ the closed disk of radius $1/k$ centred at $0$. By Runge's theorem there is a polynomial $p_k(z)$ such that $|p_k(z) - 1/z^{k}| < 1/k$ on $A_k$ and $|p_k(z)| < 1/k$ on $B_k$. Let $g_k(z) = z^{k+1} p_k(z)$. Thus $g_k$ and its first $k$ derivatives are $0$ at $0$, and $|g_k(z) - z| \le 2/k$ on $A_k \cup B_k$. Note that every point $z$ of $U$ is in $A_k \cup B_k$ for sufficiently large $k$, so $g_k(z) \to z$ pointwise.

  • $\begingroup$ No wonder I couldn't get the proof. :) Thanks. $\endgroup$ – Evan Chen Jul 8 '16 at 20:03

EDIT: I assumed uniform convergence in the following answer, which was not part of the hypothesis of the question. As pointed out in Robert Israel's answer above, the statement is false if only pointwise convergence is assumed.

Let $g(z) = \sum{a_kz^k}$ and $g_n(z) = \sum{a^{(n)}_kz^k}$. It suffices to show that $a_k = 0$ for all $k$. Choose any $0<r<1$. By Cauchy's integral formula, we have $$\left|a_k - a^{(n)}_k\right| = \left|\frac{1}{2\pi i}\int\limits_{|z|=r}{\frac{g(z)-g_n(z)}{z^{n+1}}\text{ d}z}\right|\le\frac{1}{2\pi}(2\pi r)\frac{1}{r^{n+1}}\max\limits_{|z|=r}{|g(z)-g_n(z)|}\xrightarrow{n\rightarrow\infty} 0 $$ since $g_n$ converges uniformly to $g$ on $U$. Hence, $a_k = \lim\limits_{n\rightarrow\infty}{a^{(n)}_k}$. Since $a^{(n)}_k = 0$ for $n\ge k$, the result follows.

  • $\begingroup$ "Uniformly" was not part of the hypothesis. This is pointwise convergence. $\endgroup$ – Robert Israel Jul 8 '16 at 19:16
  • $\begingroup$ @RobertIsrael I guess it would have helped to read the question. Woops. $\endgroup$ – Joey Zou Jul 8 '16 at 22:08

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