# Pointwise convergence of holomorphic functions

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 $g'$.)

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