# Uniform limit of holomorphic functions on the unit disc

my question is the following: Prove or disprove: There exists a sequence $$\{f_n\}$$ of holomorphic functions on the unit disc $$\mathbb{D} := \{z:|z| < 1\}$$ so that $$f_n$$ converges to $$\bar{z}^3$$ uniformly on the circle $$C:=\{z:|z|=\frac{1}{2}\}$$.

My attempt: I thought that the claim is wrong. My intuitive idea is to show that the uniform limit of holomorphic functions is holomorphic and $$\bar{z}^3$$ is not holomorphic. However, the question only claims uniform convergence on the circle C. As far as I understand, to say that the uniform limit of holomorphic function is holomorphic, we need uniform convergence in all compact subsets of a domain, it is not the case here.

How should I attack this problem?

There is no such sequence. If $$n\in\Bbb N$$, then, since $$f_n\colon\Bbb D\longrightarrow\Bbb C$$ is holomorphic, you have $$\int_Cz^2f_n(z)\,\mathrm dz=0$$. Therefore, since the convergence is uniform, you should also have $$\int_Cz^2\overline z^3\,\mathrm dz=0$$. But\begin{align}\int_Cz^2\overline z^3\,\mathrm dz&=\int_C|z|^4\overline z\,\mathrm dz\\&=\frac1{16}\int_C\overline z\,\mathrm dz\\&=\frac{\pi i}{32}.\end{align}
• A minor remark: One could also write $z^2 \overline z^3 = \frac{|z|^6}{z} = \frac{64}{z}$ to make the “non-zero-ness” of the integral even more obvious. Commented Jan 1, 2023 at 11:48