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?

Thanks in advance.


1 Answer 1


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}

  • 1
    $\begingroup$ 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. $\endgroup$
    – Martin R
    Commented Jan 1, 2023 at 11:48

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