Consider the (closed) unit disc {$ z \in \mathbb{C}: |z| \leq 1$}. Let $f$ be continuous function $S \to \mathbb{C}$ where $S$ is the boundary of the disk. Can this function be extended to the one continuous on the whole disk and holomorphic on $D\setminus S$?

I think that I just can write $ f(z) = \frac{1}{2\pi i}\int_S{\frac{f(\phi)\,\mathrm dz}{z - \phi}}$. Looks like all I need to write it is continuity of $f$. The rest defines by boundary values. Is that correct, have I missed something?

And here is another problem which I believe pretty close to that one above:

Let $f$ is continuous on $D$ and holomorphic on $D\setminus S$. And let $f$ is zero on some arc of $S$ I need to prove that $f$ is zero function and I think it could be done similar way, namely it's enough to show somehow that function vanishes on the whole $S$ which would prove the proposition. But I don't know how to show that.

  • $\begingroup$ I have just found out that I made a mistake. For example, the function $1/z$ satisfies the condition but cannot be extended. How to exclude such a cases? $\endgroup$
    – Invincible
    Aug 6, 2017 at 11:30

1 Answer 1


I suppose that you meant to write$$f(z)=\frac1{2\pi i}\int_S\frac{f(\phi)\,\mathrm d\phi}{\phi-z}.$$The double use of $f$ here is not a good idea. Anyway, let us see what happens when $f(z)=\overline z=\frac1z$. In this case, assuming that $|z|<1$ and that $z\neq0$,\begin{align*}\frac1{2\pi i}\int_S\frac{f(\phi)\,\mathrm d\phi}{\phi-z}&=\frac1{2\pi i}\int_S\frac{\mathrm d\phi}{\phi(\phi-z)}\\&=\operatorname{Res}_0\left(\frac1{\phi(\phi-z)}\right)+\operatorname{Res}_z\left(\frac1{\phi(\phi-z)}\right)\\&=-\frac1z+\frac1z\\&=0.\end{align*}So, as you can see, you don't get again the function that you started with.

And actually, no, the conjugation is not the restriction to $S$ of a holomorphic function.

  • $\begingroup$ Is the answer just "no" or there is a way to describe such a function which allows extention? $\endgroup$
    – Invincible
    Aug 6, 2017 at 11:39
  • $\begingroup$ @Vladislav I don't know. $\endgroup$ Aug 6, 2017 at 11:41
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    $\begingroup$ @Vladislav You must have $$\int_S f(z)\cdot z^k \,dz = 0$$ for all $k \geqslant 0$. $\endgroup$ Aug 6, 2017 at 13:38
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    $\begingroup$ The condition says that the Fourier series of $f$ has no nonzero coefficients with negative index, so $$f(e^{it}) = \sum_{k = 0}^{\infty} c_k e^{ikt}$$ (in a sense appropriate for Fourier series at least). That gives a holomorphic function on the unit disk by setting $F(z) = \sum_{k = 0}^{\infty} c_k z^k$, and if $f \in L^p(S)$ for $1 < p < \infty$, one can see that $f_r(e^{it}) = F(re^{it})$ tends to $f$ in $L^p(S)$ as $r\to 1$, so $f$ gives the boundary values of $F$ in some sense (read about Hardy spaces for more precise statements). I'm not sure about $p = 1$, I think it's also true in $\endgroup$ Aug 6, 2017 at 15:31
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    $\begingroup$ that case. It's somewhat different for $p = \infty$, then generally $f_r$ doesn't tend to $f$ in $L^{\infty}(S)$ (since the $f_r$ are continuous, the continuity of $f$ is a necessary condition for $f_r \to f$ in $L^{\infty}(S)$). But for continuous $f$, by Fejér's theorem the arithmetic means of the Fourier series converge uniformly to $f$, so their extensions to the closed unit disk converge uniformly on the closed unit disk (to $F$ on the open unit disk, to $f$ on $S$), and hence $F$ has a continuous extension to the closed unit disk, given by $f$ on the boundary. $\endgroup$ Aug 6, 2017 at 15:31

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