Let $U$ be an open , connected set in $\mathbb{C}$, the following are equivalent:

a) $f$ is holomorphic in $U$.

b) $f \in C^{1}(U)$ and for any disk/ball $\overline{B}(z_0,r) \subset U, z \in B(z_0,r)$, we have $$f(z_0) = \dfrac{1}{2\pi i}\int_{C}\dfrac{f(z)}{z-z_0}dz$$

This is quite a well known theorem in complex analysis. Just that when I was reading an old set of notes written by a previous professor, i came across $C^{1}(U)$, It is quite impossible for me to find the prof, and if anyone can guess what $C^{1}(U)$ means, it will be much appreciated.

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    $\begingroup$ $C^1(U)$ is the set of continuously differentiable functions on $U$, i.e., $f\in C^1(U)$ means $f$ has a derivative $f’$ defined on all of $U$ and $f’$ is continuous. $\endgroup$ Aug 15, 2018 at 17:45
  • $\begingroup$ @TrevorNorton No, $C^1$ absolutely does not mean what you said! If it meant that there'd be nothing to prove here. $C^1$ means the partial derivatives $f_x$ and $f_y$ are continuous. $\endgroup$ Aug 15, 2018 at 18:16
  • $\begingroup$ @DavidC.Ullrich Yeah that would make more sense. I just assumed $C^1$ had the same meaning as it did in the real-valued context. $\endgroup$ Aug 15, 2018 at 18:23
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    $\begingroup$ @TrevorNorton If you don't know what it means some would say you shouldn't tell people what it means... $\endgroup$ Aug 15, 2018 at 18:26
  • $\begingroup$ @DavidC.Ullrich I thought I knew what it meant. I just made a mistake. I'll try and be more careful in the future. $\endgroup$ Aug 15, 2018 at 18:30

1 Answer 1


Your statement of (b) makes no sense, since you don't say what $C$ is. In fact C should be the positively oriented boundary of $B(z_0,r)$.

Also the comment regarding the meaning of $C^1(U)$ is wrong. In fact saying $f\in C^1(U)$ means that $f$ and the partial derivatives $f_x, f_y$ are continuous in $U$; it certainly does not mean that $f'$ exists.

With that clarification: The result you state is false. In fact (b) is equivalent to $f$ being harmonic in $U$. A correct version of (b), that is equivalent to (a), is this:

(b') $f\in C^1(U)$, and if $B=\overline{B(z_0,r)}\subset U$ and $C$ is the positively oriented boundary of $B$ then $$f(z)=\frac1{2\pi i}\int_C\frac{f(w)}{w-z}\,dw$$for every $z\in B^o$.

The fact that (a) implies (b') is the Cauchy Integral Formula, proved in any book on complex analysis. To show that (b') implies (a) you could for example show that $f$ satisfies the Cauchy-Riemann equations by differentiating under the integral. Or probably simpler, by using Morera's theorem together with Fubini's theorem and Cauchy's theorem.


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