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As a consequence of Goursat's Theorem, we can prove that every holomorphic function on an open disk has primitive.

Question: Is it true that every continuous function $f\colon D\rightarrow \mathbb{C}$ has primitive? [D=open disc in $\mathbb{C}$]

The answer I think is "NO". But my explanation involves use of some important theorems. The example I thought is $f(z)=\overline{z}$. If this $f$ has primitive, then $f$ has to be holomorphic, a contradiction.

The problem I would concern here is the following:

Problem: Can we give an elementary argument that $f(z)=\overline{z}$ has no primitive in any open disc?

(I want to avoid the theorem that "a complex function which is once differentiable is infinitely many times differentiable").

Groups
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2 Answers2

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Take an open disc around $z_0$. Then you can find $\varepsilon>0$ such that the closed path $$ \gamma: \phi\in[0,2\pi] \mapsto z_0 + \varepsilon e^{i\phi}$$ is inside the disc.

Now note that $$\oint_\gamma \bar z dz = i \epsilon\int_0^{2\pi} (\bar z_0 + \epsilon e^{-i\phi}) e^{i\phi} d\phi = i \epsilon^2 \int_0^{2\pi} d\phi = 2\pi i \epsilon^2 \ne0.$$ However, for $\bar z$ to have a primitive this integral should be $0$ as the path is closed.

Christian Blatter
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Fabian
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The proof of @Fabian is the morally right one: indeed, to find the antiderivative of a function $f(z)$ one needs to integrate $\int^z f(\zeta) d \zeta$ since we have the complex Leibniz-Newton formula

$$F'=f \implies \int_a^b f(\zeta) d \zeta = F(b) - F(a)$$

moreover the integral must not depend on the path.

Another approach: assume that there exists $F$ with $F'(z) = f(z)$. This is a complex derivative. Now we can use the real partial derivatives and conclude:

\begin{eqnarray} \frac{\partial F}{\partial z} &=& f(z)\\ \frac{\partial F}{\partial \bar z} &=&0 \end{eqnarray}

From the equations above it follows right away that if $f$ is of class $C^k$ then $F$ is of class $C^{k+1}$. Say $f$ is of class $C^1$; we obtain

$$\frac{\partial f}{\partial \bar z} = \frac{\partial }{\partial \bar z}\frac{\partial F}{\partial z}= \frac{\partial }{\partial z}\frac{\partial F}{\partial \bar z}=0$$ because the operators $\frac{\partial }{\partial z}$, $\frac{\partial }{\partial \bar z}$ commute.

Now notice that $\frac{\partial \bar z}{\partial \bar z}\equiv 1$

orangeskid
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  • +1: for spelling out what we mean by an antiderivative and showing an alternative approach. – Fabian Feb 02 '15 at 12:18
  • @Fabian: Thanks. I am starting to understand why they wrote $\int^z f(z) dz$, with the upper limit finding its way through the domain. – orangeskid Feb 02 '15 at 20:19