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Let $G\subseteq\mathbb{C}$ be a domain and $f\colon G\to\mathbb{C}$ holomorphic. Show: $$ f'=0\mbox{ in }G\Rightarrow f\mbox{ is constant} $$

Cauchy-Riemann: $$ f'(z)=f_x(z)=u_x(x,y)+i v_x(x,y), z=x+iy=0\\ \Leftrightarrow u_x(x,y)=0~\wedge v_x(x,y)=0 $$

and $$ f'(z)=-if_y(z)=-iu_y(x,y)+v_y(x,y)=0\\\Leftrightarrow u_y(x,y)=0~\wedge~v_y(x,y)=0 $$

That means: If one derivates u and v to x, one gets 0, and therfore the x-part of these functions must be constant. The ssame for the y-part of u and v. So u and v are constant and so is then f.

Is that okay?

Willie Wong
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  • http://de.wikipedia.org/wiki/Cauchy-Riemannsche_partielle_Differentialgleichungen under "Herleitung" and I found that in a book ("Analysis 2, Königsberger") –  Jul 07 '13 at 13:15
  • What you wrote seems all true, but the last couple of sentences seem to me like avoiding very important details. You need to use the fact that $G$ is a domain (in the last couple of sentences). I don't think what you want follows that easily. – Git Gud Jul 07 '13 at 13:18
  • Perhaps it should be said -- and probably it has been said elsewhere on this site -- that the hypothesis that $f$ be complex differentiable is really too strong. For any connected open subset $U \subset \mathbb{R}^m$ and any differentiable function $f: U \rightarrow \mathbb{R}^n$, $D(f) \equiv 0 \implies f \equiv C$. – Pete L. Clark Jul 08 '13 at 08:23
  • The function $f:((-2,-1)\cup(1,2))+\mathrm i\mathbb R\to\mathbb R, x\mapsto\cases{1 & Re(x)<0\ 2 & Re(x)>0}$ is differentiable everywhere in its domain with $f'=0$, but still not constant (where I use the word "domain" in the sense it is usually used with functions, as the set of values the function is defined on; the meaning of "domain" in your question clearly is something different — I suspect it includes being connected). – celtschk Jul 09 '13 at 06:22
  • @celtschk In such a context, "domain" usually means open+connected. – Julien Jul 09 '13 at 16:36

1 Answers1

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This is not really enough. You still have to prove that zero derivative implies constant on a domain. Focusing on the real/imaginary parts and partial derivatives does not make it easier. The key point is the integral formula in grey below. You can do that locally by local convexity, but not globally since $G$ is not assumed to be convex.

Here is a standard argument. To prove $C$ is open, we could also use the series representation of a holomorphic function on a disk, but it is more elementary and direct to use that integral representation here.

Fix $z_0\in G$ and consider $$ C:=\{z\in G\,;\, f(z)=f(z_0)\}. $$ This is clearly a nonempty closed subset of $G$. We will now show it is open in $G$.

Take $z_1\in C$. Since $G$ is open, there exists $r>0$ such that $D(z_1,r)$ the open disk of radius $r$ centered at $z_1$ be contained in $G$. Now for every $z\in D(z_1,r)$, we have

$$ f(z)=f(z_1)+\int_0^1 \frac{d}{dt}f((1-t)z_1+tz)dt=f(z_0)+\int_0^1 (z-z_1)f'((1-t)z_1+tz)dt=f(z_0) $$

using that the real function $f((1-t)z_1+tz)$ is $C^1$ with derivative $(z-z_1)f'((1-t)z_1+tz)=0$. So $D(z_1,r)$ is contained in $C$, which proves that $C$ is open.

Conclusion: $C$ is non-empty, open, closed in $G$ connected, whence $C=G$, i.e. $f$ is constant $=f(z_0)$ on $G$.

Note: as mentioned by @Pete L. Clark, the whole proof works the exact same way if we only assume that $f$ is real-differentiable from $G\subseteq \mathbb{C}\simeq \mathbb{R}^2\longrightarrow \mathbb{C}\simeq \mathbb{R}^2$ with $Df\equiv 0$ on $G$. And dimension $2$ does not add anything special either. It could just be any $f:G\subseteq \mathbb{R}^n\longrightarrow \mathbb{R}^m$. Indeed, every $\mathbb{R}^n$ is locally convex, which is all we need. The only thing that changes is the integral representation, a little bit: $$ f(x)=f(x_1)+\int_0^1 \frac{d}{dt}f((1-t)x_1+tx)dt=f(x_1)+\int_0^1Df_{(1-t)x_1+tx}(x-x_1)dt=f(x_1). $$ What we really gain with the assumption that $f$ is holomorphic=complex-differentiable is that we can simply assume that $f'\equiv 0$ on some open subset of $G$. Then the argument shows that $f$ is constant on any connected component of the latter, which are open. Or simply the last part shows that $f$ is constant on some open disk in $G$. Therefore $f$ is constant on the whole connected domain $G$ by the identity principle.

Julien
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