# Constant function

1. Oct 20, 2007

### littleHilbert

Please check whether this makes sense

1. The problem statement, all variables and given/known data

If $U\subset\mathbb{C}$ open, path-connected and $f:U\longrightarrow\mathbb{C}$ differentiable with $f'(z)=0$ for all $z\in{U}$, then f is constant.

Hypotheses:
H1: U is pathconnected
H2: $f:U\longrightarrow\mathbb{C}, f'(z)=0, z\in{U}$

2. The attempt at a solution

By H1 for every $p,q\in{U},p\neq{q}$ there exists $\gamma:[a,b]\longrightarrow{U}$ such that $\gamma(a)=p,\gamma(b)=q$.
So let $\gamma$ be an arbitrary path in U and let $z\in\gamma([a,b])\subset{U}$, that is $z:=\gamma(t)$ for $t\in[a,b]$.

We shall show that f is constant on any path in U between arbitrary but fixed points p and q. Since $\gamma$ is arbitrary, it will then follow that f is constant on U.

Since f is complex-differentiable and gamma is continuous, and hence also real-differentiable, we have $(f\circ{\gamma})'(t)={\gamma}'(t)f'(\gamma(t))$. But $f'(\gamma(t))=f'(z)=0$ by H2, hence $(f\circ{\gamma})'(t)=0$, which implies that $(f\circ{\gamma})(t)=const$. In particular, $(f\circ{\gamma})(a)=(f\circ{\gamma})(b)$, that is $f({\gamma}(a))=f({\gamma}(b))$, or $f(p)=f(q)$ for all $p\neq{q}, p,q\in U$. But this is exactly the property of a constant function. It follows that f is locally, on a subset of U, constant and hence constant.

Last edited: Oct 20, 2007
2. Oct 20, 2007

### quasar987

But could you prove this?

My hint at the solution to the problem would be: path integrals.

3. Oct 21, 2007

### littleHilbert

Yes, I think I can.

We have $(f\circ{\gamma})'(t)=u'(t)+iv'(t)=0\Longrightarrow{u'(t)=0=v'(t)}$, which means that $u(t)=C_1, v(t)=C_2$ are constant functions and so $(f\circ{\gamma})(t)$ is constant.

Must I use path integrals here. Is the above argumentation incorrect or imprecise?

Last edited: Oct 21, 2007
4. Oct 21, 2007

### quasar987

Ok, now it looks good and complete!

(But with path integrals it is very fast: "Let a,b be in U and y be a path in U joining and and B. Then by the FTC for path integrals in the complex plane, the integral of f '(z) along y is both 0 and f(b)-f(a). Therefor f(b)-f(a)=0. QED)

5. Oct 21, 2007

### littleHilbert

Ok, I'll try to go into detail, because I'd like to write it down formally and clearly.

So you say that:

Constant functions are trivially continuous and holomorphic everywhere in C. U is open, so f' is holomorphic on U. Also the path gamma is continuous and hence smooth. These two statements imply that a primitive of f' on U exists and is determined up to a constant. Clearly, f' has a primitive on gamma.

Let F be a primitive. Again, differentiability of f implies that primitive F is holomorphic and F=f.

Now we apply the FTC to any path gamma between any two points a and b to conclude that:
$\displaystyle\int^{}_{\gamma} f'(z)\,dz = f(b)-f(a)$
At the same time: $\displaystyle\int^{}_{\gamma} f'(z)\,dz = 0$ by hypothesis. Hence f(b)=f(a) for all different a and b in U. Thus it follows f is constant.

Is it OK?