Prove f is Constant Function on U

[littleHilbert]

Please check whether this makes sense

Homework Statement

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.

 

[quasar987]

...hence (f\circ{\gamma})'(t)=0, which implies that (f\circ{\gamma})(t)=const.

But could you prove this?

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

 

[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?

 

[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)

 

[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?

Many thanks in advance

 

[quasar987]

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?

Many thanks in advance

Yes.