Welcome to our community

Be a part of something great, join today!

holomorphic function and an open disc

StefanM

New member
Jan 30, 2012
28
Now the function f is holomorphic in an open disc U and that Re( f ) is
constant in U. I'm trying to show that
1)f must be constant in U.
2) the essential property of the disc U that it used here
3) an example of an open set U for which the conclusion fails.

Let f=u+vi where u is a constant.Since f is holomorphic by the Cauchy–Riemann equations->
u_x=v_y and u_y=-v_x but since u is a constant u_x=u_y=0 => 0=v_y =-v_x...therefore f is constant.
The disc U has to be open,as in:U(a,r)={z:|z-a|<r}.
Is this correct?What should I do for the last part?
Thank you
 

PaulRS

Member
Jan 26, 2012
37
Note that your proof of 1) actually uses the fact that the disc is connected as well as the fact that it is open.

$U$ open implies that given $p\in U$ we must have $\varepsilon > 0$ such that $B_{\varepsilon} (p) \subseteq U$. But $B_{\varepsilon} (p)$ is convex , thus $v(x) - v(p) = \bigtriangledown v (c) \cdot (x-p) = 0$ for some $c\in B_{\varepsilon} (p)$ (*) and so $v(x) = v(p)$ for all $x\in B_{\varepsilon} (p)$. Thus $u$ is locally constant, and each set $v^{-1}\left(\{a\}\right)$ is open for $a\in \mathbb{R}$, so if $U$ were connected, $v$ can take only one value (because otherwise our connected set would be the union of 2 or more disjoint non-empty open sets, which is a contradiction).

So if you want to solve (3) look at a disconnected set (for instance, the union of 2 disjoint open balls).

Here: $B_{\varepsilon}(p) := \{z \in \mathbb{C} : |z-p| < \varepsilon\}$

(*) Mean value theorem for the function $g(t) = v\left(p\cdot (1-t) + x\cdot t\right) $ , $g: [0,1] \to \mathbb{R}$. This makes sense since we are working on the convex set $B_{\varepsilon} (p)$.
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
I want to add something to what PaulRS said.

Paul showed that f must be locally-constant. Here is a generalization of his statement.

Here is a purely topological exercise. Show that if f is locally constant on U then it must mean that f is constant on the connected component of U. In particular if U is connected then f is constant on U.
 
Last edited:

AlexYoucis

New member
Jan 26, 2012
19
College Park, Maryland
I want to add something to what PaulRS said.

Paul showed that f must be locally-constant. Here is a generalization of his statement.

Here is a purely topological exercise. Show that if f is locally constant on U then it must mean that f is constant on the connected component of U. In particular if U is connected then f is constant on U.
This is a good exercise because of how often it shows up in things like complex analysis/differential geometry. A similar exercise which is much simpler, but actually comes up even more than the exercise TPH suggested is the trivial matter that continuous maps from connected spaces to discrete spaces are constant. Useful for proving that different branches of the logarithm differ from each other by a constant multiple of $2\pi i$.