1
$\begingroup$

We discussed in class an example for the Maximum Modulus Principle:

Prove that if the subset $U\subset\mathbb{C}$ is open, connected, $f:U\rightarrow \mathbb{C}$ is holomorphic, non constant and has nonzeros in $U$, then $z\rightarrow \mid f(z)\mid$ cannot have a local min in $U$.

Taking $1/f$, it is well-defined since, by assumption, there are no zero values of $f$, but I missed the argument why this function cannot have a minimum. Could someone please explain?

$\endgroup$
2

2 Answers 2

3
$\begingroup$

Note that the map $x\mapsto 1/x$ reverses the order of the interval $(0,\infty)$. That is, the statements $x<y$ and $\frac{1}x>\frac{1}y$ are equivalent. Thus, if you want to show that $f$ does not have a local minimum, you want to show that $1/f$ does not have a local maximum - but it sounds like you already have the maximum modulus principle, which exactly says that $1/f$ does not have a local maximum.

$\endgroup$
1
  • $\begingroup$ Actually from the statement of the theorem it is pretty clear! Thanks $\endgroup$
    – FunnyBuzer
    Nov 21, 2017 at 17:14
1
$\begingroup$

Another approach is to use the open mapping theorem. I think this gives more intuition than the maximum modulus principle.

The open mapping theorem implies that $f(U)$ is open. If $f$ has no zeros in $U$ then $|f(U)|$ is an open interval that is a subset of $(0,\infty)$ and hence has no $\min$.

Note the distinction between $\min$ and $\inf$: For example, with $U=B(0,1)$ and $f(z) = 1-z$ that $|f(U)| = (0,1)$, that is, $\inf |f(U)| = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.