0
$\begingroup$

Let $f:B_2(0)\to\mathbb C$ be holomorphic with $f(1)=1$ and $f(-1)=-1$. Show that there is a $z\in B_2(0)$ and a $\varepsilon>0$ s.t. $f(z)=1-\varepsilon$.

Is it reasonable to say that since $|f|$ has a local maximum at $z=1$ the function $f$ must be constant on all of $B_2(0)$ by the maximum principle and therefore we get that for $z=0$ and $\varepsilon=1$ it holds that $f(0)=1-1=0$ which is obviously constant.

If this is not enough on its own I was considering to show that $z=1$ has an open neighborhood in $f(B_2(0))$ however I am not sure how to argue that this must be true. Do you have hints for this?

$\endgroup$

1 Answer 1

0
$\begingroup$

I'm not really sure how you get the idea that $f(1)$ is a local maximum. You have that $f(-1)=-1$ and $f(1)=1$ so clearly $f$ is not constant. By the open mapping theroem (you can apply it since $f$ is not constant) $f(B_2(0))$ is open and $1 \in f(B_2(0))$ therefore there is a $\varepsilon$-ball around 1 with $B_{\varepsilon}(1)\subset f(B_2(0))$ But this is allready enough to show the claim.

$\endgroup$
1
  • $\begingroup$ Indeed I was slightly confused and mistook constant functions with the identity for a moment. $\endgroup$ Jun 12, 2016 at 14:01

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.