11
$\begingroup$

prove that if $f:\mathbb{D}\rightarrow\mathbb{D}$ is analytic with two distinct fixed point then $f$ is identity.

i thought if one of the fixed points were zero by schwarz lemma this statement is easily proved.

but what can i do if fixed points were nonzero?

pls don't answer:ur question is duplicated .i've seen other question & answers but i'm still confused.

help pls

thnx

$\endgroup$
17
$\begingroup$

I thought if one of the fixed points were zero by the Schwarz lemma this statement is easily proved.

Good. So it remains to reduce the general case to the known special case.

Let $T$ be an automorphism of $\mathbb{D}$. What do you know about $T^{-1}\circ f \circ T$? Can you find a condition on $T$ that reduces the problem to the known case?

$\endgroup$
  • $\begingroup$ it is a function from $\mathbb{D}\rightarrow\mathbb{D}$ that fixes zero! am i right? $\endgroup$ – user115608 May 19 '14 at 17:56
  • 1
    $\begingroup$ Whether it fixes zero depends on $T$. What condition on $T$ makes it fix zero? $\endgroup$ – Daniel Fischer May 19 '14 at 18:01
  • $\begingroup$ no sorry!it is a function from $\mathbb{D}\rightarrow\mathbb{D}$ that fixes $T^{-1}$ (fixed point of f)! am i right? $\endgroup$ – user115608 May 19 '14 at 18:04
  • 1
    $\begingroup$ i want $T(0)=$ fixed point of f. $\endgroup$ – user115608 May 19 '14 at 18:08
  • 2
    $\begingroup$ @Twink $f$ has (at least) two fixed points, hence so has $h$. Let $\zeta\neq 0$ be another fixed point. Then you can deduce $c =\:?$ $\endgroup$ – Daniel Fischer Nov 3 '14 at 11:02

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.