4
$\begingroup$

Let $\gamma$ be the circle $\{z \in \mathbb{C}: \lvert z\rvert=1 \}$. Suppose $f$ is a function analytic on an open set containing $\gamma$ and its interior and that $\lvert\, f(z)\rvert<1$ for each $z$ on $\gamma$. Show that $f$ has exactly one fixed point inside $\gamma$

That is, there is exactly one $z$ in the open unit disk with $f(z)=z$.

Is this a result of Louville's theorem?

I don't know how to approach it.

$\endgroup$
13
$\begingroup$

This is a consequence of Rouché's Theorem:

Let $g(z)=-z$ and $h(z)=f(z)-z$.

Since $$ \lvert h(z)-g(z)\rvert=\lvert\, f(z)\rvert <1=\lvert g(z)\rvert, $$ for every $z$ on $\gamma$, then the functions $g$ and $h$ have the same number of zeros in the interior of $\gamma$. Clearly $g$ has exactly one zero, and hence $f$ has exactly one fixed point inside the unit circle.

$\endgroup$

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.