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$f$ holomorphic at $B(0,1)$ and $|f(z)|<1$ for every $z\in B(0,1)$ and $f(z)=z$ has at least two solutions in the domain Prove for every $z\in B(0,1)$ $f(z)=z$.

My try:
Setting $f(z)=zg(z)$ and taking $r<1$ and looking at $|z|=r$ we can deduce that $|g(z)|<\frac{1}{r}$ by taking $r\to 1$ $|g(z)|<1$ and because $f$ has at least two solutions and we can take $z_0\neq 0,f(z_0)=z_0$ we get that $z_0g(z_0)=f(z_0)=z_0$ therefore $g(z_0)=1$ in contradiction to the assumption of $g$ and from MMP we get that $g\equiv 1$ and $f(z)=z$ for all $z$ in the domain

convxy
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  • Your approach is not correct: You cannot set $f(z) = zg(z)$ because $f(0)$ is not necessarily zero. – Martin R Jun 23 '20 at 10:03
  • You can write $f(z)=zg(z)$ with a holomorphic $g$ iff $f(0)=0$. But $f(0)$ is not necessarily zero. – Aphelli Jun 23 '20 at 10:03

0 Answers0