Análisis complejo

Question

Si $f:\mathbb{C}\to\mathbb{C}$ es una función entera y tiene:

"Cada $z\in \mathbb{C}$, o $|f'(z)|\leq1$ o $|f''(z)|\leq 1$."

Luego existen $a,b,c \in \mathbb{C}$ tal que $2|a|\leq 1$ y $f(z)=az^2+bz+c$.

Answer 1

Fijar $z$ por un minuto y Supongamos que $|f'(z)|>1$. Sea $L$ la línea de $0$ $z$. Si $|f'|>1$ sobre todo de $L$ y $|f''|\le 1$ $L$, por lo tanto, $$|f'(z)|\le|f'(0)|+|z|.$$If there exists a point $p$ on $L$ where $|f'|\le 1$, then let $q$ be the closest such point to $z$; now $|f''|\le1$ on the segment from $q$ to $z$, hence $$|f'(z)|\le|f'(q)|+|z-q|\le 1+|z|.$$

Así que tenemos $$|f'(z)|\le c+|z|$$ for all $ z$ (we proved this assuming $|f'(z) | > 1 $, but it's certainly also true when $ | f'(z) |\le1$). Hence $$|f'(z)|\le 2|z|\quad(|z|>c),$$so$$g(z)=\frac{f'(z)-f'(0)}z$$is a bounded entire function. (Because if $|z| > c $ then $$|g(z)|\le 2+\frac{|f'(0)|}{|z|}\le2+\frac{|f'(0)|}c,$$while $g (z) $ is bounded for $|z|\le $ c por compacidad.)