What am I not doing in showing: If a holo fcn $f: \mathbb C \to \mathbb C$ satisfies $|f(z)| \le A|z|+B$ then $f$ is linear without using power series

Asked Oct 08 '19 at 18:15

Active Oct 08 '19 at 18:31

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We want to solve this without using power series.

I know I should use Cauchy's estimates formula, but I can't seem to get a correct proof.

We want to show that $f'(z)$ is constant and the result will then follow from Liouville's theorem.

Choose $R>0$ and consider $\overline{B_R(0)}$. Then for $z \in \partial \overline{B_R(0)}$, we have $|f(z)|\le A|z|+B \le AR+B$.

So, $$|f'(0)| \le \frac{1}{R}(AR+B) = A+\frac{B}{R}.$$

As $R \to \infty$, then $|f'(0)| \le A$.

However, I need to show that $|f'(z)|$ is bounded and thus constant.

How can I continue?

edited Oct 08 '19 at 18:31

asked Oct 08 '19 at 18:15

user5826

The linked answers use power series, no? – zhw. Oct 08 '19 at 18:29
1

Fix $w\in \mathbb C,$ and define $g(z)= f(z+w).$ Then $$|g'(0)| \le (\sup_{|z|=R} |g(z)|)/R = (\sup_{|z|=R} |f(z+w)|)/R \le (\sup_{|z|=R} (A|z|+|w|)/R = (AR+|w|)/R.$$ Let $R\to \infty$ to see $|g'(0)|\le A.$ But $g'(0)=f'(w).$ Since $w$ is arbitrary, $|f'|\le A$ everywhere, and this is what you wanted. – zhw. Oct 08 '19 at 19:03

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