If $f(i)=2i$, then what will be $f(1)$?

Question

Let $f$ be an entire function on $\Bbb{C}$ such that $|f(z)|\le 100 \ln |z|$ for $|z|\ge 2$ and if $f(i)=2i$, then what is $f(1)$?

I think $f$ will be constant i.e. $f(1)=2i$, but I'm unable to use liouville's theorem or any theorem which proves $f$ is constant.

Hint: although Liouville theorem per se is not enough, the method of proving it is. — Wojowu, Jan 15 '16 at 17:40

score 3 · Accepted Answer · answered Jan 15 '16 at 17:40

3

Consider $$g(z)=\begin{cases}\frac{f(z)-f(0)}{z}&z\ne 0\\f'(0)&z=0\end{cases} $$ which is an entire function. Then $g$ is bounded, hence constant. So $f(z)=az+b$ for some $a,b$. But then the sublinear bound shows that $a=0$. Hence $f$ is constant.

answered Jan 15 '16 at 17:40

Hagen von Eitzen

374,180

Thank you! How did you find that this is duplicate? – Savannah Jan 15 '16 at 17:57
@MathMan (Just after posting the answer) I glanced over the "Related" question list and stumbled over it ... – Hagen von Eitzen Jan 15 '16 at 18:02
okay I never observed such a list! But how do I search before posting? – Savannah Jan 15 '16 at 18:05

score 2 · Answer 2 · answered Jan 15 '16 at 17:47

For any $R > 2$ and $|z| < 1$, we have $$f'(z) = \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{f(\zeta)}{(\zeta-z)^2} d\zeta \implies |f'(z)| \le \frac{1}{2\pi}\int_{|\zeta| = R} \frac{100\log|z|}{|\zeta-z|^2} d|\zeta| \le \frac{100R\log R}{(R-1)^2}$$ Since the last term tends to zero as $R \to \infty$, $f'(z)$ vanishes on unit disk and hence on whole $\mathbb{C}$.
This implies $f(z)$ is a constant function and $f(1) = 2i$.

score 1 · Answer 3 · answered Jan 15 '16 at 17:51

1

Hint 1: Show that $g(z) = \frac{f(z)-f(0)}{z}$ is entire, and $$\lim_{z \to \infty} g(z) =0$$

Hint 2 If $g(z)$ is a continuous function and $\lim_{z \to \infty} g(z)=0$ show that $g(z)$ is bounded.

answered Jan 15 '16 at 17:51

N. S.

132,525

If $f(i)=2i$, then what will be $f(1)$?

3 Answers3