4

Let $f$ be an entire function with $|f(z)|\le 100\log|z|,\forall |z|\ge 2,f(i)=2i, \text{ Then} f(1)=?$

I have no idea how to solve this one!

$g(z)={f(z)\over \log|z|}$ Then Can I say $g$ is constant by Liouville Theorem?

Myshkin
  • 35,974
  • 27
  • 154
  • 332
  • 3
    You can't use that directly, since $\log \lvert z\rvert$ isn't holomorphic (and $\log z$ isn't entire). Use the Cauchy estimates. – Daniel Fischer Aug 02 '13 at 15:13
  • 1
    Do you know even one example of a nonconstant entire function with $|f(z)|\le 100\log|z|,\forall |z|\ge 2$? If so, perhaps you can divide the general $f$ by that function, and get somewhere...??? Or, if there is no such example the problem becomes easy... – GEdgar Aug 02 '13 at 15:23
  • @DanielFischer Can we somehow use the fact that $log|z|≤|z|$ ? – Rishi May 23 '22 at 04:37

1 Answers1

10

You can't directly use Liouville's theorem, since dividing $f$ by $\log \lvert z\rvert$ or $\log z$ doesn't produce an entire function.

But you can use the Cauchy estimates to show that that bound on $\lvert f(z)\rvert$ actually implies that $f$ is constant, for $R > 2$ and $\lvert z\rvert \leqslant \frac{R}{2}$, you can compute

$$\left\lvert f'(z) \right\rvert = \left\lvert \frac{1}{2\pi i} \int\limits_{\lvert \zeta \rvert = R} \frac{f(\zeta)}{(\zeta-z)^2}\, d\zeta\right\rvert \leqslant \frac{1}{2\pi} \int\limits_{\lvert\zeta\rvert = R} \frac{\lvert f(\zeta)\rvert}{\lvert \zeta - z\rvert^2}\, \lvert d\zeta\rvert \leqslant \frac{4R}{R^2}100\log R$$

and, since the right hand side tends to $0$ for $R \to \infty$, conclude that $f' \equiv 0$.

Daniel Fischer
  • 206,697
  • @@Daniel Fischer) One question....Where the given condition $f(i)=2i$ is needed ? – Empty Oct 24 '15 at 20:08
  • 1
    @S.Panja-1729 After having found out that $f$ is constant, that is used to determine $f(1)$. Since the question asks for $f(1)$, one value of $f$ is needed, otherwise there'd be multiple options. – Daniel Fischer Oct 24 '15 at 20:18
  • @Daniel Fischer♦ If we take $|z|\gt 2 $ and take $R$ such as the region ${z_1 : |z_{1} - z |=R}$ in the region $|z|\ge 2 $. Is that will be okay? – A learner Jan 28 '21 at 04:02
  • @Alearner I'm not quite sure what you intend. I guess it boils down to whether we must take a circle with centre $0$ or we can take a different centre. We can do that, and the contour we use need not even be a circle. But taking a circle with centre $0$ makes the estimate particularly easy, since the bound for $f$ is given in terms of $\lvert z\rvert$. – Daniel Fischer Jan 28 '21 at 11:36