1
$\begingroup$

Suppose $f(z)$ is an entire function and everywhere $|f'(z)| \leq |z^2+1|$ and further $f(0) = f'(0) = 1$. Determine $f$.

I tried using Liouville's theorem but i don't know if $f'(z)$ is an entire function, i don't think that's always the case. I also tried to to write $f'(z)$ as $(z-i)(z+i)g(z)$ with $g(z)$ holomorphic on $\mathbb{C}$.

$\endgroup$
  • 2
    $\begingroup$ The derivative of an entire function is again entire. The derivative of a holomorphic function $h$ is holomorphic on the same open set as $h$, as follows for example from Cauchy's integral formula. With your ansatz so far, what can you say about $g$? $\endgroup$ – Daniel Fischer May 25 '15 at 13:44
  • $\begingroup$ that's enough daniel fisher, i didn't know for sure i could use louisville. But if f is infinitly many times differentiable we know that f'is entire as well. $\endgroup$ – Kees Til May 25 '15 at 15:59
6
$\begingroup$

We have $\displaystyle\left|\frac{f'(z)}{z^2+1}\right|\leq 1$. A priori, $\displaystyle\frac{f'(z)}{z^2+1}$ may have poles at $\pm i$, but these are removable as $\displaystyle\frac{f'(z)}{z^2+1}$ is bounded near them. So it is a bounded entire function and thus a constant $k$, and $f'(z)=k(z^2+1)$. Plug in the data gives $\displaystyle f(z)=\frac{z^3}{3}+z+1$.

$\endgroup$
  • $\begingroup$ hmmmm nice, i only don't understand why $z = +- i$ are removable. because that term is 'bounded' near them... $\endgroup$ – Kees Til May 25 '15 at 14:09
  • 1
    $\begingroup$ @KeesTil If $\left|\frac{f'(z)}{z^2+1}\right| \le 1$ everywhere, then certainly we have $\left|\frac{f'(z)}{z^2+1}\right| \le 1$ in a neighborhood of $\pm i$. We then have this critical theorem: Riemann's Theorem. $\endgroup$ – Emily May 25 '15 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.