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

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 $\mathbb{C}$.

• 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 $g$? – Daniel Fischer May 25 '15 at 13:44
• 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. – Kees Til May 25 '15 at 15:59

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 $\displaystyle f(z)=\frac{z^3}{3}+z+1$.
• hmmmm nice, i only don't understand why $z = +- i$ are removable. because that term is 'bounded' near them... – Kees Til May 25 '15 at 14:09
• @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. – Emily May 25 '15 at 14:31