Let $f:\mathbb{C} \to \mathbb{C}$ a holomorphic function and assume that there exist $M > 0$ and $r>0$ such that $$|f(z)| \leq M |z|\ln |z|$$ $\forall z \in \mathbb{C}$ with $z \geq r$. I need to prove that $f(z) = a+bz$ with $a,b \in \mathbb{C}$.

My approach is the following: I use Cauchy's Inequalities:

$$|f^{(n)}(a)| \leq \frac{n!}{r^n} \max \{ |f(z)| : |z-a|=r \}$$

With this I find that: $$|f^{2}(0)| \leq \frac{n!}{r^n} \max \{ |f(z)| : |z|=r \} \leq \frac{2}{r^2}M r\ln r =\frac{2}{r}M \ln r$$

I need to show that this is zero, but I don't see how. I know that this question is similar to Using Cauchy's Inequality to prove a function's second derivative is zero but I don't think of it as a duplicate, since there are other assumptions.

• Let $r\to \infty$. You know the estimate holds for all $r$. Ditto for the higher derivatives $f^{(k)}(0)$. – Daniel Fischer Nov 20 '14 at 21:07
• But does the estimate hold for all $r$? Then the assumption should be: $\forall r>0$ there exists $M >0$ such that... – user54297 Nov 20 '14 at 21:09
• Ah, sorry. Use a circle of radius $R \geqslant r$ and let $R\to \infty$. – Daniel Fischer Nov 20 '14 at 21:11
• Okay I understand that, but then we have that $f^{(n)}(0)=0$ for all $n\geq 2$, but does that imply that $f^{(2)}(z) = 0$? – user54297 Nov 20 '14 at 21:20
• Look at the Taylor series of $f$ at $0$. You know that since $f$ is entire, it converges to $f$ on all of $\mathbb{C}$. – Daniel Fischer Nov 20 '14 at 21:23