$C_0$, so that $|f(z)|\leq C_0 \exp (C_0 |z|^{\rho_0})$ for all $z\in \Omega$. Suppose there are constants $C_1$, $\rho>0$ so that for all $x>1$, $|\int_1^x f'(t)/f(t)dt|\leq C_1 |x|^\rho$ and for all $x<-1$, $|\int_x^{-1} f'(t)/f(t)dt|\leq C_1 |x|^\rho$. If, in addition, $f$ does not vanish in $\Omega$, then there is a constant $C_3$ so that $|f(z)|\leq 
C_3 \exp(C_3|z|^\rho)$ for all $z\in \Omega$.

In the proof we shall denote by $C$ a constant the value of which may change from line to line without comment.

Since $f$ is nonvanishing in $\Omega$, there is a function $g$ analytic on $\Omega$ so that $\exp g(z)= f(z)$. Since $g'(z)= f'(z)/f(z)$, $$g(x)-g(1)= \int_1^x \frac
{f'(t)}{f(t)} dt\; \text{if}\; x>1$$ so that $|g(x)|\leq C |x|^\rho+|g(1)|$ when $x\geq 1$ for some constant $C$. A similar argument gives a similar bound for $x\leq -1$.

We now assume $\rho < \rho_0$ since otherwise there is nothing to prove. We give a bound on the growth of $g$ at infinity which is more than adequate to allow us to apply a version of the Phragmén-Lindellöf theorem as we will below. Since ${\operatorname{\rm Re}\nolimits}g(z)=\log |f(z)|$, for all $z\in  \Omega$, ${\operatorname{\rm Re}\nolimits}g(z) \leq  C (1+ |z|^{\rho_0})$. Applying Lemma \[l:caratheodorytype\], we find that $|g(z)|\leq C (1+|z|^{\rho_0})$ for 