disk. The estimate we obtain here is likely a crude one, but suffices for our purposes.

\[l:caratheodorytype\] Let $f$ be analytic in a neighborhood of $$\Omega_R{\stackrel{\rm{def}}{=}}\{ z\in \Complex: 
1\leq |z| \leq 
R,\; {\operatorname{\rm Im}\nolimits}z\geq 0\},$$ $\rho>0$, and for $x\in \Real \cap \Omega_R$, $|f(x)|\leq C_0|x|^{\rho}$ for some constant $C_0>0$. Set $$M=\max_{|z|=1, \; z\in \Omega_R}|f(z)|$$ and define $$A(R)= \max\left(   C_0 R^\rho,
 M R^\rho , \max_{ z\in \Omega_R }{\operatorname{\rm Re}\nolimits}f(z) \right).$$ Then if $1<r<R$ and $z\in \Omega_R$ with $|z|=r$, then $|f(z)|\leq \frac{2r^{\rho}}{R^\rho- r^\rho} A(R).$

Set $$g(z)= \frac{1}{z^\rho} \frac{f(z)}{2A(R)-f(z)}$$ which is analytic in a neighborhood of $\Omega_R$. We bound $|g|$ on the boundary of $\Omega_R$. If $z\in \Omega_R$ has $|z|=R$, then $$|g(z)|\leq \frac{1}{R^\rho} \frac{|f(z)|}{|2A(R)-f(z)|}\leq \frac{
|f(z)|}{R^\rho |f(z)|}=\frac{1}{R^\rho}.$$ Notice that if $x\in \Omega_R\cap \Real$, $$|g(x)|\leq \frac{1}{|x|^\rho}\frac{C_0 |x|^\rho}{R^\rho C_0} \leq 
\frac{1}{R^\rho}.$$ Moreover, if $z\in \Omega_R$ has $|z|=1$, since $A(R)\geq |f(z)|R^\rho$, $|g(z)|\leq 1/R^\rho$. Thus, by the maximum principle $|g(z)|\leq 1/R^\rho$ for all $z\in \Omega_R$.

Suppose $z\in \Omega_R$ with $|z|=r$, $1<r<R$. Then $$|f(z)|\leq |z|^\rho |2 A(R)-f(z)| R^{-\rho}\leq  r^\rho (2A(R)+|f(z)|) R^{-\rho}.$$ Rearranging, we find $$(R^\rho -r^\rho )|f(z)|\leq 2 r^\rho A(R),$$ or $$|f(z)|\leq \frac{ 2 r^\rho}{R^\rho -r^\rho}A(R).$$

\[l:p-lapp\] Let $\Omega=\{ z:\; {\operatorname{\rm Im}\nolimits}z\geq 0,\;|z|\geq 1\}$ and suppose $f$ is analytic in a neighborhood of $\Omega$, and there are constants $\rho_0$, 