On the third edition of Ahlfors' Complex Analysis, page 121 Lemma 3 it states:
Suppose $\phi(\zeta)$ is continuous on the arc $\gamma$. Then the function
\begin{equation*} F_n(z)= \int_{\gamma} \frac{\phi(\zeta)}{(\zeta-z)^n} d\zeta \end{equation*}
is analytic in each of the regions determined by $\gamma$ , and its derivative is $F'_n(z)=nF_{n+1}(z)$
It begins by proving $F_1$ is continuous. Let $z_0$ be a point not in $\gamma$ and we choose a neighbourhood around that point $|z-z_0| < \delta$ so that it does not meet $\gamma$. Furthermore, we have $|\zeta-z| > \frac{\delta}{2}$ for all $\zeta \in \gamma$ when $|z-z_0| < \frac{\delta}{2}$. Then a few calculations yield
\begin{equation*} |F_1(z)-F_1(z_0)| < |z-z_0| \frac{2}{\delta^2} \int_{\gamma} |\phi||d\zeta| \end{equation*}
which according to the book proves the continuity of $F_1(z)$ at $z_0$. This is where I get stuck, I understand the rest of the proof. I know that $|z-z_0| < \frac{\delta}{2}$ but we still have a $\delta$ in the denominator and because $\int_{\gamma} |\phi||d\zeta|$ is bounded this means that the right hand side will become bigger as $\delta$ becomes smaller! I can't understand how this implies continuity and I am almost sure there must be a mistake or I am missing something huge.
Don't let $\delta$ get smaller! Leave $\delta$ fixed. Say $c=\frac1{\delta^2}\int_\gamma|\phi|\,|d\zeta|.$ You have $$|F_1(z)-F_1(z_0)|<c|z-z_0|.$$ Hence $F_1(z)\to F_1(z_0)$ as $z\to z_0$. (For instance, $|z-z_0|<\epsilon/c$ implies $|F_1(z)-F_1(z_0)|<\epsilon$.)