Ahlfors proof of a substitution for differentiating the integral This is in lemma 3, page 121 of Ahlfors Complex Analysis, 3rd edition. I cannot understand some of the analytic arguments he is breezing over. The lemma says that for $\varphi(\zeta)$ a continuous function on the arc $\gamma$, the function
$$F_n(z) = \int_\gamma\frac{\varphi(\zeta)}{(\zeta - z)^n}d\zeta$$
is analytic in the regions defined by $\gamma$, and $F'_n = nF_{n+1}$. I understand the first part where we prove the continuity of $F_1$ and arrive at a formula
$$\frac{F_1(z) - F_1(z_0)}{z- z_0} = \int_\gamma \frac{\varphi(\zeta)}{(\zeta - z_0)(\zeta - z)}d\zeta$$
He then claims this quantity on the right tends to $F_2(z_0)$, which isn't totally clear to me. Do we show this by relying on the continuity of what's inside the integral?
Next we take the inductive hypothesis $F'_{n-1}(z) = (n-1)F_n(z)$. We get this nice thing for $F_n(z)  - F_n(z_0)$.
$$\Big[ \int_\gamma \frac{\varphi d\zeta}{(\zeta - z)^{n-1}(\zeta - z_0)} - \int_\gamma \frac{\varphi d\zeta}{(\zeta - z_0)^n}\Big] + (z-z_0)\int_\gamma \frac{\varphi d\zeta}{(\zeta - z)^n(\zeta - z_0)}$$
He says 'by the induction hypothesis applied to $\frac{\varphi}{\zeta  - z_0}$ the first term tends to zero for $z \rightarrow z_0$'. Further he says we can divide this by $(z-z_0)$ and the first term will tend to a derivative. All of this is unclear to me, especially how he's using the inductive hypothesis. The last term looks like it will tend to $F_{n+1}(z_0)$ to me, but that's about it. How are these derivatives/continuity being shown? Can anyone show more details?
 A: In response to your first question: yes, it looks like we are using continuity of the integrand. We can also make a more direct bounding argument. If $z_0\notin\gamma$ is in the interior of the region defined by $\gamma$, then we may find a neighborhood $N$ of $z_0$ which has nonzero (infimal) distance $D\ne 0$ from $\gamma$, and which therefore has the property that $$\Bigg|\frac{1}{(\zeta-z_0)(\zeta-z)}-\frac{1}{(\zeta-z_0)^2}\Bigg|=\Bigg|\frac{z-z_0}{(\zeta-z_0)^2(\zeta-z)}\Bigg|\leq \frac{|z-z_0|}{D^3}$$ for all $\zeta\in\gamma$ and $z\in N$.
Thus, from linearity of the integral and the triangle inequality, we may conclude that $$\Bigg|\frac{F_1(z)-F_1(z_0)}{z-z_0} - F_2(z_0)\Bigg|\leq \frac{|z-z_0|}{D^3}\int_\gamma \varphi(\zeta)d\zeta$$ and the RHS, which is an upper bound for the error, tends to zero as $z\to z_0$, since it is bounded by a constant times $|z-z_0|$. The desired result follows.
A: We have $\frac{d}{dz}\frac{\phi(\zeta)}{(\zeta - z)^{n}} = n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}$. Let $\Omega$ be the region defined by $\gamma$. By differentiating under the integral sign, for any $z_0 \in \mathbb{\Omega}$,
$$F_n'(z_0) = \int_{\gamma}n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}\,d\zeta.$$
All we need to do is justify differentiating under the integral sign. As usual, this is done using the dominated convergence theorem. Fix $z_0 \in \mathbb{C}$. As usual, to use the dominated convergence theorem to justify the differentiation, it suffices to find a dominating function $g(\zeta) \in L^1(\gamma)$ such that $|n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}| \leq g(\zeta)$ for all $z$ in some chosen neighborhood of $z_0$ and all $\zeta \in \gamma$. Here $L^1(\gamma)$ uses the arclength measure on $\gamma$. Integral with respect to this measure is $\int_{\gamma} g(\zeta) ds(\zeta) = \int_{I}g(\gamma(t))|\gamma'(t)|\,dt$. Pick a neighborhood $U$ of $z_0$ such that $d(U, \gamma) = C > 0$. Then for $z \in U$, $\zeta \in \gamma$,
$$|n\frac{\phi(\zeta)}{(\zeta - z)^{n + 1}}| \leq n\frac{|\phi(\zeta)|}{C^{n + 1}}.$$
$|\phi(\zeta)|$ is in $L^1(\gamma)$ since $\phi$ is continuous on $\gamma$, and therefore bounded since $\gamma$ is compact. So the differentiation is justified.
Edit: Above is a situation when we had to compute a derivative of an integral. We have $g(z) = \int_{\gamma}f(z, \zeta)\,d\zeta$ defined for $z$ in a neighborhood of $z_0$. We want to show that
$$g'(z_0) = \int_{\gamma}\frac{\partial f}{\partial z}(z_0, \zeta)\,d\zeta.$$
Here we assume that $f$ is complex differentiable in $z$. To show this equality, we use difference quotients:
$$\frac{g(z_0 + h) - g(z_0)}{h} = \int_{\gamma}\frac{f(z_0 + h, \zeta) - f(z_0, \zeta)}{h}\,d\zeta = \int_{a}^{b}\frac{f(z_0 + h, \gamma(t)) - f(z_0, \gamma(t))}{h}\gamma'(t)\,dt.$$
Now we want to take limits as $h \to 0$ and bring the limit inside the integral on the right to get the desired result. We do this using dominated convergence. It suffices to get a bound of the form
$$|\frac{f(z_0 + h, \gamma(t)) - f(z_0, \gamma(t))}{h}\gamma'(t)| \leq H(\gamma(t))|\gamma'(t)|\text{ with }\int_{a}^{b}H(\gamma(t))|\gamma'(t)|\,dt < \infty$$
for all $h$ in some disk $D_r(0)$ of our choice around $0$ and all $t \in (a, b)$. In other words, it suffices to find a bounding function $H \in L^1(\gamma)$ such that
$$|\frac{f(z_0 + h, \zeta) - f(z_0, \zeta)}{h}| \leq H(\zeta)\text{ for all }\zeta \in \gamma \text{ and } h \in D_r(0).$$
By the fundamental theorem of calculus, for any $h \in D_r(0)$,
$$|\frac{f(z_0 + h, \zeta) - f(z_0, \zeta)}{h}| \leq \sup_{z \in D_r(z_0)}|\frac{\partial f}{\partial z}(z, \zeta)|.$$
Thus it is enough to find $H \in L^1(\gamma)$ with $|\frac{\partial f}{\partial z}(z, \zeta)| \leq H(\zeta)$ for all $z \in D_r(z_0)$, $\zeta \in \gamma$.
