Justification of interchanging limit and integral in the proof of the derivative of Cauchy integral formula

Question

On Stein and Shakarchi's Complex Analysis book, in the course of proving that for $f$ a holomorphic (i.e. differentiable) function on an open set, and $C$ a circle whose interior is also contained in the domain of $f$, $$f^{(n)} = \frac{n!}{2\pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta$$ by induction on $n$, the author makes the following assertion: if we let $A=\frac{1}{\zeta-z-h}, B = \frac{1}{\zeta-z}$ we have that $$\frac{f^{(n-1)}(z+h)-f^{(n-1)}(z)}{h} = \frac{(n-1)!}{2\pi i}\int_C \frac{f(\zeta)}{(\zeta-z-h)(\zeta-z)}(A^{n-1}+A^{n-2}B+...+B^{n-1}) d\zeta.$$

I get everything until this point. However, I do not get why the following claim by the author implies that I can interchange limit and integral: "but observe that if $h$ is small, then $z+h$ and $z$ stay at a finite distance from the boundary circle $C$, so in the limit as $h \to 0$, we find that the quotient converges to $$\frac{n!}{2\pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta.$$

I know that we can interchange the limit and integral if the limit is uniform. However, it seems very complicated to determine that the integrand $g_m :=\frac{f(\zeta)}{(\zeta-z-h_m)(\zeta-z)}(A^{n-1}+A^{n-2}B+...+B^{n-1})$ converges uniformly to $g :=\frac{f(\zeta)}{(\zeta-z)^{n+1}}$ as $m \to \infty$ directly from the definition of uniform convergence. (Here, $\{h_m\}$ is an arbitrary sequence approaching $0$. The $h$'s in the definition of $A$ is also subtituted by $h_m$ in the definition of $g_m$.) How can the author so quickly justify the switching of the integral and the limit?

Answer 1

Define $g : C \times C^c \to \mathbb{C}$ by $$g(\zeta, z) = \frac{f(\zeta)}{(\zeta - z)^{n}}.$$ Let $g_z : C \times C^c \to \mathbb{C}$ denote the derivative of $g$ wrt $z$. Fix $z \in C^c$. You want to prove that $$\sup_{\zeta \in C}|\frac{g(\zeta, z + h) - g(\zeta, z)}{h} - g_z(\zeta, z)| \to 0 \text{ as } h \to 0.$$ By the FTC, for $\zeta \in C$, and $|h| < r := \text{dist}(z, C)$, $$\frac{g(\zeta, z + h) - g(\zeta, z)}{h} = \int_{0}^{1}g_z(\zeta, z + th)\,dt.$$ Hence for $\zeta \in C$ and $|h| < r$, \begin{align} |\frac{g(\zeta, z + h) - g(\zeta, z)}{h} - g_z(\zeta, z)| &= |\int_{0}^{1}(g_z(\zeta, z + th) - g_z(\zeta, z))\,dt| \\ &\leq \int_{0}^{1}|g_z(\zeta, z + th) - g_z(\zeta, z)|\,dt \\ &\leq \sup_{z_1 \in D_{|h|}(z)}|g_z(\zeta, z_1) - g_z(\zeta, z)| \\ &\leq \omega_{g_z, C \times \overline{D_{|h|}(z)}}(|h|), \end{align} where $\omega_{g_z, C \times \overline{D_{|h|}(z)}}$ is a modulus of continuity for the uniformly continuous function $g_z$ on $C \times \overline{D_{|h|}(z)}$. $g_z$ is indeed uniformly continuous on $C \times \overline{D_{|h|}(z)}$ because $g_z$ is continuous on $C \times C^c$, and $C \times \overline{D_{|h|}(z)}$ is compact. For $|h| < r/2$, $$\omega_{g_z, C \times \overline{D_{|h|}(z)}}(|h|) \leq \omega_{g_z, C \times \overline{D_{r/2}(z)}}(|h|) \to 0 \text{ as } h \to 0.$$ This completes the proof.

Edit: Here are some details on "modulus of continuity". I like to think of the "canonical" modulus of continuity of a function $f : X \to Y$ between two metric spaces as the map $\omega : [0, \infty) \to [0, \infty]$ defined by $$\omega(\delta) = \sup_{d(x_1, x_2) \leq \delta}d(f(x_1), f(x_2)).$$ Note that $d(x_1, x_2) \leq \delta \implies d(f(x_1), f(x_2)) \leq \omega(\delta)$. It is easy to show that $f$ is uniformly continuous if and only if $\omega(\delta) \searrow 0$ as $\delta \searrow 0$.

For this problem, we don't need to compute the modulus of continuity explicitly, i.e. we don't need hard quantitative estimates on it. For this problem, all was needed was that $g_z$ was continuous so that we could conclude that it's modulus of continuity has the limit $0$ at $0$.