Uniform convergence to swap summation and integration

Question

I am integrating a meromorphic function $f(z)$ with pole $z-z_0$, $\int_{|z-z_0|=r}f(z)dz$. Expanding this as a Laurent series we get $\int_{|z-z_0|=r}\sum_{k=-\infty}^{\infty}a_k (z-z_0)^kdz$. As, $\sum_{k=-\infty}^{\infty}a_k (z-z_0)^k$ is uniformly convergent, we can re-write the integral as $\sum_{k=-\infty}^{\infty}a_k \int_{|z-z_0|=r} (z-z_0)^kdz$. Can anyone show me the formal steps, as how this integral and sum are getting swapped using uniform convergence?

Answer 1

If $(f_n)$ converges uniformly to $f$, then you have for every finite-length path $\gamma$, that $$\left|\int_{\gamma} f_n(z) dz - \int_{\gamma} f(z) dz \right| =\left|\int_{\gamma} f_n(z) -f(z) dz \right| \leq \int_{\gamma} \left| f_n(z) -f(z) \right| dz \leq ||f_n-f||_{\infty} L(\gamma)$$

where $L(\gamma)$ denotes the length of the path $\gamma$.

Because $||f_n-f||_{\infty}$ tends to $0$, you deduce that $$\lim_{n \rightarrow +\infty} \int_{\gamma} f_n(z) dz = \int_{\gamma} f(z) dz$$

i.e.

$$\lim_{n \rightarrow +\infty} \int_{\gamma} f_n(z) dz = \int_{\gamma} \lim_{n \rightarrow +\infty} f_n(z) dz$$