Interchange of sum and integral in Cauchy integral formula

Question

I am working with the Cauchy Integral Formula for a matrix $A$ over a closed contour $C$. I have the following calculation, I believe this is correct, but I don't understand why I am allowed to interchange the sum and integral?

$f(A)=\frac{1}{2\pi i}\int\limits_Cf(z)(zI-A)^{-1}dz \\ = \frac{1}{2\pi i}\int\limits_Cf(z) \frac{1}{z} \sum_{n=0}^\infty \frac{A^n}{z^n} dz \\ = \frac{1}{2\pi i} \sum_{n=0}^\infty \left(\int\limits_Cf(z) \frac{1}{z^{n+1}} dz \right) A^n \\ = \sum_{n=0}^\infty \left( \frac{1}{2\pi i}\int\limits_Cf(z) \frac{1}{z^{n+1}} dz \right) A^n \\ = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} A^n.$

I believe it is something to do with having $||A||<|z|$ and also uniform convergence, but don't really know why it is uniformly convergent, and why that means we can interchange the sum and integral.

Any explanation would be appreciated.

Thanks.

Answer 1

It looks like your countour has the property that $\|A\|<|z|$ for all points in the curve. By compactness you can deduce that $\|A\|\leq\delta|z|$ for some $\delta$ with $0<\delta<1$. So $\|A\|/|z|<\delta$, and this implies that the series $\sum_n A^n/z^n$ is uniformly convergent, because $$ \left\|\sum_n\frac{A^n}{z^n}\right\|\leq\sum_n\frac{\|A\|^n}{|z|^n}=\sum_n\left(\frac{\|A\|}{|z|}\right)^n\leq\sum_n\delta^n $$

For a uniformly convergent power series, the integral of the series is the series of the integrals (i.e. you can interchange the sum and the integral).