Prove the functional equation $E(\lambda+\mu)=E(\lambda)E(\mu)$

Question

This equation appeared in the Proof of Theorem 10.9 in the book Functional Analysis by Rudin, p252.

The Definition of $E(\lambda)$ is: $$E(\lambda)=\sum_{n=0}^{\infty}\frac {\lambda^n}{n!}a^n.$$ Where $\lambda$ is any complex number and $a$ is an element in a Banach algebra A with norm 1.

Here is what I've tried: \begin{align*} E(\lambda+\mu) &=\sum_{n=0}^{\infty}\frac {(\lambda+\mu)^n}{n!}a^n \\ &=\sum_{n=0}^{\infty}\sum_{j=0}^{n}(^n_j)\lambda^j\mu^{n-j}a^n \end{align*}

If I can interchange the order of summation, then the proof will go through, but I don't know what conditions needs to be satisfied to interchange the order.

Answer 1

Interchange of order in a summation in a Banach space is justified whenever the sum of the norms of all the terms converges.