TITLE: Series convergence and compact space
QUESTION [7 upvotes]: Let $K$ be a compact topological Hausdorff space. $\{x_n\}_1^\infty \subset K $ such that $x_i \not= x_j, i \not=j$ and $\{a_i\}_1^\infty \subset \mathbb{K}$. Show the folowing are equivalent:

for all $f \in C(K)$ the series $\sum_{n=1}^\infty a_nf(x_n)$ is convergent
$\sum_{n=1}^\infty |a_n|<\infty$

Implication from $2$ to $1$ is ofcourse easy but what about the second one ?

REPLY [7 votes]: Let us define the functionals $\lambda_N\in C(K)^*=M(K)$ by $$ \lambda_N=\sum_{n=1}^N a_n\delta_{x_n}.$$ Your first assumption implies, through the uniform boundedness principle that $$\sup_N\|\lambda_N\|_{M(K)}<\infty. $$ 
Now recall that $$ \sup\{|\lambda_N(f)|: \|f\|_{C(K)}\le 1\}= \|\lambda_N\|_{M(K)}\le \sum_{n=1}^N|a_n|.$$ Consider the function $f$ defined on the closed set $X_N=\{x_1,\ldots, x_N\}$ by $$ f(x_j)=\left\{\begin{array}{rr} \frac{|a_j|}{a_j} & a_j\ne 0 \\
0 & a_j=0.\end{array}\right. $$
By the Tietze extension theorem we may extend $f$  to a continuous function $\tilde f$ on the whole of $K$, preserving the supremum norm, which we note to be $\|f\|_\infty \le 1$. This yields $$ \|\lambda_N\|_{M(K)}\ge |\lambda_N(\tilde f)|=\sum_{n=1}^N|a_n|.$$
The condition $$ \sup_N \|\lambda_N\|_{M(K)}<\infty $$ now exactly means $$\sum_{n=1}^\infty |a_n|<\infty.$$