Series convergence and compact space 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 ?
 A: 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 $$\sum_{n=1}^\infty |a_n|<\infty.$$
A: Scroll down (or maybe soon will be up). There is a solution and is not this one.
This is as far as I have got it, a tiny bit more than my comment above.

Restriction of the problem: Assume that the set $S\subset K$, where $\{x_n\}$ accumulates, contains only finitely many elements of $X:=\{x_n\}$. 

Suppose, in order to get a contradiction, that 
$$\sum_{n=1}^{\infty}|a_n|=\infty$$
Then there is a sequence $b_n>0$ such that $b_n\to 0$ and $$\sum_{n=1}^{\infty}|a_n|b_n=\infty$$
Let us define $$f(x_n):=\begin{cases}\text{sgn}(a_n)b_n&\text{ for }x_n\notin S\\0&\text{ for }x_n\in S\end{cases}$$
We can extend $f$ to a continuous function by putting $f(x)=0$ for $x\in S$ and to $K$ by using Tietze extension theorem.
Then $$\sum_{n=1}^{\infty}a_nf(x_n)=\sum_{n\in\mathbb{N}:\ x_n\notin S}|a_n|b_n=\sum_{n=1}^{\infty}|a_n|b_n-\text{(a finite number)}=\infty$$

Another case: Assume that $\delta: C(K)\to \mathbb{K}$, defined by $f\stackrel{\delta}{\mapsto} \sum_{n=1}^{\infty}a_nf(x_n)$ is a continuous linear functional. 

Then by Riesz-Markov-Kakutani there is a Radon measure $\mu$ on $K$, with finite total variation $|\mu|(K)<\infty$ such that $$\delta(f)=\int_K f d\mu$$
We see that $\mu=\sum_{n=1}^{\infty}a_n\delta_{x_n}$, where $\delta_{x_n}$ is the Dirac measure supported at $x_n$. Then $$\sum_{n=1}^{\infty}|a_n|=|\mu|(K)<\infty.$$
A: 1 to 2:
If $\mathbb K=\mathbb R$.(Same thought for $\mathbb K=\mathbb C$)
We are going to build a continuous function that does the job. What we want to achieve is $f(x_n)=sign (a_n)$ and then $a_n\cdot sign(a_n)=|a_n|$.
We suppose that $K$ is a connected space otherwise you can do what we will do in the connected components of $K$.
We proceed by induction.
Let $U_1$ be a region of $x_1$ and $U_2$ be a region of $x_2$ such that $U_1\cap U_2= \emptyset$.
Now let $U_{31},U_{32}$ be regions of $x_3$ and $U_{11},U_{22}$ b e regions of $x_1,x_2$ respectivly such that $U_{31}\cap U_{11}=\emptyset$ and $U_{32}\cap U_{22}=\emptyset$. We take $U_3=U_{31}\cap U_{32}$ as a region of $x_3$ and $\widetilde {U_1}=U_1\cap U_{11}$ and $\widetilde {U_2}=U_2\cap U_{22}$ the new regions of $x_1,x_2$.
We continue By finding $\widetilde{U_3}$ with which we will have now problem because $\widetilde{U_3}\subset U_3$.
Let  $f_1:K\to \mathbb K$ such that $f_1(x)=sign(a_1)$ if $x\in \widetilde{U_1}$ and $0$ otherwise.
Let  $f_2:K\to \mathbb K$ such that $f_2(x)=sign(a_2)$ if $x\in \widetilde {U_2}$ and $0$ otherwise.
 and 
we let $f_3:K\to \mathbb K$ such that $f_3(x)=sign(a_3)$ if $x\in \widetilde {U_3}$ and $0$ otherwise.
and so on...
Now if we let $\widetilde{f_i} $ to be the "smoothness" of $f_i$, meaning to "glue" the jump from $1$(or $-1$) to $0$, we can see that $\widetilde{f_i}$ are continuous.
Let $f=\cup \widetilde{f_i}$ and because $K$ is compact $f$ is continuous with the property $f(x_n)=sign(a_n)$.
A quick thought of mine.
Try to do the gluing if you like:)
