Disjointly supported functions

Let $(f_n)_n$ be sequence of real-valued continuous functions on a compact, Hausdorff space $K$ with pairwise disjoint (closed) supports satisfying $$0<\inf_n \|f_n\|\leq \sup_n\|f_n\|<\infty.$$

Take a sequence $(a_n)\in c_0$, that is, with $\lim_{n\to \infty}a_n=0$. Does the formula

$$\sum_{n=1}^\infty a_n f_n$$

define an element of $C(K)$? The convergence of the series above is supposed to be uniform, that is in the norm of $C(K)$.

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Yes, and the condition $0<\inf_n\|f_n\|$ isn't needed. Let $f$ denote the limit. Then for each $N\in \mathbb N$, $$\left\|f-\sum\limits_{n=1}^N a_nf_n\right\|=\left\|\sum\limits_{n={N+1}}^\infty a_nf_n\right\|=\sup\limits_{n\geq N+1} |a_n|\|f_n\|,$$

and this converges to $0$ as $N\to\infty$ because $(f_n)_n$ is uniformly bounded and $(a_n)$ converges to $0$. (Note how disjointness of the supports is used to infer that these series are well-defined and that the last equality is true.)

Answer to the original version, before the condition of uniform boundedness was added (I had posted after the edit but before I saw the edit):

Not in general. The series need not even represent a continuous function, let alone converge uniformly. E.g. let $K=\{0\}\cup\{\frac{1}{n}:n\in\mathbb Z^{>0}\}$ as a subspace of $\mathbb R$ with the usual topology, let $f_n=n^2\chi_{\{1/n\}}$, and let $a_n=\frac{1}{n}$.

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Thanks, but how do we know $f$ exists? –  Jan Veselý Jan 25 '12 at 17:28
Let $x\in K$. Let $A_x=\{n\in\mathbb N:f_n(x)\neq 0\}$. Note that $|A_x|\leq 1$. If $|A_x|=0$, then $f(x)=0$. If $A_x=\{n\}$, then $f(x)=a_nf_n(x)$. That is, each $x$ in $K$ is in at most one of the supports, so the series simply evaluates according to that one support, or is $0$ if $x$ is not in any of the supports. –  Jonas Meyer Jan 25 '12 at 17:31
Thank you very much. –  Jan Veselý Jan 25 '12 at 17:39