# When is a sum of continous functions continous?

Let $$(f_{\alpha})_{\alpha \in A}$$ be a family of real valued continous functions on a topological space $$X$$ such that for each $$x\in X$$ $$f_\alpha (x)\neq 0$$ for only finitely many $$\alpha\in A$$. Then we can define $$f=\sum_{\alpha \in A}f_\alpha$$ .

In general we can can not expect $$f$$ to be continous, for example if $$X$$ consists only of discrete points together with one limit point. Are there any conditions on $$X$$ that will guarantee the continouity of $$f$$?

Any help appreciated!

I do not have the complete answer as of yet. But maybe a partial answer will already help you.

I noticed, that this is not even true if X = R together with the standard topology, since you can take any strictly monotonous convergent sequence $$x_n$$. Then construct a sequence of neighborhoods $$U_n$$ around each $$x_n$$ with $$U_n\cap U_m$$ is the empty set whenever $$m\neq n$$ (Construct this for example by taking the open balls around $$x_n$$ with diameter smaller then $$\min(x_n-x_{n-1},x_{n+1}-x_n)$$) Now one can construct a sequence of continuous $$f_n$$ such that $$f_n(x_n) = n$$ and the support of $$f_n$$ is restricted to $$U_n$$. Then for every point of the real line, there is at most one n such that $$f_n\neq 0$$ Therefore I can write the sum. Furthermore $$f(x)=0$$ but there exists a convergent sequence under which the images diverge. Therefore the limit of the sum of $$f_n$$ can not be continuous.

I think this construction can be modified to include at least all benach spaces.

PS: You could change your Assumptions to: For each x in X there exists a neighborhood $$U_x$$ such that $$f_\alpha(U_x))={0}$$ only for finitely many $$\alpha$$. Then my as well as your example fail to comply, and this might be true for at least any Banach space, since then any sequence converging to x would be at some point in $$U_x$$ and in $$U_x$$ we can restrict the sum to a finite sum.

The standard condition (used in paracompactness considerations, e.g.) is that the family $$(f_\alpha)^{-1}[(0,\rightarrow)]$$, $$\alpha \in A$$ is locally finite. Every point of $$X$$ then has a neighbourhood $$U_x$$ on which all but finitely many $$f_\alpha$$ vanish. This clearly implies continuity.

If $$X$$ is Hausdorff, the counterexample you give is essentially the only one: the implication holds iff the set of isolated points of $$X$$ has no limit point. Indeed, such continuous $$f_\alpha$$ have to be supported on a finite set of isolated points, and a discontinuity of $$\sum f_\alpha$$ can only happen at a limit point of isolated points.

For general $$X$$, the condition you need is that there is no disjoint union of finite open sets that has a limit point.