# Continuous functions vanishing at infinity on a non-locally-compact space

Let $$X$$ be a topological space which is not locally-compact (e.g., an infinite-dimensional Hilbert space). Let $$C_{0}(X)$$ denote the space of complex-valued, continuous functions vanishing at infinity on $$X$$, that is, an element $$f\in C_{0}(X)$$ is a complex-valued, continuous function on $$X$$ such that,for every $$\epsilon>0$$, there exists a compact $$K\subset X$$ such that $$|f(x)|\leq \epsilon$$ outside $$K$$. It is my understanding that $$C_{0}(X)$$ is a Banach space just as $$C_{0}(Y)$$ with $$Y$$ a locally-compact space.

However, in the second page of this article, it is stated that $$C_{0}(X)$$ only contains the zero function when $$X$$ is not locally-compact, but the statement is not proved. On the other hand, at the end of the first page of this article, it is said that $$C_{0}(X)$$ may be very small (which, I guess, means that there are non-vanishing elements in it), but the statement is also not proved. Furthermore, in the accepted answer of this question, it is given an example of a non-vanishing functions vanishing at infinity on a non-locally compact space, but nothing is said about its continuity.

Since I have not a strong background in functional analysis, I really do not know where to start to prove/disprove the previous statements, thus I would appreciate any hint, or any suggestion about references dealing with these matters explicitely (that is, by giving explicit proofs).

• Consider the function $f\colon \ell^2\to \mathbb R$ given by $$f(x_1, x_2, x_3, \ldots)=e^{-x_1^2}.$$ Isn't this function in $C_0(\ell^2)$? – Giuseppe Negro Jan 10 at 11:00
• I was thinking to a similar example, however, if we fix $\epsilon$, then $|f(x)|\geq\epsilon$ whenever $x=(x_{1},x_{2},x_{3},...)$ is such that $x_{1}^{2}\leq \ln(\epsilon)$ and $x_{j}$ is arbitrary for $j>1$, and I do not know if this is a compact set. – SepulzioNori Jan 10 at 11:30
• Of course that set is not compact. You are totally right. Sorry, my example is stupid; even the function $$f(x_1, x_2)=e^{-x_1^2}$$ is not in $C_0(\mathbb R^2)$. The right example would be $$f(x_1, x_2, \ldots) = e^{-x_1^2 -x_2^2 -\ldots}.$$ And now it is not so obvious that this is in $C_0(\ell^2)$. Actually, I am quite sure it is not. – Giuseppe Negro Jan 10 at 11:49

The space $$C_0(X)$$ can certainly be non-trivial even if $$X$$ is not locally compact. Just take a locally compact space $$Y$$, a space $$Z$$ that is not locally compact and let $$X$$ be the disjoint union of $$Y$$ and $$Z$$. Every $$f\in C_0(Y)$$ induces a a function $$F\in C_0(X)$$ by setting $$F|_Y=f$$, $$F|_Z=0$$.
That $$C_0(X)$$ may be small if $$X$$ is not locally compact is not so much a rigorous statement, but gives a good intuition. If $$X$$ lacks compact sets, then the condition $$f\in C_0(X)$$ is quite rigid. In the example above there are still enough compact subsets to produce some functions in $$C_0(X)$$.
In the special case of infinite-dimensional normed spaces, $$C_0$$ is indeed trivial. To see this, let $$f\in C_0(X)$$ and $$\epsilon>0$$. By definition there exists $$K\subset X$$ compact such that $$|f|\leq \epsilon$$ on $$X\setminus K$$. By Riesz's lemma, $$K$$ has empty interior. Thus $$|f|\leq \epsilon$$ everywhere. Since $$\epsilon>0$$ was arbitrary, we conclude $$f=0$$.