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).
Best Answer
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 $f=0$.