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).

  • $\begingroup$ 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 $C_0(\ell^2)$? $\endgroup$ – Giuseppe Negro Jan 10 at 11:00
  • $\begingroup$ 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. $\endgroup$ – SepulzioNori Jan 10 at 11:30
  • $\begingroup$ 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. $\endgroup$ – 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 $f=0$.

  • $\begingroup$ A very nice and exhaustive answer, thank you very much. $\endgroup$ – SepulzioNori Jan 10 at 15:53
  • $\begingroup$ I agree, great answer, thank you. $\endgroup$ – Giuseppe Negro Jan 10 at 17:20

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