What happens to a Banach space of functions when the domain is not compact?

Question

Let $X$ be a noncompact topological space. Let $V$ be a normed vector space. Consider the set of bounded continuous functions $C_B(X,V)$ from $X$ to $V$. Define a norm on $C_B(X,V)$ by $\|f\|_{\sup} = \sup_{x\in X} \| f(x) \|$ If $X$ is compact then this space is complete (and thus Banach). That is every cauchy sequence converges to a point in $C_B(X,V)$. I suspect that if $X$ is not compact then the space is no longer complete but I am not sure how to show it. I cannot think of an example of a cauchy sequence which does not converge because $X$ is not compact.

Answer 1

Notice that you need $V$ to be a Banach space (even in the case when $X$ is compact).

In this case it is complete. If you have a Cauchy sequence, $f_n$, in $C_B(X,V)$ then it is straightforward to see that it is pointwise cauchy (ie that for each $x\in X$, $f_n(x)$ is cauchy in $V$. Then since $V$ is a Banach space you get that there is a pointwise limit, and it is a standard $\epsilon/3$ argument to show that the limit is continuous.

Bounded is a little trickier. To see this note that in any metric space, cauchy sequences are bounded. Thus there is an $M$ such that $\|f_n\|\leq M$ for all $n$. By the definition of the norm this means that $|f_n(x)|\leq M$ for all $n$ and all $x$, thus $\lim_{n\rightarrow\infty}f_n(x)\leq M$ for all $x$.