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.

  • $\begingroup$ If $X$ is not compact then your "norm" is potentially not a norm because it can take the value $+\infty$. You can consider instead the space of bounded continuous functions from $X$ to $V$; that will be complete if $V$ is a Banach space. (If $V$ is not complete then neither is $C(X,V)$, even if $X$ is compact.) $\endgroup$ – Nate Eldredge Feb 13 '17 at 19:24
  • $\begingroup$ If $X$ isn't compact $\|f\|_{\sup}$ may not even be a norm. $\endgroup$ – Umberto P. Feb 13 '17 at 19:24
  • $\begingroup$ Ok I edited it to the space of bounded functions $\endgroup$ – edenstar Feb 13 '17 at 19:27
  • $\begingroup$ A solution is to look at things like $C_c(X,V)=\{ f: X\to V\mid \mathrm{supp}(f)\text{ is compact}\}$, $C_0(X,V)=\{f: X\to V\mid \forall\epsilon>0 \exists K\subset X \text{ compact s.t.} \|f(x)\|<\epsilon \forall x\notin K\}$. The first space is not complete, the second is. $\endgroup$ – s.harp Feb 13 '17 at 19:27
  • $\begingroup$ Also if $X$ is sigma compact (with $\bigcup_n K_n = X$) you can give $C(X,V)$ a complete metric via $d(f,g)=\sum_n 2^{-n}\frac{\|f-g\|_{K_n}}{1+\|f-g\|_{K_n}}$. (This and my previous comment suppose that $V$ is Banach) $\endgroup$ – s.harp Feb 13 '17 at 19:36

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 $x$.

  • $\begingroup$ What is the argument that the pointwise limit is bounded? $\endgroup$ – s.harp Feb 13 '17 at 19:32
  • $\begingroup$ (Nevermind it is clear, if $N$ is so that $\|f_n-f_m\|<1$ for all $n,m>N$ and $\|f_n\|<C$ then $\|f_m\|≤\|f_m-f_n\|+\|f_n\|≤C+1$ for any $m>N$. It follows the pointwise limits must be bounded by $C+1$.) $\endgroup$ – s.harp Feb 13 '17 at 19:39
  • $\begingroup$ @NateEldredge $C$ is the bound of a specific $f_n$, if I replace $1$ by $\epsilon$ I may not have $\|f_m-f_n\|<\epsilon$ for all $m$ greater than some $N$ and this same specific $n$ from before. $\endgroup$ – s.harp Feb 13 '17 at 19:45
  • $\begingroup$ This. A banach space should have a normed topology provided the right cauchy parameters. $\endgroup$ – McTaffy Feb 13 '17 at 20:03
  • $\begingroup$ @s.harp: Oops, yes. I misread the quantifiers. $\endgroup$ – Nate Eldredge Feb 13 '17 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.