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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.

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  • $\begingroup$ 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
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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 $x$.

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  • $\begingroup$ 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

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