Question from Real Analysis by Haaser and Sullivan

Let X be the set of all continuous functions from [a,b] into $R^n$ and let $d$ be defined by $$d(f,g)=max(|f(t)-g(t)|:t\in[a,b]) $$ Show that (X,d) is a complete metric space.

What I need to show is that for every Cauchy sequence in X then the cauchy sequence converges to a point in X. Now I know $R^n$ is a complete metric space. So if $f_n(t)$ is a Cauchy sequence then $f_n(t)$ converges to some point say $f(t)\in R^n$. Now this means that $$d(f_n(t),f(t))\lt\partial.$$ Since $f$ continuous and a $0\lt\partial$ exists then $\forall \epsilon \gt 0$ $$d(f_n,f)\lt\epsilon.$$ Hence (X,d) is a complete metric space.

If someone could explain the difference between uniformly continuous and continuous? Also in general how one shows uniform continuity vs. continuity?

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    $\begingroup$ Your equation defining $d(x,y)$ doesn't have $x$ or $y$ appearing in the right hand side... I assume you meant $d(f,g)$? $\endgroup$ – Zach L. Nov 29 '12 at 6:17
  • $\begingroup$ @ZachL. Yes thank you for noting that. $\endgroup$ – drew Nov 29 '12 at 17:40
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    $\begingroup$ The last paragraph is not related to the rest and should be postponed to another question. About the proof you propose: nothing guarantees a priori that the limit $f$ is continuous, in fact to show that it is is the heart of the proof you are asked to write down. $\endgroup$ – Did Dec 2 '12 at 12:50

The question is almost a duplicate of Space of bounded continuous functions is complete but it's easier because continuous functions on $[a,b]$ are automatically bounded. So, I repeat the proof by Matt N. with slight modifications.

Given a Cauchy sequence $(f_n)$ in $C([a,b];\mathbb R^n)$, we first show that the sequence has a pointwise limit. For this we note that because $f_n$ is Cauchy with respect to the supremum norm, it follows that $f_n(x)$ is a Cauchy sequence in $\mathbb{R}^n$ for any $x$ in $[a,b]$. But $\mathbb{R}^n$ is complete and hence the limit $\lim_{n \to \infty} f_n (x)$ exists in $\mathbb{R}^n$. Let $f(x)$ denote this limit.

Now we want to show that $f_n$ converges to $f$ uniformly, that is $\sup_{[a,b]} \| f - f_n |\ \to 0$. Given $\varepsilon > 0$, we have $N$ such that for $n,m \geq N$, $\|f_n(x) - f_m(x) \| < \frac{\varepsilon}{2}$, again because $f_n$ is Cauchy. Passing to the limit $m\to\infty$ we get $\|f_n(x) - f (x) \| \le \frac{\varepsilon}{2}<\epsilon$.

Finally, now that we have convergence in norm, we can apply the uniform limit theorem (proof here) to get that $f$ is continuous and hence $f$ is in $C([a,b];\mathbb R^n)$.


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