Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Your equation defining $d(x,y)$ doesn't have $x$ or $y$ appearing in the right hand side... I assume you meant $d(f,g)$? – Zach L. Nov 29 '12 at 6:17
@ZachL. Yes thank you for noting that. – drew Nov 29 '12 at 17:40
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. – 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 $C([a,b];\mathbb R^n)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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