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

enter image description here

No, it is not complete metric space: by Stone-Weierstrass thm we know that $|x|$ can be uniformly approximated by sequence of polynomials which are clearly $\mathcal{C}^1[0,1]$, but $|x|$ is not $\mathcal{C}^1$. Is my argument correct?

share|cite|improve this question
As Pete points out in his answer the idea can be made into an argument (although I myself wouldn't apply Stone-Weierstrass, that's quite an overkill for this question). However, you should pick another function, e.g. $x \mapsto |x-1/2|$ (which you can approximate uniformly by explicit polynomials -- how?) Note that the function $|x| = x$ on $[0,1]$ is $C^1$. – t.b. Jul 22 '12 at 13:36
Note that, uniform convergence preserves continuity, but it does not preserve differentiability. – Mhenni Benghorbal Jul 22 '12 at 13:39
so my argument is incorrect in $[0,1]$ :( but I dont know how to prove the result Mr. Pete has written in his answer – Un Chien Andalou Jul 22 '12 at 13:45
@Patience: Switching from $|x|$ to (say) $|x-\frac{1}{2}|$ seems like a very minor adjustment. Is there something else in my answer that you don't understand? – Pete L. Clark Jul 22 '12 at 13:58
how to show $C[0,1]$ is dense in $C^1[0,1]$? – Un Chien Andalou Jul 22 '12 at 14:08
up vote 2 down vote accepted

[Added: By $C[0,1]$ I mean the set of continuous functions $f: [0,1] \rightarrow \mathbb{R}$ endowed with the metric $d(f,g) = \max_{x \in [0,1]} |f(x)-g(x)|$. This is a complete metric space by the Cauchy Criterion for Uniform Convergence.]

Yes. To recap it: you have a complete metric space, $\mathcal{C}[0,1]$, and a subspace, $\mathcal{C}^1[0,1]$, which is not closed (rather, it is proper and dense). Therefore $\mathcal{C}^1[0,1]$ cannot be complete.

Added: As t.b. points out, the absolute value function is $C^1$ on the interval $[0,1]$, so you should pick something else (e.g. what t.b. says). I also agree that Weierstrass Approximation is much more than you need here. For instance, in Example 7 of these notes I show -- in an intentionally clunky, hands-on fashion -- that the absolute value function is a uniform limit of $C^1$-functions on $[-1,1]$.

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.