Completion of this metric space

Question

Let $d'$ be a metric on $C^1([0,1])$ as follows: $$ d':\ C^{1}\left([0,1]\right)^{2}:\ |f(0)-g(0)| + \sup\left\{ |f'(x)-g'(x)| \mid x\in[0,1] \right\} $$

I've already managed to prove that this is indeed a metric and that the metric space $C^1([0,1]), d'$ is not complete: $$ f_n:\ [0,1]\rightarrow \mathbb{R}:\ x \mapsto \sqrt{\left(x-\frac{1}{2}\right)^2 + \frac{1}{n^2}}$$

I'm a little stuck on finding the completion. I think the completion will be based on $C([0,1])$ but I don't quite know how to adapt $d'$ to work for non-differentiable functions like $\left|x-\frac{1}{2}\right|$.

Just throwing away the second part of the metric is at least not good enough. Any hints?

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

Actually, the space $C^1([0,1])$ is complete with this metric. Your sequence $f_n$ is not Cauchy in this metric. Indeed, if it was, the sequence $f_n'$ would converge uniformly, but the pointwise limit of this sequence is a discontinuous function.

To prove completeness, show that $d'$ is comparable (with fixed multiplicative constants) to the usual metric, the completeness of which is discussed in Prove that $C^1([a,b])$ with the $C^1$- norm is a Banach Space.

Indeed, on one hand $$ d'(f,g) \le \sup|f-g| + \sup|f'-g'| $$ while on the other, integration yields $$ \sup |f-g| \le |f(0)-g(0)| + \int_0^1 |f'-g'| \le d'(f,g) $$ which shows $$ \sup|f-g| + \sup|f'-g'| \le 2d'(f,g) $$

edited Apr 13 '17 at 12:21

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answered Jun 14 '15 at 22:26

Thank you very much for catching my error! Could you prove convergence for any Cauchy sequence (of this metric)? – Syd Kerckhove Jun 14 '15 at 22:28
See the expanded answer. – Jun 14 '15 at 22:30
Amazing! I wish I could upvote this answer more than once – Syd Kerckhove Jun 14 '15 at 22:31

Completion of this metric space

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