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Let $\mathcal{D}$ be the subspace of $C[0,1]$ (with uniform metric) consisting of the continuous functions $[0,1]\to \mathbb{R}$ that are differentiable on $(0,1)$. Is $\mathcal{D}$ complete?

I'm thinking no, because I'm thinking that $\mathcal{D}$ is not closed, but I don't know how to show this.

Could somebody please let me know whether I'm right in my reasoning, and if so, how to show this?

If I'm not correct in my reasoning, then if you could please steer me in the right direction, I would very much appreciate it.

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  • $\begingroup$ A metric space is ALWAYS closed (since you say subspace I assume you mean subspace topology), so that doesn't apply here. $\endgroup$ – user223391 Dec 3 '15 at 5:47
  • $\begingroup$ If uniform metric means $||f||_0 = \sup_x |f(x)|$ then the space is, in fact, not complete. An example of a Cauchy-Sequence which does not converge can be constructed by searching for a sequence the derivatives of which are not bounded. There should be several questions here already addressing this (including answers). $\endgroup$ – Thomas Dec 3 '15 at 5:49
  • $\begingroup$ @Thomas, what should I search for, specifically? $\endgroup$ – ALannister Dec 3 '15 at 5:52
  • $\begingroup$ If you happen to know that the polynomials are dense in $C[0,1]$ (do you?) then the answer is immediate. Finding one example is less interesting and more messy. $\endgroup$ – B. S. Thomson Dec 3 '15 at 17:12
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Hint: Consider $f_n(x) = |x-1/2|^{1+1/n}, \,n = 1,2,\dots$

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  • $\begingroup$ you messed up the exponent. Can you please edit your answer and fix it? Also, I'm supposed to show that that converges to what, exactly? $\endgroup$ – ALannister Dec 3 '15 at 6:29
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    $\begingroup$ @JessyCat converges to $|x-1/2|$ which is not differentiable at $1/2$. $\endgroup$ – Mirko Dec 3 '15 at 6:34
  • $\begingroup$ @zhw and Mirko, thank you both! $\endgroup$ – ALannister Dec 3 '15 at 6:35
  • $\begingroup$ @zhw, one more thing...are you sure that $f_{n}(x)=\vert x - 1/2 \vert ^{1+1/n}$ is differentiable on $(0,1)$? $\endgroup$ – ALannister Dec 3 '15 at 16:20
  • $\begingroup$ Yes, certainly. It's differentiable everywhere on $\mathbb R.$ Just use the definition of the derivative at $1/2.$ The uniform convergence of $f_n$ to $|x-1/2|$ requires a little more work. $\endgroup$ – zhw. Dec 3 '15 at 18:25

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