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Let $a,b\in\mathbb{R}$ and $C^1([a,b])$ the space of the function that are continuously differentiable on the interval $[a,b]$. For $f,g\in C^1([a,b])$, let $$ d_{\infty}(f,g):=\lVert f-g\rVert_{\infty}:=\sup_{x\in [a,b]}\lvert f(x)-g(x)\rvert. $$

Show that $C^1([a,b]),d_{\infty})$ is not complete.

My idea is to use $$ C^1([a,b])\subset C([a,b]), $$ i.e. to find a sequence $(f_n)$ in $C^1([a,b])$ that converges to some $f\in C([a,b])$ with respect to the supremums norm $\lVert\cdot\rVert_{\infty}$ but which is not continously differentiable.

Then, $(f_n)$ should be a Cauchy sequence in $C^1([a,b])$ that does not converge in $C^1([a,b])$ (otherwise $f$ would be continously differentiable).

Now, I do have to find such a sequence... and did not have an idea yet.

EDIT

Maybe on $[-1,1]$ consider $$ f_n(x)=\sqrt{x^2+\frac{1}{n}},~~f(x)=\sqrt{x^2}. $$

$f(x)$ is not differentiable in $x=0$, hence not continuously differentiable on $[-1,1]$. The functions $f_n$ are continously differentiable on $[-1,1]$ and uniformly converge to $f$.

H. Hawks
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    That example works on $[a,b]$ after appropriate translations! I suggest you to post your - modified- edit as an answer to this question and then accept it! – Eman Yalpsid Apr 27 '16 at 16:48
  • What do you mean with appropriate translations? – H. Hawks Apr 27 '16 at 16:50
  • The example is OK if $a=-1, b=1$. You should now find a way to generalize it to an arbitrary interval. HINT: An affine change of variables $x=\alpha X+\beta $ transports your example sequence $f_n(x)$ on $[-1,1]$ into a sequence on $[\beta-\alpha , \alpha+\beta]$. – Giuseppe Negro Apr 27 '16 at 16:52
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    What I mean is that your $f(x)$ is not differentiable at $0$. If $0 \in [a,b] $, then this is a great example. Otherwise you will have to modify your $f$ such that this point of non-differentiability is moved into $[a,b]$. The change of variables @GiuseppeNegro suggested should work. – Eman Yalpsid Apr 27 '16 at 16:54
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    @Andrew Got it, thanks. – H. Hawks Apr 27 '16 at 17:04
  • I think I have to show that $(f_n)$ is a Cauchy-sequence. But that should be rather easy: Let $\varepsilon >0, \tilde{\varepsilon}=\varepsilon/2$. Then $\lvert f_n(x)-f_m(x)\rvert=\lvert f_n(x)-f(x)+f(x)-f_m(x)\rvert\leq\lvert f_n(x)-f(x)\rvert + \lvert f_m(x)-f(x)\rvert$. Moreover, $\lvert f_n(x)-f(x)\rvert\leq \lvert 1/n\rvert$ and, similarly, $\lvert f_m(x)-f(x)\rvert\leq\lvert 1/m\rvert$. Hence, for $N$ large enough, we have $\lvert f_n(x)-f_m(x)\rvert\leq 2\tilde{\varepsilon}=\varepsilon$ for all $n,m\geq N$. – H. Hawks Apr 27 '16 at 17:18
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    $\lvert f_n(x)-f(x)\rvert\leq \lvert 1/n\rvert \le 1/n$ is false – zhw. Apr 27 '16 at 19:19
  • @zhw.Why is that false? $\lvert f_n(x)-f(x)\rvert = \lvert \sqrt{x^2+\frac{1}{n}}-\sqrt{x^2}\rvert\leq\lvert\sqrt{x^2}+\sqrt{1/n}-\sqrt{x^2}\rvert=\lvert\sqrt{1/n}\rvert\leq\lvert 1/n\rvert=1/n$ – H. Hawks Apr 27 '16 at 19:31
  • @zhw Dont see why this is false. – H. Hawks Apr 28 '16 at 09:35
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    $\sqrt {1/n} > 1/n, n = 2,3,\dots $ But note $\sqrt {1/n}$ will give the result. – zhw. Apr 28 '16 at 15:38
  • You are right. Then I just write that $\lvert f_n(x)-f_m(x)\rvert\leq \lvert\sqrt{1/n}\rvert + \lvert\sqrt{1/m}\rvert\leq 2\tilde{\varepsilon}<\varepsilon$ for $n,m$ large enough. That should be correct then. – H. Hawks Apr 28 '16 at 15:44

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