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Let $C^1([a,b])$ denote the space of all continously differentiable functions on $[a,b], a,b\in\mathbb{R}$.

On this space, define the following metric: $$ d(f,g)=d_{\infty}(f,g)+d_{\infty}(f',g'), $$ where $$ d_{\infty}(f,g):=\lVert f-g\rVert_{\infty}:=\sup_{x\in [a,b]}\lvert f(x)-g(x)\rvert $$ and, similarly, $d(f',g')_{\infty}$ is defined.

So I have to show that each Cauchy sequence $(f_n)$ in $C^{1}([a,b])$ converges in $C^1([a,b])$. Don't really have an idea, but at least I can write down what Cauchy sequence means.

Let $(f_n)$ be some Cauchy-sequence in $C^1([a,b])$, i.e. for all $\varepsilon >0$ there is some $N\in\mathbb{N}$ such that $$ \lVert f_n-f_m\rVert_{\infty}+\lVert f_n'-f_m'\rVert_{\infty}\leq\varepsilon~~\forall n,m\geq N. $$

Now, I have to find some $f\in C^1([a,b])$ such that $f_n\to f$, i.e. for all $\varepsilon >0$ there exists some $N\in\mathbb{N}$ such that $$ \lVert f_n-f\rVert_{\infty}+\lVert f_n'-f'\rVert_{\infty}\leq\varepsilon~~\forall n\geq N. $$

H. Hawks
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  • You can try to adapt the proof that $C^{0}([a,b])$ is complete w.r.t to $||.||_{\infty}$ – Dark Apr 27 '16 at 17:59
  • I would say that you need to think a bit differently. The sequence "clearly" converges to some unique function, and you need to show that that function is $C^1$. So the problem is not so much about finding a function as it is about trying to figure out things about a function you don't know so much about. – Arthur Apr 27 '16 at 18:00
  • Suppose $f_n$ is Cauchy with the above metric. Then show that $f_n $ converges to some $f$ and $f'_n $ converges to some $g$. All that remains is to show that $g=f'$. – copper.hat Apr 27 '16 at 18:19

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