Prove that $C^1([a,b])$ with the $C^1$- norm is a Banach Space. Consider the space of continuously differentiable functions,
$$C^1([a,b]) = \{f:[a,b]\rightarrow \mathbb{R}\mid f \text{ differentiable with }f' \text{ continuous}\}$$ with the $C^1$-norm
$$\lVert f\rVert := \sup_{a\leq x\leq b}|f(x)|+\sup_{a\leq x\leq b}|f'(x)|.$$
Prove that $C^1([a,b])$ is a Banach Space.
This was the proof we were given:
Assuming $C^1([a,b])$ is a normed linear space all we need to show is completeness. Let $(f_n)$ be a Cauchy Sequence in $C^1([a,b])$ with respect to the $C^1$-norm. Then each $f_n,f'_n\in (C([a,b]),\|\cdot\|_{\sup})$. We know that $C([a,b])$ is complete and thus there exists $f,g\in C([a,b])$ such that $f_n\rightarrow f$, and $f'_n\rightarrow g$ (uniformly) with respect to $\|\cdot\|_{\sup}$. If we let
$$ F_n(x) = \int_a^x f_n(t)dt, \hspace{2mm}  F(x) = \int_a^x f(t)dt $$
then $F_n\rightarrow F$ uniformly because
$$\lVert F_n-F\rVert_{\sup}\leq \sup_{a\leq x\leq b}\int_a^x|f_n(t)-f(t)|dt\leq \lVert f_n-f\rVert_{\sup}<\epsilon.$$
From the fundamental theorem of calculus:
$$f_n(x)-f_n(a) = \int_a^x f'_n(t)dt $$
Since $f'_n\rightarrow g$ uniformly then
$$ \int_a^xf'_n(t)dt\rightarrow \int_a^x g(t)dt $$
Since we know that $f_n\rightarrow f$ uniformly,
$$f(x)-f(a) = \int_a^x g(t) dt $$
which by the fundamental theorem of calculues implies $f'=g$. So we know have $f_n\rightarrow f$ and $f'_n\rightarrow g=f'$ which mean $f_n\rightarrow f\in C^1([a,b])$ with respect to $C^1$-norm. So every cauchy sequence converges. Hence $C^1([a,b])$ is a Banach Space.
So I understand most of the proof. Where I get confused is that how did we actually show this satisfies the $C^1$-norm? Maybe I don't understand what this norm actually does.
Thank you for any help, comments and advice!
 A: When $(f_n)_{n\geq1}$ is a Cauchy sequence with respect to the $C^1$-norm, then given an $\epsilon>0$ there is an $n_0$ with
$$\eqalign{|f_m(x)-f_n(x)|+|f_m'(x)-f_n'(x)|&\leq \sup_t|f_m(t)-f_n(t)|+\sup_t|f_m'(t)-f_n'(t)|\cr &=\|f_m-f_n\|_{C^1}<\epsilon\cr}$$
for all $x\in[a,b]$ and all $m$, $n>n_0$. It follows that both $(f_n)_{n\geq1}$ and $(f_n')_{n\geq1}$ are Cauchy sequences with respect to the $\sup$-norm and so converge  uniformly to functions $f$ and $g\in C\bigl([a,b]\bigr)$. Furthermore we know  that under the given circumstances the limit function $f$ is differentiable and that  $f'=g$.
It remains to prove that the given sequence $(f_n)_{n\geq1}$ converges to $f$ with respect to the $C^1$-norm. To this end let an $\epsilon>0$ be given. Since the $f_n$ and the $f_n'$ converge uniformly to $f$ and $f'$ there is an $n_0$ with
$$\|f_n-f\|_\sup<{\epsilon\over2},\qquad\|f_n'-f'\|_\sup<{\epsilon\over2}\qquad \forall\ n>n_0\ ,$$
and this implies
$$\|f_n-f\|_{C^1}<\epsilon \qquad \forall\ n>n_0\ .$$
