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.

• What I mean is that we have to show this Cauchy sequence converges with respect to the $C^1$-norm. How was this achieved? Sep 27, 2013 at 20:45
• I know the $C^1$-norm is a sum of the $\sup$-norms, I see that. How in particular was convergence with $\sup$-norm shown for $f'_n$? Sep 27, 2013 at 20:47
• We used completeness of $C([a,b])$ to argue that the $f_n'$ must converge uniformly to some function, call it $g$. Then the fundamental theorem of calculus was used to show that in fact $g$ is precisely the function $f'$. Sep 27, 2013 at 20:51
• I whats happening now. It's because $f_n,f'_n$ are in $C([a,b])$ that they converge uniformly to $f,g$ respective with respect to $||\cdot||_{sup}$. The fundamental theorem is simply used to show $f'=g$. Sep 27, 2013 at 21:01
• I see whats happening now. It's because $f_n,f'_n\in C([a,b])$ that we know they converge to $f,g$ WRT the $||\cdot||_{sup}$. Fundamental theorem is used simply to show $f'=g$. I was failing to notice that it was already converging uniformly WRT $||\cdot||_{sup}$ because they are in $C([a,b])$. Sep 27, 2013 at 21:04
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 $$\|f_n-f\|_{C^1}<\epsilon \qquad \forall\ n>n_0\ .$$