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!

  • $\begingroup$ What I mean is that we have to show this Cauchy sequence converges with respect to the $C^1$-norm. How was this achieved? $\endgroup$
    – RDizzl3
    Sep 27, 2013 at 20:45
  • $\begingroup$ 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 $f'_n$? $\endgroup$
    – RDizzl3
    Sep 27, 2013 at 20:47
  • 2
    $\begingroup$ 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 $f'$. $\endgroup$ Sep 27, 2013 at 20:51
  • $\begingroup$ 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 $f'=g$. $\endgroup$
    – RDizzl3
    Sep 27, 2013 at 21:01
  • $\begingroup$ 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 $C([a,b])$. $\endgroup$
    – RDizzl3
    Sep 27, 2013 at 21:04

1 Answer 1


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 $$\|f_n-f\|_{C^1}<\epsilon \qquad \forall\ n>n_0\ .$$


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