I'm working on a exercice where I need to show that, for
- $E = \mathcal{C}^1([-1,1])$
- with the norm $\|f\| = \|f\|_\infty + \|f'\|_\infty$,
$(E, \| \cdot\|)$ is Banach.
I have already shown that $(\mathcal{C^0}([-1,1]),\|\cdot\|_\infty)$ is Banach and that $(E, \| \cdot\|)$ is a normed vector space. I am mostly interested in proving its completeness. I have already seen a different proof but I was wondering if the following is correct.
Here is what I have:
for a Cauchy sequence $(f_n)_{n\geq0} \subset E$, there exist
- $f \in \mathcal{C^0}([-1,1])$ such that $f_n\rightarrow f$ and
- $g \in \mathcal{C^0}([-1,1])$ such that $f_n'\rightarrow g$
when $n\rightarrow \infty $ since $(f_n)$ and $(f_n')$ are both in $\mathcal{C^0}([-1,1])$. So I want to show that $f' = g$.
Let $t\in[-1,1]$ and $\epsilon >0$, then by the convergence $\exists N$ such that
- $\epsilon>|f_n'(t)-g(t)| = \Bigl|\frac{f_n(t+h)-f_n(t)-R(h)}{h} - g(t)\Bigr|$ $\forall n \geq N$ and for some small enough $h$ .
and where $\lim_{h\rightarrow 0}\frac{R(h)}{h} = 0$.
If we let $n \rightarrow \infty$ (convergence of $f_n$ plus continuity of modulus), then we obtain that $\epsilon>\Bigl|\frac{f(t+h)-f(t)-R(h)}{h} - g(t)\Bigr|$ and then by letting $h \rightarrow 0 $ we obtain that $\epsilon>|f'(t) - g(t)|$.
Since $\epsilon$ and $t$ were arbitrarily chosen, we have that $f' = g$ and with a couple of extra steps we can go on to conclude that the space is complete.
I am unsure about the step where $h \rightarrow 0$. Is that legal? If not, could you please provide some explanation as to why?
Thanks in advance!