proof of a Banach space Let X be a Banach space with the norm $||.||$ and f be a vector function which acts from $\mathbb                              {R}$ into the Banach space X.
Can you please show that the following space  is a Banach space ?
$BC^1[\mathbb{R},X]=\{x:\mathbb{R} \rightarrow X: x\in C^1 (\mathbb{R},X), |||x|||=\sup\left\{||x||, ||x'(t)||\}< \infty \right\}$
,i.e., the space of continously differentiable functions on $\mathbb{R}$ and bounded together with their derivatives.
My attempt is as follows:
Let $X$ be a Cauchy sequence in $BC^1[R,X]$ then, $|||f_n−f_m|||<ϵ,$ $n,m>N$ that is, $||f_n(x)−f_m(x)||<ϵ,$ and $||f_n′(x)−f_m′(x)||<ϵ$, ∀ϵ>0. I showed that  $f_n(x)\rightarrow f(x)\in C[R,X]$ but i could not showed that $f'_n(x)→g(x)\in C^{1}[R,X].$
Note that this is not a homework problem. I would be appreciate if you could help me. 
 A: Here is an idea that might work. I post it as an answer because it's too long for a comment. Let $(f_n)$ be a Cauchy sequence in $BC^1[\mathbb{R},X]$. As $X$ is Banach, there are functions $f,\ g$ such that $\lim f_n=f$ and $\lim f'_n=g.$ These are pointwise limits. On the other hand, for any $\epsilon>0,\ \||f_m-f_n\||<\epsilon$ for sufficiently large $m,n$ so if we let $m\to \infty$ we see that $f$ is bounded. And since $\|f(t)-f(x)\|\le \|f(t)-f_m(t)\|+\|f_m(t)-f_m(x)\|+\|f_m(x)-f(x)\|,\ f\ $ is continuous. Similar results hold for $g$. 
Define for each $f\in BC^1[\mathbb{R},X],\  F(t)=\|f(t)\|.$ The reverse triangle inequality shows that $F_n$ and $F'_n$ are continuous. Therefore, $\int_{a}^{x}F_n'dt=F_n(x)-F_n(a).$ But an application of the Dominated Convergence Theorem gives $\lim \int_{a}^{x}F_n'dt=\int_{a}^{x}\lim F_n'dt=\int_{a}^{x}Gdt$ from which we infer that $F(x)-F(a)=\int_{a}^{x}Gdt$ and so $F'(x)=G(x)$ and therefore $f'=g.$
A: Let $f_{n}$ be a such Cauchy sequence then $\forall$ $\epsilon$ $>$ 0 $\exists$ M such that $\forall$ m,n $\geq$ M such that $\forall$ t $\in$ $\mathbb{R}$ 
both $\left \| f_{n}(t)-f_{m}(t)  \right \|$ $<$ $\epsilon$ and $\left \| f^{'}_{n}(t)-f^{'}_{m}(t)  \right \|$ $<$ $\epsilon$ then there exist two functions a:$\mathbb{R}$ $\rightarrow $ $X$ and b:$\mathbb{R}$ $\rightarrow $ $X$   such that for every t $\in$ $\mathbb{R}$ $f_{n}(t)$ converge to $a(t)$ and $f^{'}_{n}(t)$ converge to $b(t)$ , and because $\forall$ $\epsilon$ $>$ 0 $\exists$ M such that $\forall$ m,n $\geq$ M such that $\forall$ t $\in$ $\mathbb{R}$ 
both $\left \| f_{n}(t)-f_{m}(t)  \right \|$ $<$ $\epsilon$ and  $\left \| f^{'}_{n}(t)-f^{'}_{m}(t)  \right \|$ $<$ $\epsilon$ we fix n and let m go to $\infty$   so we wille have a and b such that $\forall$$\epsilon$ $>$ 0 $\exists$ N $\forall$ n $\geq$ N $\left \| f^{'}_{n}(t)-b(t)  \right \|$ $<$ $\epsilon$  and $\left \| f_{n}(t)-a(t)  \right \|$ $<$ $\epsilon$ if  we can show that b=$a^{'}$ (i didn't try )  we are done
