Let $E=\{f\in C^1([0,\infty[, R), \lim_{t\to\infty}\frac{f(t)}{1+t}=\lim_{t\to\infty}f'(t)=0\}$ with the norm $$||f||=\max\left(\sup\limits_{t\geq 0}\dfrac{|f(t)|}{1+t}, \sup\limits_{t\geq 0}|f'(t)|\right). $$ Prove that $E$ is a Banach space.
I started by let $(u_n)$ a Cauchy sequence that is
$$\forall \varepsilon>0, \exists n_0\in \mathbb{N}, \forall p,q\in \mathbb{N}; p>q\geq n_0\Rightarrow ||u_p-u_q||<\varepsilon $$
that is
$$\forall \varepsilon>0, \exists n_0\in \mathbb{N}, \forall p,q\in \mathbb{N}; p>q\geq n_0\Rightarrow \dfrac{|u_p(t)-u_q(t)|}{1+t}<\varepsilon \, \text{and}\, |u'_p(t)-u'_q(t)|<\varepsilon $$
that is $u_n'(t)$ and $\frac{u_n(t)}{1+t}$ are a Cauchy sequences in the complete $(\Bbb R,|\cdot|)$ so $u_n'(t)$ converge to $v(t)$. and $\frac{u_n(t)}{1+t}$ converge to $w(t)$
How to prove that $(1+t)w(t)$ is derivable??
Please
$\frac {u_n(t)} {1+t}=\frac {u_n(0)} {1+0}+\int_0^{t} \frac {(1+s)u_n'(s)-u_n(s)} {(1+s)^{2}}\, ds$. Taking limits we get $w(t)=c+\int_0^{t} [\frac {v(s)} {1+s}-\frac {w(s)} {1+s}]\, ds$ for some constant $c$. This implies that $w$ is differentiable and $w'(t)=\frac {v(t)} {1+t}-\frac {w(t)} {1+t}$ [ When we take limits in the first equation observe that $\frac {u_n'(s)} {1+s} \to \frac {v(s)} {1+s}$ uniformly and $\frac {u_n(s)} {(1+s)^{2}} \to \frac {w(s)} {1+s}$ uniformly on $[0,t]$].
We have $|\frac {u_n(t)} {1+t} -\frac {u_m(t)} {1+t}| <\epsilon$ for all $t$ if $n$ and $m \geq n_0$ for some $n_0$. Let $m \to \infty$ and conclude that $|\frac {u_{n_0}(t)} {1+t} -\frac {w(t)} {1+t}| \leq \epsilon$ for all $t$. Hence $|\frac {w(t)} {1+t}| \leq \epsilon + |\frac {u_{n_0}(t)} {1+t}| $. But the second term is less than $\epsilon$ for $t$ sufficiently large. This proves that $\frac {w(t)} {1+t} \to 0$ as $t \to \infty$. A similar argument can be given to prove that $w'(t) \to 0$ as $t \to \infty$.