How to prove that is a Banach space

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??

$$\frac {u_n(t)} {1+t}=\frac {u_n(0)} {1+0}+\int_0^{t} \frac {(1+s)u_n'(s)-u_n(s)} {(1+s)^{2}}\, ds$$. 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 $$t \to \infty$$.
• @PolineSandra $w$ is differentiable at every point $t \geq 0$. That is what I have proved. – Kabo Murphy Jun 23 at 7:07
• @kavi Rama MURTHY How to prove that $\lim_{t\to\infty}w(t)=\lim ((1+t)w(t))'=0$ ? – Poline Sandra Jun 23 at 7:08