Show that $\mathbb{X}$ is complete Let $\mathbb{X}=\{f:[t_0,t_0+\delta]\rightarrow[f_0-r,f_0+r]\}$ now we consider the distance $d_\lambda$ defined as $$d_\lambda:=\sup\{|u(t)-v(t)|e^{-\lambda(t-t_0)} :t \in [t_0,t_0+\delta]\}$$ Prove that $\mathbb{X}$ is complete with the distance $d_\lambda$. 
How can I handle this proof? I don't know where to start, can anyone give some hints? 
 A: We have to check that if $\left(u_n\right)_{n\geqslant 1}$ is a Cauchy sequence for $d_\lambda$, then there exists $u\in\mathbb X$ such that $d_\lambda(u_n,u)\to 0$ as $n\to \infty$.
This can be done with the following steps:


*

*Take $t\in[t_0,t_0 +\delta]$ and show that the sequence $\left(u_n(t)\right)_{\mathbb R}   $ is Cauchy in $\mathbb R$.  

*Call $u(t)$ the limiting number. Check that $u(t)$ lies in the interval $[f_0-r,f_0+r]$. 

*Now we have to prove that $d_\lambda(u_n,u)\to 0$. Fix $\varepsilon$. By defining of a Cauchy sequence, there is an integer $N$ such that if $m,n\geqslant N$, then 
$$\forall t\in[t_0,t_0+\delta],\quad \left|u_m(t)-u_n(t)\right|\exp\left(-\lambda(t-t_0)\right)\leqslant \varepsilon.$$
Now take the $\limsup_{m\to  +\infty}$.  
A: I find out another solution: let consider the distance $$d:=sup\{|u(t)-v(t)|:t \in [t_0,t_0+\delta]\}$$ the metric space$\mathbb{X}$ with the distance $d$ is complete. If $\lambda>0\Rightarrow d_\lambda(u,v)\leq d(u,v)$ and that proves the equivalence of the two distances. We can say that the space with the distance $d_\lambda$ is complete.
