$M[0,2] = \{x_n=x_n(t): sup|x_n(t)| < \infty , 0\leq t\leq2 \}$
The norm is $||x_n|| = sup |x_n(t)|, 0\leq t \leq 2$
I must investigate the size $sup |x_n(t)-f(t)| = sup |e^{-nt}-f(t)|, 0\leq t \leq 2$. How to find a function $f(t)$?
I think that $x_n$ not converge in $M[0,2]$.
Thank You.
Note that convergence with respect to the supremum norm clearly implies point wise converges. Thus, if the $x_n$'s converge in $M[0,2]$ then they converge to the point wise limit which is precisely $$ f(t) = \begin{cases} 1 &t=0\\ 0 & 0<t\leq 2 \end{cases} $$ Now, we check if $x_n$ also converges to $f$ with respect to the supremum norm. First note that for each $n\in\mathbb{N}$, $$ x_n-f(t) = \begin{cases} 0 &t=0\\ e^{-nt} & 0<t\leq 2 \end{cases} $$ Therefore, $$ \lVert x_n -f\rVert = \sup_{0\leq t\leq 2} \lvert x_n(t) - f(t) \rvert =\sup_{0< t\leq 2} e^{-nt} = 1 $$ Hence, the $x_n$'s do not converge since they do not tend to their point wise limit.