 Solution

First $w\in {\ell }^2$, since ${\sum }_{n\ge 1}{4}^{-n}=1/3$. Any finite linear dependence among $w$ and the $e_m$ has the form $aw+{\sum }_{m\in F}{b}_m{e}_m=0$ for a finite index set $F$. At an index outside $F$, its coordinate equals $a{2}^{-n}$, so $a=0$. Each remaining coordinate then gives $b_m=0$. Therefore the set is linearly independent, and the prescribed values define a unique linear map on its algebraic span.

Take the tailsThey belong to $V$, satisfy $f(v_N)=1$, and haveThus $v_N\to 0$ while $f(v_N)$ does not approach $f(0)=0$. The map is not continuous, and part (b) also shows it is unbounded. This is a discontinuous functional detected by vanishing sequence tails, illustrating why the finite-dimensional conclusion in part (a) cannot be extended to arbitrary normed spaces.