= Solution
First $w\in\ell^2$, since $\sum_{n\geq1}4^{-n}=1/3$. Any finite <linear dependence> among $w$ and the $e_m$ has the form $aw+\sum_{m\in F}b_me_m=0$ for a finite index set $F$. At an index outside $F$, its coordinate equals $a2^{-n}$, 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 tails
$$
v_N=w-\sum_{m=1}^N2^{-m}e_m.
$$
They belong to $V$, satisfy $f(v_N)=1$, and have
$$
\boxed{\|v_N\|_2^2=\sum_{n>N}4^{-n}=\frac{4^{-N}}3\longrightarrow0.}
$$
Thus $v_N\to0$ while $f(v_N)$ does not approach $f(0)=0$. \b[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.
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