= Solution
The <continuous dual space> of a normed vector space $V$ is
$$
V^*=\{f:V\to\mathbb F:f\text{ is linear and continuous}\},
$$
with the <operator norm>
$$
\lVert f\rVert=\sup_{\lVert v\rVert\leq1}|f(v)|.
$$
Let $(f_n)$ be a <Cauchy sequence> in this norm. For each $v\in V$,
$$
|f_n(v)-f_m(v)|\leq\lVert f_n-f_m\rVert\,\lVert v\rVert,
$$
so $(f_n(v))$ is Cauchy in the scalar field. Define $f(v)=\lim_n f_n(v)$. Passing to limits in the linearity identities shows that $f$ is linear.
A norm-Cauchy sequence is bounded, say $\lVert f_n\rVert\leq M$. Hence $|f(v)|\leq M\lVert v\rVert$, so $f\in V^*$. Given $\varepsilon>0$, choose $N$ such that $\lVert f_n-f_m\rVert<\varepsilon$ for $m,n\geq N$. Letting $m\to\infty$ gives
$$
|f_n(v)-f(v)|\leq\varepsilon\lVert v\rVert
$$
for every $v$, and therefore $\lVert f_n-f\rVert\leq\varepsilon$. Thus $f_n\to f$ in operator norm, proving <completeness of the dual space>: $V^*$ is Banach.
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