= Solution
{c}
The <vector subspace> contains zero, whose approximation error is $\|u\|$. For any $q$ with $\|q\|>2\|u\|$, the <triangle inequality> gives $\|u-q\|\geq\|q\|-\|u\|>\|u\|$, so such points cannot improve on zero.
Restrict to $K=\{q\in A:\|q\|\leq2\|u\|+1\}$. This is nonempty, closed and bounded in a finite-dimensional <normed space>, hence compact. The <continuous function> $q\mapsto\|u-q\|$ attains a <minimum> on $K$ by part (a). Outside $K$ its value exceeds the error at zero, so that minimizer is best on all of $A$:
$$
\boxed{\|u-p\|\leq\|u-q\|\quad(q\in A).}
$$
Completeness of the ambient space is unnecessary; <compactness> is used only in the finite-dimensional <vector subspace>.
Back to article page
