= Solution
A <strictly convex normed space> is a <normed vector space> whose unit sphere contains no nontrivial line segment. Equivalently, for distinct unit vectors $x,y$,
$$
\left\|\frac{x+y}{2}\right\|<1.
$$
Suppose $u,v\in U$ are both elements of <best approximation in a normed space> to $f$, and let $d=\inf_{w\in U}\|f-w\|$. If $d=0$ then $u=v=f$. If $d>0$ and $u\ne v$, the vectors $(f-u)/d$ and $(f-v)/d$ are distinct unit vectors. Strict convexity gives
$$
\left\|f-\frac{u+v}{2}\right\|
=d\left\|\frac{(f-u)/d+(f-v)/d}{2}\right\|<d.
$$
But $(u+v)/2\in U$, contradicting the definition of $d$. Thus \b[there is at most one best approximant]. This proves <uniqueness of best approximation in a strictly convex space>, and does not assert existence for an arbitrary, possibly nonclosed, <linear subspace>.
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