 Solution

A strictly convex normed space is a normed vector space whose unit sphere contains no nontrivial line segment. Equivalently, for distinct unit vectors $x,y$,Suppose $u,v\in U$ are both elements of best approximation in a normed space to $f$, and let $d={\inf }_{w\in U}\Vert f-w\Vert$. 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 givesBut $(u+v)/2\in U$, contradicting the definition of $d$. Thus 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.