Strictly convex normed space ($\Vert x\Vert =\Vert y\Vert =1,x\ne y\Rightarrow \Vert (x+y)/2\Vert <1$)

A normed vector space is strictly convex when distinct unit vectors have their midpoint strictly inside the unit ball. The Lp space is strictly convex for $1<p<\infty$: strict convexity of $|z{|}^p$ gives a strict integrated midpoint inequality for distinct unit functions. The endpoint spaces $L^1$ and $L^{\infty }$ need not have this property.

If two distinct elements $u,v$ of a linear subspace both minimize distance to $f$, normalize their residuals by the common positive minimum $d$. Strictly convex normed space geometry gives $\Vert f-(u+v)/2\Vert <d$, a contradiction. If $d=0$, both elements equal $f$. This proves uniqueness when a best approximant exists, without asserting existence for a nonclosed subspace.

A strictly convex norm permits at most one best approximation from a linear subspace, but does not force the distance infimum to be attained. In the Hilbert space ${\ell }^2$, the finite-support vector subspace $c_{00}$ is dense and omits $u=(2^{-j}{)}_{j\ge 1}$. Its distance from $u$ is zero and no member attains that distance. Finite-dimensionality restores existence by compactness of bounded sets in the vector subspace. If $Y$ is a bounded projection onto a vector subspace, the estimate $\Vert u-Yu\Vert \le (1+\Vert Y\Vert ){\inf }_q\Vert u-q\Vert$ follows directly by writing $u-Yu=(I-Y)(u-q)$ for each candidate $q$, even when the infimum is not attained.