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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.

For real functions, let $r=f-p_{\ast }$ and put $\mathrm{sign}(0)=0$. If $\int v\mathrm{sign}(r)=0$ for every $v\in U$, then $p_{\ast }$ is a best approximation in a normed space using the $L^1$ norm. The proof is the pointwise inequality $|r-v|\ge |r|-v\mathrm{sign}(r)$, followed by integration. This is a sufficient criterion even if the residual vanishes on a positive-measure set; a general necessary criterion may choose any subgradient in $[-1,1]$ on that zero set.