Best approximation theorem

Theorem

Let X be an inner product space with induced norm, and $Asubseteq X$ a non-empty, complete convex subset. Then, for all $xin X$ , there exists a unique best approximation a₀ to x in A.

Proof

Suppose x = 0 (if not the case, consider A − {x} instead) and let $d=d(0,A)=inf_{ain A} ||a||$ . There exists a sequence (a_n) in Asuch that

$||a_n||to d$ .

We now prove that (a_n) is a Cauchy sequence. By the parallelogram rule, we get

$||frac{a_n-a_m}{2}||^2+||frac{a_n+a_m}{2}||^2=frac{1}{2}(||a_n||^2+||a_m||^2)$ .

Since A is convex, $frac{a_n+a_m}{2}in A$ so

$underset{m,nin mathbb N}{forall }; ||frac{a_n+a_m}{2}||geq d$ .

Hence

$||frac{a_n-a_m}{2}||^2leq frac{1}{2}(||a_n||^2+||a_m||^2)-d^2to 0$ as $m,nto infty$

which implies $||a_n-a_m||to 0$ as $m,nto infty$ . In other words, (a_n) is a Cauchy sequence. Since A is complete,

$underset{a_0in A}{exists }; a_nto a_0$ .

Since $a_0in A$ , $||a_0||geq d$ . Furthermore

$||a_0||leq ||a_0-a_n||+||a_n||to d$ as $nto infty$ ,

which proves | | a₀ | | = d. Existence is thus proved. We now prove uniqueness. Suppose there were two distinct best approximations a₀and a₀‘ to x (which would imply | | a₀ | | = | | a₀‘ | | = d). By the parallelogram rule we would have

$||frac{a_0+a_0'}{2}||^2+||frac{a_0-a_0'}{2}||^2=frac{1}{2}(||a_0||^2+||a_0'||^2)=d^2$ .

Then

$||frac{a_0+a_0'}{2}||^2<d^2$

which cannot happen since A is convex, and as such $frac{a_0+a_0'}{2}in A$ , which means $||frac{a_0+a_0'}{2}||^2geq d^2$ , thus completing the proof.