Best approximation theorem

Theorem

Let X be an inner product space with induced norm, and A\subseteq X a non-emptycomplete convex subset. Then, for all x\in 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_{a\in A} ||a||. There exists a sequence (an) in Asuch that
  • ||a_n||\to d.
We now prove that (an) 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,n\in \mathbb N}{\forall }\; ||\frac{a_n+a_m}{2}||\geq d.
Hence
  • ||\frac{a_n-a_m}{2}||^2\leq \frac{1}{2}(||a_n||^2+||a_m||^2)-d^2\to 0 as m,n\to \infty
which implies ||a_n-a_m||\to 0 as m,n\to \infty. In other words, (an) is a Cauchy sequence. Since A is complete,
  • \underset{a_0\in A}{\exists }\; a_n\to a_0.
Since a_0\in A||a_0||\geq d. Furthermore
  • ||a_0||\leq ||a_0-a_n||+||a_n||\to d as n\to \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}||^2\geq d^2, thus completing the proof.
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