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Best approximation theorem

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 (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,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, (an) 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.