# 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
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.