proof of existence and uniqueness of best approximations

- Proof of the theorem on existence and uniqueness of - ( entry (http://planetmath.org/BestApproximationInInnerProductSpaces))

Existence : Without loss of generality we can suppose $x=0$ (we could simply translate by $-x$ the set $A$).

Let $d=d(0,A)=inf\{\parallel a\parallel :a\in A\}$ be the distance of $A$ to the origin. By defintion of infimum there exists a sequence $(a_n)$ in $A$ such that

$$\parallel a_n\parallel ⟶d$$

Let us see that $(a_n)$ is a Cauchy sequence. By the parallelogram law we have

$${\parallel \frac{a_n-a_m}{2}\parallel }^2+{\parallel \frac{a_n+a_m}{2}\parallel }^2=\frac{1}{2}{\parallel a_n\parallel }^2+\frac{1}{2}{\parallel a_m\parallel }^2$$

i.e.

$${\parallel \frac{a_n-a_m}{2}\parallel }^2=\frac{1}{2}{\parallel a_n\parallel }^2+\frac{1}{2}{\parallel a_m\parallel }^2-{\parallel \frac{a_n+a_m}{2}\parallel }^2$$

As $A$ is convex, ${\displaystyle \frac{a_n+a_m}{2}}\in A$, and therefore

$$\parallel \frac{a_n+a_m}{2}\parallel \ge d$$

So we see that

$${\parallel \frac{a_n-a_m}{2}\parallel }^2\le \frac{1}{2}{\parallel a_n\parallel }^2+\frac{1}{2}{\parallel a_m\parallel }^2-d^2⟶0\mathit{\hspace{1em}\hspace{0.5em} }\text{when}m,n\to \mathrm{\infty }$$

which means that $\parallel a_n-a_m\parallel ⟶0$ when $m,n\to \mathrm{\infty }$, i.e. $(a_n)$ is a Cauchy sequence.

Since $A$ is complete (http://planetmath.org/Complete), $a_n⟶a_0$ for some $a_0\in A$.

As $a_0\in A$ its norm must be $\parallel a_0\parallel \ge d$. But also

$$\parallel a_0\parallel \le \parallel a_0-a_n\parallel +\parallel a_n\parallel ⟶d$$

which shows that $\parallel a_0\parallel =d$. We have thus proven the existence of best approximations (http://planetmath.org/BestApproximationInInnerProductSpaces).

Uniqueness : Suppose there were $a_0,b_0\in A$ such that $\parallel a_0\parallel =\parallel b_0\parallel =d$. Then, by the parallelogram law

$${\parallel \frac{a_0-b_0}{2}\parallel }^2+{\parallel \frac{a_n+b_0}{2}\parallel }^2=\frac{1}{2}{\parallel a_0\parallel }^2+\frac{1}{2}{\parallel b_0\parallel }^2=d^2$$

If $a_0-b_0\ne 0$ then we would have , which is contradiction since ${\displaystyle \frac{a_0+b_0}{2}}\in A$ ($A$ is convex).

Therefore $a_0=b_0$, which proves the uniqueness of the . $\mathrm{\square }$