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

 $\|a_{n}\|\longrightarrow d$

Let us see that $(a_{n})$ is a Cauchy sequence. By the parallelogram law we have

 $\left\|\frac{a_{n}-a_{m}}{2}\right\|^{2}+\left\|\frac{a_{n}+a_{m}}{2}\right\|^% {2}=\frac{1}{2}\|a_{n}\|^{2}+\frac{1}{2}\|a_{m}\|^{2}$

i.e.

 $\left\|\frac{a_{n}-a_{m}}{2}\right\|^{2}=\frac{1}{2}\|a_{n}\|^{2}+\frac{1}{2}% \|a_{m}\|^{2}-\left\|\frac{a_{n}+a_{m}}{2}\right\|^{2}$

As $A$ is convex, $\displaystyle\frac{a_{n}+a_{m}}{2}\in A$, and therefore

 $\left\|\frac{a_{n}+a_{m}}{2}\right\|\geq d$

So we see that

 $\left\|\frac{a_{n}-a_{m}}{2}\right\|^{2}\leq\frac{1}{2}\|a_{n}\|^{2}+\frac{1}{% 2}\|a_{m}\|^{2}-d^{2}\longrightarrow 0\;\;\;\;\;\;\text{when}\;\;m,n\rightarrow\infty$

which means that $\|a_{n}-a_{m}\|\longrightarrow 0$ when $m,n\rightarrow\infty$, i.e. $(a_{n})$ is a Cauchy sequence.

Since $A$ is complete (http://planetmath.org/Complete), $a_{n}\longrightarrow a_{0}$ for some $a_{0}\in A$.

As $a_{0}\in A$ its norm must be $\|a_{0}\|\geq d$. But also

 $\|a_{0}\|\leq\|a_{0}-a_{n}\|+\|a_{n}\|\longrightarrow d$

which shows that $\|a_{0}\|=d$. 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 $\|a_{0}\|=\|b_{0}\|=d$. Then, by the parallelogram law

 $\left\|\frac{a_{0}-b_{0}}{2}\right\|^{2}+\left\|\frac{a_{n}+b_{0}}{2}\right\|^% {2}=\frac{1}{2}\|a_{0}\|^{2}+\frac{1}{2}\|b_{0}\|^{2}=d^{2}$

If $a_{0}-b_{0}\neq 0$ then we would have $\displaystyle\left\|\frac{a_{0}+b_{0}}{2}\right\|^{2}, 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 . $\square$

Title proof of existence and uniqueness of best approximations ProofOfExistenceAndUniquenessOfBestApproximations 2013-03-22 17:32:22 2013-03-22 17:32:22 asteroid (17536) asteroid (17536) 6 asteroid (17536) Proof msc 49J27 msc 46N10 msc 46C05 msc 41A65 msc 41A52 msc 41A50