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[parent] proof of existence and uniqueness of best approximations (Proof)

- Proof of the theorem on existence and uniqueness of best approximations on inner product spaces - (parent entry)

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\{\Vert a\Vert : 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

$\displaystyle \Vert a_n\Vert \longrightarrow d $

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

$\displaystyle \left \Vert\frac{a_n-a_m}{2}\right \Vert^2+\left\Vert\frac{a_n+a_m}{2}\right\Vert^2 = \frac{1}{2}\Vert a_n\Vert^2 + \frac{1}{2}\Vert a_m\Vert^2 $
i.e.
$\displaystyle \left\Vert\frac{a_n-a_m}{2}\right\Vert^2 = \frac{1}{2}\Vert a_n\Vert^2 + \frac{1}{2}\Vert a_m\Vert^2-\left\Vert\frac{a_n+a_m}{2}\right\Vert^2 $

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

$\displaystyle \left\Vert\frac{a_n +a_m}{2}\right\Vert \geq d $

So we see that

$\displaystyle \left\Vert\frac{a_n-a_m}{2}\right\Vert^2 \leq \frac{1}{2}\Vert a_n\Vert^2 + \frac{1}{2}\Vert a_m\Vert^2-d^2 \longrightarrow 0\;\;\;\;\;\;$   when$\displaystyle \;\; m,n \rightarrow \infty $
which means that $ \Vert a_n - a_m\Vert \longrightarrow 0$ when $ m,n \rightarrow \infty$, i.e. $ (a_n)$ is a Cauchy sequence.

Since $ A$ is complete, $ a_n \longrightarrow a_0$ for some $ a_0 \in A$.

As $ a_0 \in A$ its norm must be $ \Vert a_0\Vert \geq d$. But also

$\displaystyle \Vert a_0\Vert \leq \Vert a_0 - a_n \Vert + \Vert a_n\Vert \longrightarrow d $

which shows that $ \Vert a_0\Vert = d$. We have thus proven the existence of best approximations.

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

$\displaystyle \left \Vert\frac{a_0-b_0}{2}\right \Vert^2+\left\Vert\frac{a_n+b_... ...}\right\Vert^2 = \frac{1}{2}\Vert a_0\Vert^2 + \frac{1}{2}\Vert b_0\Vert^2=d^2 $

If $ a_0 - b_0 \neq 0$ then we would have $ \displaystyle \left \Vert\frac{a_0 + b_0}{2} \right\Vert^2 < d^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 best approximation. $ \square$



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Cross-references: contradiction, norm, convex, parallelogram law, Cauchy sequence, sequence, infimum, origin, distance, translate, without loss of generality, proof
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This is version 3 of proof of existence and uniqueness of best approximations, born on 2007-09-14, modified 2007-09-15.
Object id is 9937, canonical name is ProofOfExistenceAndUniquenessOfBestApproximations.
Accessed 575 times total.

Classification:
AMS MSC: 41A50 (Approximations and expansions :: Best approximation, Chebyshev systems)
 41A52 (Approximations and expansions :: Uniqueness of best approximation)
 41A65 (Approximations and expansions :: Abstract approximation theory )
 46C05 (Functional analysis :: Inner product spaces and their generalizations, Hilbert spaces :: Hilbert and pre-Hilbert spaces: geometry and topology )
 46N10 (Functional analysis :: Miscellaneous applications of functional analysis :: Applications in optimization, convex analysis, mathematical programming, economics)
 49J27 (Calculus of variations and optimal control; optimization :: Existence theories :: Problems in abstract spaces)
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