Weierstrass Approximation Theorem/Proof 3

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $f$ be a real function which is continuous on the closed interval $\Bbb I = \closedint a b$.


Then $f$ can be uniformly approximated on $\Bbb I$ by a polynomial function to any given degree of accuracy.


Proof

Let $\AA \subseteq \map C {\Bbb I, \R}$ be the set of real polynomial functions.

$\AA$ is a subalgebra of $\map C {\Bbb I, \R}$ because polynomials over $\R$ form an algebra over $\R$.

Let $I$ denote the identity mapping on $\Bbb I$, i.e.:

$\forall x \in \Bbb I : \map I x = x$

Then $I \in \AA$.

Thus $\AA$ separates the points of $\Bbb I$, since trivially:

$\forall x,y \in \Bbb I : x \ne y \implies \map I x \ne \map I y$

It is also clear that $1 \in \AA$.

Therefore the claim follows from Stone-Weierstrass Theorem.

$\blacksquare$


Source of Name

This entry was named for Karl Weierstrass.