Weierstrass Approximation Theorem/Proof 3
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.