# Weierstrass Approximation

• Nov 19th 2007, 08:21 PM
ThePerfectHacker
Weierstrass Approximation
If the elegant Weierstrass approximation theorem true for complex functions? Meaning, if $\displaystyle f(z)$ be a continous function on a compact set $\displaystyle S$ (non-empty) then for any $\displaystyle \epsilon > 0$ there exists a polynomial $\displaystyle g(z)$ such that $\displaystyle |f(z)-g(z)| < \epsilon$.
• Nov 20th 2007, 02:37 AM
Opalg
Quote:

Originally Posted by ThePerfectHacker
If the elegant Weierstrass approximation theorem true for complex functions? Meaning, if $\displaystyle f(z)$ be a continous function on a compact set $\displaystyle S$ (non-empty) then for any $\displaystyle \epsilon > 0$ there exists a polynomial $\displaystyle g(z)$ such that $\displaystyle |f(z)-g(z)| < \epsilon$.

That depends on what you want to take for the set S. If S is a subset of the complex plane then the result is false. For example, on the unit circle the complex conjugate function $\displaystyle f(z) = \bar{z}$ is not a uniform limit of polynomials (because a uniform limit of analytic functions has to be analytic).

However, if S is a subset of the real line, then the result is true. This is an immediate consequence of the Stone–Weierstrass theorem, a beautiful and very far-reaching generalisation of the Weierstrass theorem. It relies on the fact that if p(x) is a (complex-valued) polynomial in the real variable x, then the complex conjugate of p(x) is also a polynomial in x.