If the elegant Weierstrass approximation theorem true for complex functions? Meaning, if be a continous function on a compact set (non-empty) then for any there exists a polynomial such that .

Printable View

- Nov 19th 2007, 09:21 PMThePerfectHackerWeierstrass Approximation
If the elegant Weierstrass approximation theorem true for complex functions? Meaning, if be a continous function on a compact set (non-empty) then for any there exists a polynomial such that .

- Nov 20th 2007, 03:37 AMOpalg
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 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.