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