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Weierstrass’s theorem states that a continuous function on [0,1] can be uniformly approximated by polynomials. Can every continuous function on the closed unit disc be approximated uniformly by polynomials in the variable z?

i think its false

is the function f(z)=conjugate z a counterexample? if no i would be grateful if someone tell me a counterexample or a proof if it is correct.

thanks

user115608
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  • You are correct. Uniform approximation by polynomials (which are holomorphic) on a compact set, means that the limit must be holomorphic too. –  Apr 03 '14 at 07:59
  • yr right but is this condition(on a compact set) really nessecary?because i think the theorem is this:if a sequence of holomorphic functions on set A converges uniformly to function f on all compact subsets of A then f is holomorphic. – user115608 Apr 03 '14 at 08:21

1 Answers1

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In the complex case an additional condition is required (c.f. v.g. Baby Rudin): the algebra of approximating functions must be self-adjoint, i.e. $$p\in\mathcal A\implies \bar p\in\mathcal A.$$ This is false for complex polynomials: $p(z)=z\in\Bbb C[z]$ but $\bar p(z)=\bar z\not\in\Bbb C[z]$.

And if polynomial is interpreted as $p(x,y)\in\Bbb C[x,y]$...

Martín-Blas Pérez Pinilla
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