2
$\begingroup$

Let $X:=\{z \in \mathbb C : |z|\le 1\}$ , is it true that any element in $C(X , \mathbb C)$ can be uniformly approximated by polynomials , in $z, \bar z$ , with real co-efficients ? If we wanted uniform approximation by polynomials in $z, \bar z$ , with complex co-efficients then I know it would be true by Complex version of Stone-Weierstrass theorem . But I have no idea what happens if we want real-coefficients ( I can't see whether the set $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ is a subalgebra over $\mathbb C$ or not ) . Please help . Thanks in advance

$\endgroup$
  • 1
    $\begingroup$ Unless I'm mistaken, the function $p(z, \bar{z}) = iz$ is a good example to have in mind. $\endgroup$ – Cameron Williams May 10 '17 at 11:39
  • $\begingroup$ @CameronWilliams : what to do with that polynomial ? $\endgroup$ – user228168 May 10 '17 at 13:46
  • $\begingroup$ Can you approximate that function? $\endgroup$ – Cameron Williams May 10 '17 at 14:09
1
$\begingroup$

To expand on the counterexample $p(z, \bar z)=iz$ given by Cameron Williams, note that when restricted to a subinterval of real line, such as $[-1,1]$, the set $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ reduces to $\mathbb{R}[x]$. Clearly, any function one can approximate by the elements of $\mathbb{R}[x]$ must be real-valued on $[-1,1]$, and this does not include $iz$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy