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Consider a subset $K\subset \mathbb C^2$ consisting of pairs $(z,\bar z)$ such that $|z|=1$. Is there an easy way to see that continuous functions on $K$ can be uniformly approximated by polynomials in two variables?

The problem is not so trivial because if you take the set $L\subset \mathbb C^2$ of pairs $(z,0)$ with $|z|=1$ then $K$ is homeomorphic $L$, but it is clear that it is impossible to approximate the continuous function $\frac{1}{z}$ on $L$ by polynomials because of the Maximum Principle.

Thank you.

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    $\begingroup$ Sounds like the Stone-Weierstrass theorem plus. $\endgroup$ – Raskolnikov Feb 11 '11 at 11:29
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    $\begingroup$ Note that $1/z=\bar{z}$ when $|z|=1$. So it is not a problem on $K$. $\endgroup$ – Raskolnikov Feb 11 '11 at 13:29
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If $f$ is a continuous function on $K$ then $g(t):=f(e^{i t}, e^{-it})$ is a continuous $2\pi$-periodic function of the real variable $t$. It follows that $g$ can be approximated on ${\mathbb R}$ by trigonometric polynomials $\sum_{k=-N}^N a_k e^{i k t}$. Therefore one has $f(z,\bar z)\doteq \sum_{k=1}^N a_{-k}\bar z^k + \sum_{k=0}^N a_{k} z^k$ on $K$.

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