Consider the algebra $ T $ of trigonometric polynomials (i.e. functions like $ \sum_{n=-N}^{N}c_{n}e^{inz} $ where $ c_n \in \mathbb{C} $) on the closed disc $ D $ of radius $ \frac{1}{3} $. $ T $ is self-adjoint, it separates points in the disc, and it vanishes nowhere on the disc. Thus, by Stone-Weierstrass theorem, any continuous function on $ D $ can be approximated uniformly by a trigonometric polynomial on $ D $. This contradicts with the fact that some continuous functions on $ D $ are not holomorphic. Could you help me find what I did wrong? Thanks very much
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4$\begingroup$ Why is $T$ self-adjoint? $\endgroup$– Jose27Feb 18, 2015 at 6:41
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1 Answer
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To expand on @Jose27's comment:
$\overline{e^{inz}}$ is not $e^{-inz}$, since $z$ is not assumed to be real.
In fact with $z=x+iy$, $$ e^{inz} = e^{in(x+iy)} = e^{inx}e^{-ny} $$ so $$ \overline{e^{inz}} = e^{-inx}e^{-ny} $$ but $$ e^{-inz} = e^{-in(x+iy)} = e^{-inx}e^{ny}. $$ In other words, $\overline{e^{inz}}$ is not a member of $T$ (it's not holomorphic when $n\neq 0$).
