# Stone-Weierstrass applied to trigonometric polynomials on a disc

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

• Why is $T$ self-adjoint? Feb 18, 2015 at 6:41

$\overline{e^{inz}}$ 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 $n\neq 0$).