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\begin{document}







\vspace*{.3in}
\noindent{\bf Rudin, page 165/21:}
The identify function $e^{i\theta}\mapsto e^{i\theta}$ belongs to 
$\mathcal{A}$, vanishes nowhere and is injective.  Hence $\mathcal{A}$
is nowhere vanishing and separates points.  Nevertheless,
I claim that the function $f(e^{i\theta}) = 1/e^{i\theta} = e^{-i\theta}$
is not in the uniform closure of $\mathcal{A}$.  

\proof
Note first that
$$
\int_0^{2\pi} f(e^{i\theta})e^{i\theta}\,d\theta = 2\pi.
$$
Now suppose in order to obtain a contradiction that $f$ is in the uniform 
closure of $\mathcal{A}$.  Then for any $\epsilon>0$ we could find an
element $g\in\mathcal{A}$ such that 
$$
|g(e^{i\theta}) - f(e^{i\theta})| < \epsilon
$$
for every $e^{i\theta}$.  
$$
\left|\int_0^{2\pi} g(e^{i\theta})e^{i\theta}\,d\theta -
      \int_0^{2\pi} f(e^{i\theta})e^{i\theta}\,d\theta\right| \leq
\int_0^{2\pi} |g(e^{i\theta}) - f(e^{i\theta})||e^{i\theta}|\,d\theta
\leq 
2\pi\epsilon < 2\pi.
$$
provided we choose $\epsilon<1$.  In particular, 
$$
\int_0^{2\pi} g(e^{i\theta})e^{i\theta}\neq 0.
$$
On the other hand $g(\theta) = \sum_{n=0}^N c_n e^{in\theta}$, so
$$
\int_0^{2\pi} g(e^{i\theta})\,d\theta
=
\sum_{n=0}^N c_n\int_0^{2\pi} e^{i{n+1}\theta}\,d\theta 
=
\sum_{k=1}^{N+1} c_{k-1}\left(\int_0^{2\pi} cos(k\theta)\,d\theta
                       + i\int_0^{2\pi} sin(k\theta)\,d\theta\right)
= 0.
$$
This contradicts the above and proves that $f$ is
not in the uniform closure of $\mathcal{A}$.
\qed
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