Exercise 7.21

Question

\def\ints{\mathbb{Z}}
Exercise 21: 
    Let $K$ be the unit circle in the complex plane and let $\mathscr A$ be the algebra of all functions of the form $$f(e^{i	heta})=\sum_{n=0}^Nc_ne^{in	heta}\quad\quad(	heta\hbox{ real}).$$
    Then $\mathscr A$ separates points on $K$ and $\mathscr A$ vanishes at no point of $K$, but nevertheless there are continuous functions on $K$ which are not in the uniform closure of $\mathscr A$.

analambanomenos · Accepted Answer

\def\ints{\mathbb{Z}}

Since $e^{in	heta}e^{im	heta}=e^{i(m+n)	heta}$, it is clear that $\mathscr A$ is an algebra, and since it contains the identity function $f(e^{i	heta})=e^{i	heta}$, $\mathscr A$ separates
    points on $K$ and vanishes at no point of $K$. Following the hint, for $f=\sum_0^Nc_ne^{in	heta}\in\mathscr A$ we have
    \begin{align*}
        \int_0^{2\pi}f(e^{i	heta})e^{i	heta}\,d	heta &= \sum_{n=0}^Nc_n\int_0^{2\pi}e^{i(n+1)	heta}\,d	heta \
        &= \sum_{n=0}^Nc_n\int_0^{2\pi}\cos\big((n+1)	heta\bigr)\,d	heta + \sum_{n=0}^Nic_n\int_0^{2\pi}\sin\big((n+1)	heta\bigr)\,d	heta \
        &= 0.
    \end{align*}
    Hence if $g$ is in the uniform closure of $\mathscr A$, we have by the same reasoning used in Exercise 7.20 that $\int_0^{2\pi}g(e^{i	heta})e^{i	heta}\,d	heta=0$. However, $e^{-i	heta}$ is a
    continuous function on $K$ such that $\int_0^{2\pi}e^{-i	heta}e^{i	heta}\,d	heta=2\pi$, so that $e^{-i	heta}$ is not in the uniform closure of $\mathscr A$.

Exercise 7.21

Answers

Comments

Exercise 7.21

Answers

Comments

Add answer