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Fall 2012 - Problem 2

density argument

Suppose $\diff\mu$ is a Borel probability measure on the unit circle in the complex plane such that

\lim_{n\to\infty} \int_{\set{\abs{z}=1}} z^n \,\diff\mu\p{z} = 0.

For $f \in L^1\p{\diff\mu}$ , show that

\lim_{n\to\infty} \int_{\set{\abs{z}=1}} z^nf\p{z} \,\diff\mu\p{z} = 0.

Solution.

Notice that if $p = a_0 + \cdots + a_mz^m$ is a polynomial, then

\lim_{n\to\infty} \int z^n p\p{z} \,\diff\mu\p{z} = \lim_{n\to\infty} \sum_{k=0}^m a_k \int z^{k+n} \,\diff\mu\p{z} = 0.

To generalize to $f \in L^1\p{S^1}$ , first observe that the unit circle $S^1$ is compact, so we may apply Stone-Weierstrass. Let $A = \set{\sum_{k=0}^n a_kz^k \mid n \in \N, a_k \in \C}$ . It's clear that $A$ is closed under scaling by $\C$ , addition, and multiplication, so $A$ is an algebra. $1 \in A$ , so $A$ vanishes nowhere. Finally, $z \in A$ , so $A$ separates points. Thus, $A$ is dense in $C\p{S^1}$ with respect to the uniform norm, hence with respect to the $L^1$ norm. Moreover, $C\p{S^1}$ is dense in $L^1\p{\diff\mu}$ , so given $\epsilon > 0$ and $f \in L^1\p{\diff\mu}$ , there exists $p = a_0 + \cdots + a_mz^m$ such that $\norm{f - p}_{L^1} < \epsilon$ . Thus, if $n$ is large enough, then $\abs{\int z^np} < \epsilon$ and so

\begin{aligned} \abs{\int z^nf\p{z} \,\diff\mu\p{z}} &\leq \abs{\int z^np\p{z} \,\diff\mu\p{z}} + \int \abs{z^n\p{f\p{z} - p\p{z}}} \,\diff\mu\p{z} \\ &\leq \epsilon + \int \abs{f\p{z} - p\p{z}} \,\diff\mu\p{z} && (\text{we're integrating over } \abs{z} = 1) \\ &\leq 2\epsilon. \end{aligned}

Sending $\epsilon \to 0$ , we get the claim.