Analysis Problem of the Day 75

Today’s problem appeared as Problem 2 on the UCLA Fall 2012 Analysis Qual:

Problem 7. Let $\mu$ be a probability measure on the unit circle such that

$\lim_{n \to \infty}\int_{|z|=1} z^n d\mu(z) =0.$

Show that for all $f \in L^1(d\mu),$ one has

$\lim_{n \to \infty} \int_{|z|=1} z^n f(z) d\mu(z)=0.$

Solution: Notice that one may parametrize the unit circle as $z=e^{-i \theta},$ so that the given assumption may be equivalently reformulated as $\widehat{\mu}(n) \to 0$ as $n \to \infty,$ where $\widehat{\mu}(n):=\lim_{n \to \infty}\int_{|z|=1} e^{-in \theta} d\mu(z)$ is the Fourier transform of the measure $\mu.$ Then, we attempt a density argument using continuous functions. Note that the set $A$ of $f$ satisfying $\lim_{n \to \infty} \int_{|z|=1} z^n f(z) d\mu(z)=0$ by assumption includes all trigonometric polynomials. In particular, such a set is a subalgebra of continuous functions that separates points, and therefore by Stone-Weierstrass is dense in the continuous functions on the unit circle in the supremum norm. This implies that

$\left|\int_{|z|=1} z^n (f(z)-p(z)+p(z)) d\mu(z)\right| \leq \int_{|z|=1} |f(z)-p(z)| d\mu(z) + \epsilon \leq 2\epsilon$

for a trigonometric polynomial $p$ such that $\|f-p\|_\infty<\epsilon$ and large enough $n.$ It follows that $\widehat{f}(n) \to 0$ as $n \to \infty$ for all continuous functions. Then, since continuous functions are dense in $L^1,$ one obtains

$\left|\int_{|z|=1} z^n f(z) d\mu(z)\right| \leq \int_{|z|=1} |f(z)-g(z)| d\mu(z) < \epsilon$

for $\|f-g\|_1 < \epsilon,$ $g$ continuous, and large enough $n,$ so from this we deduce the given claim.

Leave a Reply Cancel reply