Past exam of the real and harmonic analysis course of the University of Oxford / 2007 / 3

The Riesz-Markov-Kakutani representation theorem identifies the continuous dual of $C(\mathbb{T})$ with the finite complex Radon measures on $\mathbb{T}$: every bounded complex-linear functional $L$ has a unique representationThe right side is the total variation norm of the measure. On this compact metric space Radon measures are regular Borel measures. A functional is positive precisely when its representing measure is positive; then $\Vert L\Vert =L(1)=\mu (\mathbb{T})$.

Let $m$ be normalized arclength on the unit circle, so in angular coordinates $dm=dt/(2\pi )$. It is a finite Radon measure. The given identity says that $\mu$ and $m$ represent the same continuous linear functional on $C(\mathbb{T})$. Uniqueness in the Riesz-Markov-Kakutani representation theorem givesThe printed phrase “Borel sets in $E$” is interpreted as Borel subsets $E$ of the circle.

Write $A_N(f)=(N+1{)}^{-1}{\sum }_{n=0}^{N}f(e^{in\xi })$. For a monomial $z^k$, $k\in \mathbb{Z}$, the geometric series givesThus every such average converges, and by linearity so does the average of every trigonometric polynomial. Also $|A_N(f)|\le \Vert f{\Vert }_{\infty }$ uniformly in $N$.

Given $\epsilon >0$, choose a trigonometric polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<\epsilon$. ThenThe last term tends to zero as $N,M\to \infty$. The scalar sequence is Cauchy and therefore the limit exists for every continuous $f$ and every real $\xi$.

Define $L_{\xi }(f)={\lim }_N{A}_N(f)$. Limits preserve linearity and positivity, and $|L_{\xi }(f)|\le \Vert f{\Vert }_{\infty }$, $L_{\xi }(1)=1$. The Riesz-Markov-Kakutani representation theorem supplies a unique Radon probability measure ${\mu }^{\xi }$. By its definition,which is exactly weak-star convergence in $C(\mathbb{T}{)}^{\prime }$.

If $\xi /(2\pi )=p/q$ in lowest terms, the orbit has exactly $q$ points. Averaging a periodic sequence givesThis includes $q=1$. If $\xi /\pi$ is irrational, then $e^{ik\xi }\ne 1$ for every nonzero integer $k$. The preceding calculation gives $L_{\xi }(z^k)=0$ for $k\ne 0$ and $L_{\xi }(1)=1$, exactly the integrals against normalized Haar measure $m$. Density of trigonometric polynomials implies equality on all continuous functions, and uniqueness givesThese are the periodic-orbit and irrational-rotation cases of equidistribution.

For each $n\ge 0$, choose an open arc $I_n$ about $e^{in}$ of arclength $\pi 2^{-n-2}$, and put $G^1={\bigcup }_{n\ge 0}{I}_n$. It contains every point used by every empirical measure, whereasThus ${\mu }_N^1(G^1)=1$ for all $N$. Let $K=\mathbb{T}\setminus G^1$. This is closed and ${\mu }_N^1(K)=0$ for all $N$. But $1/\pi$ is irrational, so the weak-star limit is normalized arclength andTherefore the proposed implication is false. Weak-star convergence tests continuous functions and does not assert convergence of the indicator function of an arbitrary closed set.