Past exam of the real and harmonic analysis course of the University of Oxford / 2008 / 3

For a complex measure, its total variation measure isIts norm is $\Vert \mu \Vert =|\mu |(\mathbb{T})$. For Radon measures on the compact circle this is also ${\sup }_{\Vert f{\Vert }_{\infty }\le 1,f\in C(\mathbb{T})}|\int fd\mu |$. The stated convergence is weak-star convergence against continuous tests:It is the weak-star topology from the identification of these measures with the dual of $C(\mathbb{T})$, rather than convergence of their values on every Borel set.

Set $C={\sup }_k\Vert {\mu }_k\Vert$. Given $f\in C(\mathbb{T})$, choose $g\in X$ with $\Vert f-g{\Vert }_{\infty }<\epsilon$. ThenTake the upper limit in $k$ and then $\epsilon \downarrow 0$. Thus convergence on the dense test family suffices; no assumption that $X$ is a vector space is needed.

For nonempty compact $K$, put $f_j(x)=\max \{0,1-j\mathrm{dist}(x,K)\}$. These continuous functions decrease to $1_K$ and satisfy ${\mu }_k(K)\le \int f_{j}d{\mu }_k$. Consequently ${\mathrm{limsup}}_k{\mu }_k(K)\le \int f_{j}d\mu$ for every $j$. Dominated convergence for the finite positive measure $\mu$ yieldsThe empty-set case is immediate. Positivity is essential for the comparison with these cutoff functions.

Write $m(dx)=dx/(2\pi )$ for the normalized Haar measure. The identity ${\sin }^2(kx)=\frac{1}{2}-\frac{1}{2}\cos (2kx)$ gives ${\mu }_k=\frac{1}{2}m-\frac{1}{2}\cos (2kx)m$. For any fixed trigonometric polynomial, orthogonality of complex exponentials makes the oscillatory integral zero once $2k$ exceeds its largest frequency. The norms satisfy $\Vert {\mu }_k\Vert ={\mu }_k(\mathbb{T})=1/2$, so the dense-test criterion applies. ThereforeIn particular the limit has mass one half; treating these measures as probabilities would introduce a factor-of-two error.

The Fourier coefficient of a measure on the circle and convolution of measures on the circle use the angular group law:Addition is modulo $2\pi$. This is the pushforward of the product complex measure under addition; equivalently its integral against a continuous $f$ is $\iint f(x+y)d\mu (x)d\nu (y)$. The integrands are bounded and the variation measures finite, so Fubini applies. In particular,Thus $\widehat{\mu \ast \nu }(n)=\hat{\mu }(n)\hat{\nu }(n)$. These measure coefficients use no additional Haar normalization factor: a probability has zeroth coefficient one.

Read the coefficient inside the limit as $\widehat{{\mu }_k}(n)$; the printed omission of its index would otherwise make that condition independent of the sequence. Weak-star convergence immediately implies convergence of every coefficient by testing $e^{-inx}$. Conversely coefficient convergence gives convergence against every trigonometric polynomial. Since $\Vert {\mu }_k\Vert =1$, the dense-test argument yields weak-star convergence to $\mu$.

The limit is necessarily a probability. Testing the constant function gives $\mu (\mathbb{T})=1$. For every nonnegative continuous $f$, $\int fd\mu ={\lim }_k\int fd{\mu }_k\ge 0$. A Radon measure whose integrals against all such functions are nonnegative is positive, by the positive case of the Riesz representation theorem (or continuous approximation of Borel indicators). Thus $\mu$ is nonnegative and has total mass one.

Convolution preserves probability measures: its defining pushforward is nonnegative and its total mass is the product of the two unit masses. For every fixed integer $n$ the previous coefficient identity givesThe probability version of the Fourier coefficient criterion therefore provesThe fixed compact underlying group and the uniform probability masses supply precisely the bounds needed in the dense-test argument.