Szegő recurrence and Verblunsky coefficients

Remarks

For part 1, the polynomial $\Phi_{n+1} - z\Phi_n$ has degree at most $n$. It is orthogonal to $z, z^2, \dots, z^n$, because $\Phi_{n+1}$ is, and multiplication by $z$ is unitary on $L^2(\mu)$ and maps $\Phi_n \perp \set{1, \dots, z^{n-1}}$ onto $z\Phi_n \perp \set{z, \dots, z^n}$. The polynomials of degree at most $n$ orthogonal to $z, \dots, z^n$ form a one-dimensional space spanned by $\Phi_n^*$, which gives the recurrence. Taking norms, with $\norm{z\Phi_n} = \norm{\Phi_n^*} = \norm{\Phi_n}$, gives part 2, and $\norm{\Phi_{n+1}} > 0$ forces $\abs{\alpha_n} < 1$. Part 3 builds $\mu$ from any sequence in $\mathbb{D}$ by running the recurrence and taking a weak limit of the measures $\abs{\varphi_n(e^{i\theta})}^{-2} d\theta/2\pi$.

Example: for $d\mu = d\theta/2\pi$, $\Phi_n(z) = z^n$ and $\Phi_n^* = 1$, so $\Phi_{n+1} = z\Phi_n$ and all $\alpha_n = 0$. Conversely, by part 3, the uniform measure is the only one with all Verblunsky coefficients zero.

Infinite support is needed. If $\mu$ is supported on $N$ points, then $L^2(\mu)$ has dimension $N$ and $\Phi_N$ has norm $0$; the recurrence ends with $\abs{\alpha_{N-1}} = 1$.

Going deeper

For a student

The recurrence turns questions about the measure $\mu$ into questions about the sequence $(\alpha_n)$. For example, Szegő's theorem says that $\sum \abs{\alpha_n}^2 < \infty$ exactly when $\log$ of the density of $\mu$ is integrable. The $\alpha_n$ are the entries of the CMV matrix, the unitary analogue of the Jacobi matrix, which represents multiplication by $z$ on $L^2(\mu)$. In prediction theory and signal processing the same numbers appear as reflection coefficients, computed by the Levinson algorithm from the moments of $\mu$.