# Dominated convergence theorem for log-integrable rational functions

Let $\{f_n\}_{n\ge 0}$ be a sequence of rational functions $f_n\colon \mathbb{C}\to\mathbb{R}_+$ which are non-negative and analytic on the unit circle and assume that $\lim_{n\to\infty} f_n = f$ pointwise, where $f$ is again a rational function non-negative and analytic on the unit circle. Notice that $f_n$ can possess zeros on the unit circle. It is well known however that the latter zeros represent log-integrable singularities. Indeed it holds that $\log\,f_n$ is integrable on the unit circle, i.e. $$\left|\int_{-\pi}^{\pi} \log\, f_n(e^{j\theta})\, \mathrm{d}\theta\right|<\infty.$$

My question concerns the limit of the integral of the sequence $\{\log\,f_n\}_{n\ge 0}$, namely $$\lim_{n\to \infty}\int_{-\pi}^{\pi}\, \log\, f_n(e^{j\theta}) \,\mathrm{d}\theta \overset{?}{=} \int_{-\pi}^{\pi} \log\, f(e^{j\theta})\, \mathrm{d}\theta.$$

My attempt was to apply Lebesgue’s dominated convergence theorem. However in this case the dominating function $g$ such that $|\log\, f_n|\le \log\,g$ can have (at most) a countable number of zeros on the unit circle (corresponding to all the possible zeros on the unit circle of $f_n$’s) and I’m not completely sure whether the logarithm of such a function is integrable.

Any help is really appreciated.

• You are denoting the unit circle by $\mathbb C?$ – zhw. Jul 20 '17 at 5:56
• @zhw. No, $f_n\colon z\mapsto f_n(z)$ are rational functions defined on the complex plane $\mathbb{C}$ which are analytic on the unit circle $\{e^{j\theta}, \theta\in[-\pi,\pi]\}$. – Ludwig Jul 20 '17 at 6:02
• What's an example of a rational function that is nonnegative on the unit circle (besides a constant)? – zhw. Jul 20 '17 at 6:13
• @zhw. For instance $f(z)=(z+2)(z^{-1}+2)$ (and, in general, any spectral density of a linear time invariant stochastic process). – Ludwig Jul 20 '17 at 6:16
• Thanks. In what sense do the $f_n$'s converge? (btw, $\bar f$ usuallly represents the conjugate of $f$ in complex analysis) – zhw. Jul 20 '17 at 6:22

If $p$ is a polynomial, then on the circle $|p|^2$ equals a rational function of the kind under discussion. But such $|p|^2$ are dense in the continuous nonnegative functions on the circle. (This follows from Stone-Weierstrass.) This allows for bending these $|p|^2$ around quite a bit.
For $n=1,2,\dots$ define $f_n$ on the circle by visualizing it on $[0,2\pi].$ There it is the piecewise linear function whose graph is connects the points $(0,1),(1/n,1),(2/n,e^n),(3/n,e^n),(4/n,1),(2\pi,1).$ (Good to draw a picture.) So $f_n$ is a gliding hump that equals $1$ except on the interval $[1/n,4/n],$ where it spikes up to the value $e^n$ on the interval $[2/n,3/n].$ Note that $f_n \to 1$ pointwise everywhere on $[0,2\pi].$
We can find polynomials $p_n$ such that $||p_n|^2-f_n|<1/n$ on the circle. Thus $|p_n|^2\to 1$ pointwise everywhere on the cirlce. However,
$$\int_0^{2\pi}\log |p_n|^2 \ge \int_{2/n}^{3/n}\log |p_n|^2 \ge \int_{2/n}^{3/n}\log (f_n -1/n) = \int_{2/n}^{3/n}\log (e^n -1/n).$$
The last expression $\to 1$ as $n\to \infty,$ so certainly we don't have convergence to $\int_0^{2\pi} \log 1 = 0.$