Exercise 4.5.1

Answers

When $F (λ) = \log λ$ , we have $F (A (𝜃)) = \log A (𝜃)$ , from equation (3), we have

\begin{array}{l} \lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{n - 1} F (λ_{k}) & = \lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{n - 1} \log λ_{k} \\ = \lim_{n \to \infty} \frac{1}{n} \log \prod_{k = 0}^{n - 1} λ_{k} \\ = \lim_{n \to \infty} \frac{1}{n} \log (\det A) \\ = \frac{1}{2 π} \int_{0}^{2 π} F (A (𝜃)) d 𝜃 \\ = \frac{1}{2 π} \int_{0}^{2 π} \log (A (𝜃)) d 𝜃 . \end{array}

This is exactly equation (2)

niuers

2020-03-20 00:00