# Szego limit theorems for Toeplitz matrix

Let matrix $$A$$ to be a auto-correlation matrix for a stationary signal $$x(t)$$, hence matrix $$A$$ is symmetric Toeplitz matrix such that $$$$A:=\begin{bmatrix}\phi(0) & \phi(-1) & \cdots & \phi(1-N) \\ \phi(1) & \phi(0) & \cdots & \phi(2-N)\\ \vdots & \vdots & \ddots & \vdots\\ \phi(N-1) & \phi(N-2) & \cdots & \phi(0) \end{bmatrix}$$$$ It is easy to proof that $$$$\lim_{N\rightarrow\infty}\frac{\lambda_0 + \lambda_1 + \cdots + \lambda_{N-1}}{N} = \phi(0) = \int^{f_N}_{-f_N}S(f)df, \tag1$$$$ where $$f_N$$ is the Nyquist frequency; $$\lambda_i$$ are eigenvalues of matrix $$A$$, $$S(f)$$ and $$\phi(t)$$ are Fourier Transform pairs such that $$$$\phi(t) = \int^{f_N}_{-f_N} S(f)e^{i2\pi f\frac{t}{f_N}}df. \tag2$$$$ Now, I would like to prove that $$$$\lim_{N\rightarrow\infty}\frac{\log(\lambda_0) + \log(\lambda_1) + \cdots + \log(\lambda_{N-1})}{N} = \frac{1}{2f_N}\int^{f_N}_{-f_N}\log[2f_N S(f)]df, \tag3$$$$ Does anyone know how to prove it? Many thanks!

• If $\,S(f):=c\,$ where $\,c\,$ is a constant, then $\,A\,$ is a multiple of an identity matrix with $\,\lambda_k=\phi(0)\,$ for all $\,k.$ What happens to your equations (1) and (3) in this special case? Sep 18, 2019 at 1:03

The answer is fairly easy, because $$$$\lim_{N\rightarrow\infty}\frac{\lambda_0 + \lambda_1 + \cdots + \lambda_{N-1}}{N} = \phi(0) = \int^{f_N}_{-f_N}S(f)df = \frac{1}{2f_N} \int^{f_N}_{-f_N}[2f_N S(f)]df,$$$$ which is the Arithmetic mean of continuous function $$2f_N S(f)$$, as well as $$\{\lambda_i\}_{0}^\infty$$. Hence, the Geometric mean of $$\{\lambda_i\}_{0}^\infty$$ can be written as $$$$\lim_{N\rightarrow\infty}\left(\prod_{i=1}^N \lambda_i\right)^{\frac{1}{N}} = e^{\frac{1}{2f_N}\int^{f_N}_{-f_N}\log[2f_N S(f)]df}.$$$$ Finally, we have \begin{aligned} \lim_{N\rightarrow\infty}\frac{\log(\lambda_0) + \cdots + \log(\lambda_{N-1})}{N} &= \lim_{N\rightarrow\infty}\log\left(\prod_{i=1}^N \lambda_i\right)^{\frac{1}{N}}\\ &=\frac{1}{2f_N}\int^{f_N}_{-f_N}\log[2f_N S(f)]df. \end{aligned}