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Denote: $f(e^{i\theta})$ is continuous and strictly positive on the interval $ 0 \le \theta \le 2\pi$ with Fourier coefficients

$$ t_j = \frac{1}{2\pi}\int_0^{2\pi}f(e^{i\theta})e^{-ij\theta} \quad T_k(f) = \begin{bmatrix} t_0 & t_{-1} & \cdots &t_{-k} \\ t_1 & t_0 & \cdots & t_{1-k}\\ \vdots & \ddots & \ddots & \vdots \\ t_k & \cdots & t_1 & t_0 \end{bmatrix}$$ for $k = 0,1, ..$ The matrix $T_k(f)$ is constant on diagonals; such matrices are called Toeplitz matrices.

I've already shown that for $b=\begin{bmatrix}b_0&\cdots&b_n\end{bmatrix}^T$ we have $$\frac{1}{2\pi}\int_0^{2\pi}\left|\sum_{j=0}^nb_je^{ij\theta}\right|^2f(e^{i\theta})d\theta = b^HT_nb\,.$$

And also have shown that if $T_n\succ0$ and $v^H=\begin{bmatrix}t_1&\cdots & t_n\end{bmatrix}$, then $$T_n=\begin{bmatrix}1&v^HT_{n-1}^{-1}\\0&1\end{bmatrix} \begin{bmatrix}\rho_n & 0^H\\0&T_{n-1}\end{bmatrix} \begin{bmatrix}1&0^H\\ T^{-1}_{n-1}v & I_{n}\end{bmatrix} \,.$$

Where $\rho_n^{-1} = (T_n^{-1})_{00}$. [Note: This is a special case of a connection between the blocks of a matrix and the blocks of its inverse that may be obtained by Schur complements.

Where (Need to prove) $$\min\left\{ \frac{1}{2\pi}\int_0^{2\pi} \left|1-\sum_{j=1}^nc_je^{ij\theta}\right|^2f(e^{i\theta})d\theta: c_1,\ldots,c_n\in\mathbb{C} \right\} = \rho_n \,.\tag{*}$$

Question: I'm struggling to show $(*)$.

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  • $\begingroup$ I had a little mistake, I've edited the question :) $\endgroup$ – Ilan Aizelman WS Jun 8 at 13:16

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