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I am reviewing a controversial paper, and the main result, a revolution within my field, comes down to whether or not the following is true. I strongly believe it is not, but would need confirmation.

Let $X_n$ be an $n \times n$ Toeplitz matrix with associated function $f(\omega)$ where $f(\omega)$ is band limited, i.e., $f(\omega)=0$ for $|\omega|>\beta \pi$, for some $\beta < 1$. Thus, element (k,l) in $X_n$ is $$\int_{-\pi}^{\pi}f(\omega)\exp(-(k-l)i \omega)d\omega$$.

Is $X_n$ invertible as $n\to \infty$ ?

For any finite $n$ it is, but to me it seems as Szegö's theorem, which gives the asymptotic eigenvalue distribution, will kick so the the inverse does not exist as $n \to \infty $.&=

Using a particular construction, which requires $X_n$ to be invertible, the authors can prove some remarkable things, as $n\to \infty$. I believe that the mistake is exactly that they assume that $X_n$ is invertible as $n\to \infty$….

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  • $\begingroup$ Is your situation finitely many bands? I don't understand your notation.. $\endgroup$ Mar 25, 2016 at 12:37
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    $\begingroup$ Could you define the "associated function" to a Toplitz matrix ? $\endgroup$ Mar 25, 2016 at 19:15
  • $\begingroup$ I don't see any reason why $X_n$ would be invertible (as you claim it is) for general $f$. Is your question about a specific $f$ then? $\endgroup$ Mar 26, 2016 at 0:13
  • $\begingroup$ $X_n$ (finite n) is perhaps not invertible in general, but for the particular $f(\omega)$ that the authors deal with, it is. You may for example assume that $f(\omega)=1,\, |\omega|<\beta \pi$ and 0 otherwise. $\endgroup$
    – Max Hamper
    Mar 26, 2016 at 7:57

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