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In fact from the Herglotz representation for ''n'' > 0
:<math> a_n =2\int_0^{2\pi} e^{-in\theta}\, d\mu(\theta).</
Hence
:<math>\sum_m\sum_n a_{m-n} \lambda_m\overline{\lambda_n} =\int_0^{2\pi} \left|\sum_{n=}^N \lambda_n e^{-in\theta}\right|^2 \, d\mu(\theta) \ge 0.</math>
Conversely, setting λ<sub>''n''</sub> =''z''<sup>''n''</sup>,
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