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:<math> a_0=2,\,\,\, a_{-m} =\overline{a_m}</math>
for ''m'' > 0.
In fact from the Herglotz representation for ''n'' > 0
:<math> a_n =2\int_0^{2\pi} e^{-in\theta}\, d\mu(\theta).</ref>
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}|^2 \, d\mu(\theta) \ge 0.</math>
Conversely, setting λ<sub>''n''</sub> =''z''<sup>''n''</sup>,
:<math>\sum_{m=0}^\infty\sum_{n=0}^\infty a_{m-n} \lambda_m\overline{\lambda_n} = 2(1-|z|^2) \,\Re\, f(z).</math>
==References==
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