Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 3 / c / Solution

On the real interval, $h(x)$ is real and $H(x)=h(x{)}^2\ge 0$. Expanding and integrating the finite sum yieldsApply part (b)(iv) to the entire function $H=h^2$. Since $|H(e^{i\theta })|=|h(e^{i\theta }){|}^2$, orthogonality of complex exponentials givesConsequentlyThe quadratic form is actually nonnegative, so its absolute value can be dropped. This shifted Hilbert quadratic-form bound comes from combining the logarithmic contour estimate with a finite Parseval identity.