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

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- Solution c

Solution

On the real interval,

h (x)

is real and

H (x) = h (x)^{2} \geq 0

. Expanding and integrating the finite sum yields

\sum_{r, s = 1}^{n} \frac{a _{r} a _{s}}{r + s + 1} = \int_{0}^{1} h (x)^{2} d x .

Apply part (b)(iv) to the entire function

H = h^{2}

. Since

∣ H (e^{i θ}) ∣ = ∣ h (e^{i θ}) ∣^{2}

, orthogonality of complex exponentials gives

\int_{0}^{2 π} ∣ h (e^{i θ}) ∣^{2} d θ = \sum_{r, s} a_{r} a_{s} \int_{0}^{2 π} e^{i (r - s) θ} d θ = 2 π \sum_{r = 1}^{n} a_{r}^{2} .

Consequently

r, s = 1 \sum n \frac{a _{r} a _{s}}{r + s + 1} \leq π r = 1 \sum n a_{r}^{2} .

The 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.

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