Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 3 / c

Table of contents
- Solution c

Solution

Let

S = \sum_{T < t \leq 2 T} \sum_{R < r \leq 2 R} b_{t} c_{r} e (α r^{2} t^{2}) .

Applying the Cauchy-Schwarz inequality in

t

gives

∣ S ∣^{2} \leq ∥ b ∥_{2}^{2} \sum_{t} \sum_{r} c_{r} e (α r^{2} t^{2})^{2} = ∥ b ∥_{2}^{2} \sum_{r_{1}, r_{2}} c_{r_{1}} \overline{c_{r_{2}}} \sum_{t} e (α t^{2} (r_{1}^{2} - r_{2}^{2})) .

Take absolute values and apply Cauchy-Schwarz to

(r_{1}, r_{2})

. Since

\sum_{r_{1}, r_{2}} ∣ c_{r_{1}} c_{r_{2}} ∣^{2} = ∥ c ∥_{2}^{4}

, expanding the remaining square gives

∣ S ∣^{4} \leq ∥ b ∥_{2}^{4} ∥ c ∥_{2}^{4} \sum_{T < t_{1}, t_{2} \leq 2 T R < r_{1}, r_{2} \leq 2 R} e (α (t_{1}^{2} - t_{2}^{2}) (r_{1}^{2} - r_{2}^{2})) .

Taking fourth roots proves the claimed bilinear quadratic exponential sum fourth-moment bound.

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