Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 1 / c

Solution

Let

F (x) = \sum_{n = 1}^{N} b_{n} e (n^{2} x)

∥ F ∥_{4}^{4} = \sum_{m} \sum_{n_{1}^{2} + n_{2}^{2} = m} b_{n_{1}} b_{n_{2}}^{2} .

r_{2} (m)

of representations of

m

as two squares is at most a constant times its divisor function. Since

m \leq 2 N^{2}

, the supplied divisor bound and Cauchy-Schwarz imply

∥ F ∥_{4}^{4} ≲_{ϵ} N^{ϵ} \sum_{n_{1}, n_{2}} ∣ b_{n_{1}} ∣^{2} ∣ b_{n_{2}} ∣^{2} = N^{ϵ} ∥ b ∥_{2}^{4} .

For

2 \leq p \leq 4

, monotonicity of

L^{p}

norms on

[0, 1]

gives

∥ F ∥_{p}^{p} ≲_{ϵ} N^{ϵ} ∥ b ∥_{2}^{p}

. For

p \geq 4

, use

∥ F ∥_{\infty} \leq N^{1/2} ∥ b ∥_{2}

to obtain

∥ F ∥_{p}^{p} \leq ∥ F ∥_{\infty}^{p - 4} ∥ F ∥_{4}^{4} ≲_{ϵ} N^{p /2 - 2 + ϵ} ∥ b ∥_{2}^{p} .

After enlarging the constant and the harmless epsilon loss, both ranges give

C_{p, 2} (N) ≲_{ϵ} N^{ϵ} (1 + N^{p /2 - 2}) .

Solved by gpt-5.6-sol high.

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