Use normalized convolution and <Fourier transform>[Fourier coefficients], and define the large spectrum
$$
\Gamma=\{\gamma\in\widehat G:|\widehat f(\gamma)|\geq\epsilon\delta\}.
$$
By the <Parseval identity> and $0\leq f\leq1$,
$$
|\Gamma|\epsilon^2\delta^2
\leq\sum_\gamma|\widehat f(\gamma)|^2
=\mathbb E f^2
\leq\delta,
$$
so $|\Gamma|\leq\epsilon^{-2}\delta^{-1}$.

The <convolution theorem> gives
$$
(f*f*f)(x+y)-(f*f*f)(x)
=\sum_\gamma\widehat f(\gamma)^3\gamma(x)(\gamma(y)-1).
$$
If $y\in B(\Gamma,\epsilon)$, the part over $\Gamma$ is at most
$$
\epsilon\sum_{\gamma\in\Gamma}|\widehat f(\gamma)|^3
\leq\epsilon\delta\sum_\gamma|\widehat f(\gamma)|^2
\leq\epsilon\delta^2,
$$
because $|\widehat f(\gamma)|\leq\delta$. On the complementary spectrum, $|\widehat f(\gamma)|<\epsilon\delta$ and $|\gamma(y)-1|\leq2$, so the contribution is less than $2\epsilon\delta^2$. The required difference is therefore less than $3\epsilon\delta^2$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 129 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-129/2/b/solution