Decompose $f=\sum_{j=0}^n f^{(=j)}$ into its homogeneous <Fourier-Walsh transform>[Fourier levels]. The <Bonami lemma> and the <triangle inequality> give
$$
\lVert T_{1/\sqrt3}f\rVert_4
\leq\sum_j3^{-j/2}\lVert f^{(=j)}\rVert_4
\leq\sum_j\lVert f^{(=j)}\rVert_2
\leq\sqrt{n+1}\,\lVert f\rVert_2.
$$
Apply this estimate to the $m$-fold tensor power $f^{\otimes m}$. Tensor products multiply both relevant norms and commute with the <noise operator on the Boolean hypercube>, so
$$
\lVert T_{1/\sqrt3}f\rVert_4^m
\leq\sqrt{mn+1}\,\lVert f\rVert_2^m.
$$
Taking $m$th roots and the <limit> $m\to\infty$ proves the <hypercontractive inequality on the Boolean hypercube>
$$
\lVert T_{1/\sqrt3}f\rVert_4\leq\lVert f\rVert_2.
$$

The noise operators are self-adjoint and satisfy $T_\rho T_\sigma=T_{\rho\sigma}$. By the <duality of Lp spaces>,
$$
\lVert T_{1/\sqrt3}f\rVert_2
=\sup_{\lVert g\rVert_2=1}|\langle f,T_{1/\sqrt3}g\rangle|
\leq\lVert f\rVert_{4/3}.
$$
Consequently
$$
\operatorname{Stab}_{1/3}(f)
=\langle f,T_{1/3}f\rangle
=\lVert T_{1/\sqrt3}f\rVert_2^2
\leq\lVert f\rVert_{4/3}^2.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 168 / 2 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-168/2/i/solution