= Solution
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.
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