Let \(s\subseteq I\) and \(t\subseteq J\) be finite sets, let \(r\ge 1\), and let \(a:I\times J\to \mathbb {R}\) satisfy \(a(i,j)\ge 0\) for \(i\in s\), \(j\in t\). Then \((\sum _{j\in t}(\sum _{i\in s}a(i,j))^r)^{1/r}\le \sum _{i\in s}(\sum _{j\in t}a(i,j)^r)^{1/r}\).

Let \(0\lt q\le p\) and \(F:\{ 0,1\} ^n\times \{ 0,1\} \to [0,\infty )\). Writing \(M_{r,z}u=(\mathbb {E}_z u^r)^{1/r}\) for uniform means, \(M_{q,x}(M_{p,b}F(x,b))\ge M_{p,b}(M_{q,x}F(x,b))\).

Let \(q\lt 1\), \(0\le \rho \le \sigma \le 1\), and \(f:\{ 0,1\} ^n\to [0,\infty )\). Then \(M_q(T_\rho f)\ge M_q(T_\sigma f)\). For nonnegative functions \(u\), the extended mean is \(M_r(u)=(\mathbb {E}u^r)^{1/r}\) when \(r\gt 0\); the same formula applies for \(r\lt 0\) when \(u\) is positive everywhere. For positive \(u\), \(M_0(u)=\exp (\mathbb {E}\log u)\); if \(r\le 0\) and \(u\) has a zero, \(M_r(u)=0\). All expectations are uniform. For \(d\in \mathbb N\), \(h:\{ 0,1\} ^d\to \mathbb {R}\), and \(\tau \in \mathbb {R}\), the spectral noise operator is \(T_\tau h(x)=\sum _{S\subseteq [d]}\tau ^{|S|}\widehat h(S)\chi _S(x)\), where \(\chi _S(x)=(-1)^{\sum _{i\in S}x_i}\) and \(\widehat h(S)=\mathbb {E}[h\chi _S]\) under the uniform measure.