3 Unbalancing

Lemma 3.1 ✓

For any function \(f:G\to \mathbb {R}\) and integer \(k\geq 0\)

\[ \mathbb {E}_x f\circ f(x)^k\geq 0. \]

Proof ▼

Test.

Lemma 3.2 ✓

Let \(\epsilon \in (0,1)\) and \(\nu :G\to \mathbb {R}_{\geq 0}\) be some probability measure such that \(\widehat{\nu }\geq 0\). Let \(f:G\to \mathbb {R}\). If \(\lVert f\circ f\rVert _{p(\nu )}\geq \epsilon \) for some \(p\geq 1\) then \(\lVert f\circ f+1\rVert _{p'(\nu )}\geq 1+\tfrac {1}{2}\epsilon \) for \(p'=120\epsilon ^{-1}\log (3/\epsilon )\).

Proof ▼

Up to gaining a factor of 5 in \(p'\), we can assume that \(p\geq 5\) is an odd integer. Since the Fourier transforms of \(f\) and \(\nu \) are non-negative,

\[ \mathbb {E}\nu f^{p}= \widehat{\nu }\ast \widehat{f}^{(p)}(0_{\widehat{G}})\geq 0. \]

It follows that, since \(2\max (x,0)=x+\left\lvert x\right\rvert \) for \(x\in \mathbb {R}\),

\[ 2\langle \max (f,0),f^{p-1}\rangle _\nu =\mathbb {E}\nu f^{p}+\langle \left\lvert f\right\rvert ,f^{p-1}\rangle _\nu \geq \lVert f\rVert _{p(\nu )}^p\geq \epsilon ^p. \]

Therefore, if \(P=\{ x : f(x) \geq 0\} \), then \(\langle 1_{P},f^{p}\rangle _\nu \geq \frac{1}{2}\epsilon ^{p}\). Furthermore, if \(T=\{ x\in P : f(x) \geq \tfrac {3}{4}\epsilon \} \) then \(\langle 1_{P\backslash T},f^p\rangle _\nu \leq \tfrac {1}{4}\epsilon ^{p}\), and hence by the Cauchy-Schwarz inequality,

\[ \nu (T)^{1/2}\lVert f\rVert _{2p(\nu )}^{p}\geq \langle 1_{T}, f^{p}\rangle _\nu \geq \tfrac {1}{4}\epsilon ^{p}. \]

On the other hand, by the triangle inequality

\[ \lVert f\rVert _{2p(\nu )}\leq 1+\lVert f+1\rVert _{2p(\nu )}\leq 3, \]

or else we are done, with \(p'=2p\). Hence \(\nu (T)\geq (\epsilon /3)^{3p}\). It follows that, for any \(p'\geq 1\),

\[ \lVert f+1\rVert _{p'(\nu )}\geq \langle 1_{T},\left\lvert f+1\right\rvert ^{p'}\rangle _\nu ^{1/p'}\geq (1+\tfrac {3}{4}\epsilon )(\epsilon /3)^{3p/p'}. \]

The desired bound now follows if we choose \(p'=24\epsilon ^{-1}\log (3/\epsilon )p\), using \(1-x\leq e^{-x}\).