TCSLib

311 Boolean Analysis — Centered

311.1 Overview

Absolute-power remainders yield a uniform scalar reflected contraction for every exponent at least two. Its affine form gives norm contraction for centered random variables on probability spaces.

311.2 Declarations

Theorem 311.1 Quadratic remainder lower bound
✓

For every \(q\ge 2\) and \(x\in \mathbb R\), \(x^2\le |1+x|^q-1-qx\).

Theorem 311.2 Integer remainder upper bound
✓

For every \(m\in \mathbb N\) and \(x\in \mathbb R\) with \(|x|\le 1\), \((1+x)^m-1-mx\le (2^m-1-m)x^2\).

Theorem 311.3 Local real-power remainder bound
✓

For every \(q\ge 2\), an integer \(m\ge q\), and \(|x|\le 1\), \(|1+x|^q-1-qx\le 2^m x^2\).

Theorem 311.4 Global remainder upper bound
✓

For every real \(q\ge 2\), there is \(M\gt 0\) such that \(|1+x|^q-1-qx\le M(x^2+|x|^q)\) for every real \(x\).

Theorem 311.5 Translated absolute-power bound
✓

For every \(q\ge 0\) and \(x\in \mathbb R\), \(|x|^q\le 2^q(|1+x|^q+1)\).

Theorem 311.6 Power controlled by affine remainder
✓

For every \(q\ge 2\) and \(x\in \mathbb R\), \(|x|^q\le 2^q(q+3)(|1+x|^q-1-qx)\).

Theorem 311.7 Uniform scalar reflected contraction
✓

For every \(q\ge 2\) there is \(0\lt c\le 1\) such that \(|1-cz|^q\le |1+z|^q-q(1+c)z\) for every real \(z\).

Theorem 311.8 Reflected contraction with quartic constant
✓

For every \(q\ge 2\) there is \(0\lt c\le 1\), with \(c=2/5\) when \(q=4\), such that \(|1-cz|^q\le |1+z|^q-q(1+c)z\) for every real \(z\).

Theorem 311.9 Affine scalar reflected contraction
✓

Let \(q\gt 0\) and \(0\le c\le 1\) satisfy \(|1-cz|^q\le |1+z|^q-q(1+c)z\) for every real \(z\). Then for all real \(a,x\), \(|a-cx|^q\le |a+x|^q-[q(1+c)|a|^q/a]x\), with the quotient interpreted as zero when \(a=0\).

Theorem 311.10 Centered random-variable contraction
✓

Let \((\Omega ,\mu )\) be a probability space, \(q\ge 2\), and \(0\le c\le 1\) satisfy \(|1-cz|^q\le |1+z|^q-q(1+c)z\) for every real \(z\). If \(X\in L^q(\mu )\) is real with \(\int X\, d\mu =0\), then for every \(a\in \mathbb R\), \(\| a-cX\| _{L^q(\mu )}\le \| a+X\| _{L^q(\mu )}\).