Questioning a Proof of Khinchine Inequality

Question

[Khinchine Inequality] Let $a_1,\ldots,a_n\in R$, $\varepsilon_1,\ldots,\varepsilon_n$ be i.i.d. Rademacher random variables: $P(\varepsilon_i=1)=P(\varepsilon_i=-1)=0.5$, and $0<p<\infty$. Then

\begin{equation} A_p(\sum^{n}_{i=1}\mid a_i\mid^2)^{1/2}\leq (E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^p)^{1/p}\leq B_p(\sum^{n}_{i=1}\mid a_i\mid^2)^{1/2} \end{equation} for some constants $A_p$ and $B_p$ depending on $p$.

Proof. Let $\sum \mid a_i\mid^2=1$ WLOG. Then \begin{align} E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^p&=\int^{\infty}_{0}P(\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^p\geq s^p)ds^p\\ &=\int^{\infty}_{0}P(\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid\geq s)ps^{p-1}ds\\ &\leq \int^{\infty}_{0} 2\exp(-s^2/2)ps^{p-1}ds \quad \textbf{Hoeffding Inequality}\\ & =(B_p)^p , \quad \text{when} \quad p\geq 2. \end{align} When $0<p<2$, \begin{align} E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^p&\leq E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^2\\ &=E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^{\frac{2}{3}p+(2-\frac{2}{3}p)}\\ &\leq (E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^p)^{2/3} (E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^{6-2p})^{1/3}\quad \textbf{Holder Inequality}\\ &\leq (B_{6-2p})^{2-\frac{2}{3}p}(E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^{p})^{2/3} \end{align} Thus $E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^p\leq (B_{6-2p})^{6-2p}$, completing the proof.

MY QUESTION: When $0<p<2$, why is it necessary that $E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^p\leq E\mid \sum^{n}_{i=1}a_i\varepsilon_i\mid^2$?

Answer 1

Since the functions $\varepsilon_i$ are mutually orthogonal in $L^2$ and have unit norm, it follows that $$E\left|\sum a_i \varepsilon_i \right|^2 = \sum |a_i|^2=1$$

By Jensen's inequality, for any $r\ge 1$ and any nonnegative measurable function $f$ on a probability space $\Omega$ we have $$\left( \int_\Omega f\right )^r\le \int_\Omega f^r$$ Applying this with $r=2/p$ and $f(\varepsilon)=|\sum a_i \varepsilon_i|^p$, we get $$\left( E\left|\sum a_i \varepsilon_i \right|^p \right)^{2/p}\le E\left|\sum a_i \varepsilon_i \right|^{2}=1 $$ Raise both sides to power $p/2$ to get $$ E\left|\sum a_i \varepsilon_i \right|^p \le 1 = E\left|\sum a_i \varepsilon_i \right|^{2} $$

Answer 2

Your proof works for $p>2$ but does not allow to conclude for $p<2$: the answer to your question is that your inequality is a consequence of Jensen Inequality because the right hand side is $1$, as already pointed out. But it does not prove the theorem. In this case the second inequality in Khintchine Inequality comes directly from Jensen and you want to prove the first one. The classical proof consists in using Hölder Inequality to write $$\sum a_i^2=E|\sum a_i\varepsilon_i|^2\leq\left(\leq E|\sum a_i\varepsilon_i|^p\right)^{2/p}\left(\leq E|\sum a_i\varepsilon_i|^{2s}\right)^{1/s}$$ with $s$ the conjugate exponent of $2/p$. Since $2s>2$, we can use Khintchine Inequality for exponents larger than $2$ to conclude that the second factor is smaller than $B_s^2 (\sum a_i^2)^{1/s}$. You then pass this quantity in the left hand side to conclude.