Khintchine’s inequality

It was recently suggested to me that the proof of Khintchine’s inequality is one which I should think about deeply, and I would especially like to explore the relationship between Khintchine’s and Rudin’s inequality. This will be the first of a series of short posts about these inequalities.

We begin with Khintchine’s inequality. We have a set of real numbers $a_n$ and a corresponding set of independent random variables $\epsilon_n$ taking the values $\pm1$ with equal probability. Khintchine’s inequality shows that we can control the sum $\sum \epsilon_na_n$ in any $L^p$ norm by the $L^2$ norm of the coefficients $a_n$. The slogan here is “changing the signs of a sum is well-behaved on average”.

More precisely, for any $0, we have the following bound

$\mathbb{E}\lvert\sum\epsilon_na_n\rvert^p\leq C_p\left(\sum a_n^2\right)^{p/2}.$

The constant $C_p$ depends only on $p$, and we shall obtain an explicit value below. The inequality is perhaps more suggestive if we observe that $\sum a_n^2=\| \sum\epsilon_na_n\|_2^2$, where the $L^2$ norm is taken over the probability space, so that we can write the inequality as

$\| \sum\epsilon_na_n\|_p\ll_p \|\sum\epsilon_na_n\|_2.$

Hence for random variables of this shape we have very good control over all $L^p$ norms.

We now give the proof, which is a surprisingly straightforward combination of exponential means and Markov’s inequality, polished off with a simple integral. First note that the independence of the $\epsilon_n$ combined with the elementary inequality $e^x+e^{-x}\leq 2e^{x^2/2}$ gives

$\mathbb{E}\left( e^{t\sum\epsilon_na_n}\right)\leq e^{\frac{t^2}{2}\sum a_n^2}.$

For any $\lambda>0$ applying Markov’s inequality and setting $t=\lambda/\sum a_n^2$ gives the inequality

$\mathbb{P}\left(\lvert \sum\epsilon_na_n\rvert\geq \lambda\right)\leq 2e^{-\lambda^2/2\sum a_n^2}.$

Recalling the distributional definition of $L^p$ norms, that is,

$\mathbb{E}\lvert f\rvert^p=p\int \lambda^{p-1}\mathbb{P}(\lvert f\rvert\geq \lambda)\,\textrm{d}\lambda,$

and making the substitution $\lambda=\left(2\sum a_n^2x\right)^{1/2}$ to do the integration, we compute that

$\mathbb{E}\left(\lvert \sum \epsilon_na_n\rvert^p\right)\leq 2^{p/2}p\Gamma(\frac{p}{2})\left(\sum a_n^2\right)^{p/2}$

and the proof is complete. Let me just record for future use the following explicit form which follows from the proof above together with Stirling’s formula:

$\| \sum\epsilon_na_n\|_p\leq C\sqrt{p}\|\sum\epsilon_na_n\|_2$

where $C$ is some absolute constant.

The proof above is taken from Thomas Wolff’s excellent lecture notes on harmonic analysis. Of course, the proof also holds for complex $a_n$ with appropriate modulus signs scattered about. Using duality and the fact that equality holds for $p=2$ we also get a similar lower bound.