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<p<\infty, 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.

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1 Response to Khintchine’s inequality

  1. andré says:

    I could not proof your elementrary inequality

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