Since the paper was published, later it became clear that the key identity

valid for all functions , where is an arbitrary normed space, implies several important results on the hamming cube. I will briefly list some of these results together with proofs, hints, and references about how they follow from the identity.

**Enflo’s problem**: in an arbitrary Banach space , and an arbitrary , the inequality holds with some finite constant and all , if and only if the inequality holds for linear functions .

*Reference: see the paper.***Pisier’s inequality with dimension free constant**: In an arbitrary Banach space , and any , Pisier’s inequality

holds with a finite constant for all and all if and only if has finite cotype.

*Reference: see the paper*.**Pisier’s inequality with constant**: in an arbitrary Banach space , the inequality holds

holds for all and all .

*Reference*: see my comment to this post. (Thanks to A. Eskenazis for this remark)

*Remark: is sharp, the example is due to Talagrand. For the inequality holds without factor (see this post).* **L1 Poincaré inequality on the Hamming cube: **for any we have

*Reference:* *here is the proof* *from the identity*

.

*Remark:* *one can also get Poincaré inequality with constant using the identity, however, this does not give the correct growth on . The correct grow is of order , and this grow can be obtained using the identity in a much more subtle way. ***Talagrand’s conjecture**: There exists a universal finite constant such that for a we have

.

*Reference: see Ramon van Handel’s lecture. *

Hint: Notice a wonderful equality for all . Indeed, since takes values 0 or 1 you can write and now take the expectation. Next estimate the first term by just by integrating the identity from 0 to t, and assuming . By martingale arguments and hypercontractivity the second term can be estimated as . Now choose for some positive constants . **Kahn-Kalai-Linial inequality (KKL inequality)**: For any boolean function one has

.

*Remark 1: In a literature the objects are called influences in variable and are denoted as . KKL inequality has several important corollaries. One of them says that any boolean function with nonzero variance has an influential variable . To verify the corollary, it is a calculus problem to show that if for some then .* * In fact, this corollary is called KKL theorem, however, it is not hard to see that the theorem follows from the techniques of the original paper of KKL: one starts with the equality , and “amplifies” it by applying the equality to instead of *, *after which the fourier coefficients change as . Finally, one integrates the both sides of the equality with respect to the measure on , and applies the hypercontractivity to get the inequality *

*Remark 2: another nice corollary of KKL theorem is that if you have a monotone function such that , then one can find about variables such that if we assign them values then the average of with respect to the rest of the variables will become at least . In social choice this means that in a monotone election, a candidate can bribe around votes to win the election. To obtain this corollary, first switch the most influential variable to , and repeat these steps times. Notice that on each step we get a new function , and it follows from monotonicity that . *

*Reference: One can get corollary on the lower bound in KKL inequality from Talagrand’s conjecture, however, this looks like to be a long path. So here is the proof of the KKL inequality via the identity. Since is boolean for any . Apply the identity to instead of , integrate over in , and take the norm of both sides. We obtain *

.

*Remark: * The advantage of this “*more involved*” proof I presented above is that it extends to Banach space valued functions having Rademacher type 2. See this paper. Such an extension cannot be obtained with original arguments because, otherwise, this would solve Enflo’s problem which was open for a long time.