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.