=== PAGE 1 ===
arXiv:2407.04835v1  [math.PR]  5 Jul 2024
SHARPENING THE GAP BETWEEN L1 AND L2 NORMS
PAATA IVANISVILI AND YONATHAN STONE
Abstract. We reﬁne the classical Cauchy–Schwartz inequality ∥X∥1 ≤∥X∥2
by demonstrating that for any p and q with q > p > 2, there exists a constant
C = C(p, q) such that
∥X∥1 ≤1 −C ·

∥X∥p
p −1
 q−2
q−p

∥X∥q
q −1
 p−2
q−p
holds true for all Borel measurable random variables X with ∥X∥2 = 1 and
∥X∥p < ∞.
We illustrate two applications of this result:
one for biased
Rademacher sums and another for exponential sums.
1. Introduction
1.1. Separating ﬁrst and second moments. Let F be a family of Borel mea-
surable random variables. For any p > 0, deﬁne ∥X∥p := (E|X|p)1/p.
Assum-
ing ∥X∥2 = 1 for any X ∈F, then the Cauchy–Schwarz inequality tells us that
∥X∥1 ≤1. In this paper, our goal is to understand under what conditions on X
there exists ε > 0 such that
∥X∥1 ≤1 −ε
(1.1)
holds for all X ∈F.
The conditions on X should be such that one can easily verify them. There
are two typical examples we keep in mind: F = {Pn
j=1 e2πj2iθ, n ≥2} with θ ∼
unif([0, 1]), and F = {Pn
j=1 ajξj, n ≥1, aj ∈R}, where ξj are i.i.d.
random
variables. Notice that if ∥X∥1 ≤1 −ε, then it follows from H¨older’s inequality
that 1 ≤∥X∥θ
1∥X∥1−θ
p
for any p > 2, where θ =
p−2
2(p−1). Hence, we arrive at the
necessary condition ∥X∥p ≥1 + C(p)ε for some positive constant C(p) > 0, which
in practice can be veriﬁed for speciﬁc values of p. However, random variables X
taking two values show that this condition is not suﬃcient for (1.1).
The main result in this paper shows that if the necessary condition holds for
some p > 2 and we also have good control on the growth of ∥X∥q for some q > p,
then we get (1.1).
Theorem 1. Let q > p > 2 be ﬁnite. Then there exists a constant C = C(p, q) > 0
such that
∥X∥1 ≤1 −C

∥X∥p
p −1
 q−2
q−p

∥X∥q
q −1
 p−2
q−p
holds for any Borel measurable random variable X with ∥X∥2 = 1 and ∥X∥p < ∞.

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2
PAATA IVANISVILI AND YONATHAN STONE
Remark 1.1. We will see from the proof that
C(p, q) =
inf
a,c∈(0,1)
(cq−2(aq −1) + cq(a2 −aq) + 1 −a2)
p−2
q−p
(cp−2(ap −1) + cp(a2 −ap) + 1 −a2)
q−2
q−p · (1 −c)(1 −a)(1 −ac),
and this result is sharp, i.e., the sharpness of the constant C(p, q) can be veriﬁed
using random variables taking on two values. Our proof will show
C(p, q) ≥(min{1, q −2})
p−2
q−p
p
2(q−2)
q−p
> 0
however, sometimes the constant C(p, q) can be explicitly computed for given pow-
ers p and q. For example, we will see that C(4, 6) = 1/3.
Remark 1.2. One could use arguments from the classical moment problem, specif-
ically the positive semideﬁniteness of two matrices whose entries are the moments
of X chosen in a particular way (see page 781 in [2]), to prove Theorem 1 when p
and q are integers. This approach seems feasible for small integers p and q, such
as p = 4 and q = 6. However, the computational complexity may increase with
larger values of p and q. Our method is diﬀerent and does not require p and q to
be integers.
1.2. An application to exponential sums. In [3] Bourgain asked the following
question: does there exist ε > 0 such that for any ﬁnite subset S ⊂Z of distinct
integers with |S| > 1 we have
Z 1
0

1
p
|S|
X
j∈S
e2πijθ

dθ ≤1 −ε.
(1.2)
It was proved in [1] that if such ε > 0 exists then it must be at most 1−
√π
2 . Denote
XS =
1
p
|S|
X
j∈S
e2πijθ,
where θ ∼unif[0, 1]. The best result in (1.2) is due to Bourgain [3], who showed
existence of a constant c > 0 such that
∥XS∥1 ≤1 −clog(|S|)
|S|
.
Bourgain’s question is already nontrivial for squares of integers, i.e.,
Q := {j2 | j = 1, . . . , m},
and to the best of our knowledges the inequality (1.2) is open in this case. In this
case Bourgain [4] showed that for any q > 4 we have
∥XQ∥q
q ≤C(q)|Q|
q
2 −2
for some
C(q) < ∞,
(1.3)
∥XQ∥4
4 ≥C log(|Q|)
with some positive C > 0.
(1.4)
Combining the estimates (1.3) and (1.4) together with Theorem 1 applied with
p = 4 we obtain
∥XQ∥1 ≤1 −C′(q)log
q−2
q−4 (|Q|)
|Q|

=== PAGE 3 ===
SHARPENING THE GAP BETWEEN L1 AND L2 NORMS
3
holds with some constant C′(q) > 0. Given any N > 1 we can choose q > 4 so that
q−2
q−4 = N, thus we obtain
Corollary 1. For any N > 0 there exists a constant c(N) > 0 such that
∥XQ∥1 ≤1 −c(N)logN(|Q|)
|Q|
.
1.3. An application to L1 Poincar´e inequality on the hypercube. Fix p ∈
(0, 1), and let ξ1, . . . , ξn be i.i.d. Bernoulli random variables such that
ξ1 =



q
1−p
p
with probability
p,
−
q
p
1−p
with probability
1 −p.
(1.5)
Clearly Eξ1 = 0, and E|ξ1|2 = 1. The following theorem was proved in [8].
Theorem 2. There exists ε > 0 such that
sup
n≥1
sup
a2
1+...+a2
n=1
E

n
X
j=1
ajξj

< 1
holds for any p ∈(3/4 −ε, 3/4 + ε).
Notice that the conclusion of the theorem does not hold for p = 1/2. The theorem
was one of the technical steps in proving existence of a small constant δ > 0 such
that
E|f(x) −Ef(x)| ≤(π
2 −δ)E|∇f|(x)
(1.6)
holds for any f : {−1, 1}n 7→R, all n ≥1, where x = (x1, . . . , xn) ∼{−1, 1}n,
|∇f|(x) =


n
X
j=1
|Djf(x)|2


1/2
,
and
Djf(x) := f(x) −f(Sj(x))
2
,
here
Sj(x) = (x1, . . . , xj−1, −xj, xj+1, . . . , xn).
The estimate (1.6) was particularly surprising due to the variety of proofs leading
to the same bound E|f(x) −Ef(x)| ≤π
2 E|∇f|(x). Notably, one of these proofs
employed a non-commutative approach by L. Ben-Efraim and F. Lust-Piquard [5].
The consistent appearance of the constant π/2 across these proofs suggested that
the Cheeger constant in the L1 Poincar´e inequality might be π/2. However, this
conjecture was refuted in [8] by establishing (1.6).
Recently R. van Handel [7] obtained more quantitative bound on δ by showing
∥X∥1 ≤
s
1 −(∥X∥4
4 −1)2
32∥X∥6
6
hold for all random variables X with ∥X∥2 = 1, and ∥X∥4 < ∞, which allowed him
to obtain
sup
n≥1
sup
a2
1+...+a2n=1
E

n
X
j=1
ajξj

≤
r
1 −min {4p3(1 −p)3, (2p −1)4p(1 −p)}
480
.
(1.7)

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4
PAATA IVANISVILI AND YONATHAN STONE
The inequality (1.7) combined together with techniques from [9] gives δ ≈0.000006....
As an immediate application of Theorem 1 we will show the following
Corollary 2. We have
sup
n≥1
sup
a2
1+...+a2n=1
E

n
X
j=1
ajξj

≤1 −min

4p3(1 −p)3, (2p −1)4p(1 −p)
	
45 −3(p(1 −p))3
(1.8)
holds for all p ∈[0, 1].
The corollary combined with arguments from [9] gives the following
Corollary 3. For any f : {−1, 1}n 7→R we have that
E|f(x) −Ef(x)| ≤
π
2 −δ

E|∇f|(x)
(1.9)
holds with δ ≈0.00013...
Remark 1.3. It seems to us that the best δ one can choose in (1.9) is π
2 −p π
2 ≈
0.31748..., at least this is the case for an analogous question in Gauss space, however,
to prove such a statement one needs to come up with techniques diﬀerent from
Section 2.4. Indeed, let X = Pn
j=1 ajξj be a biased Rademacher sum. Let R =
R(p) ∈(0, 1) be the best constant, for each p ∈(0, 1), such that ∥X∥1/∥X∥2 ≤R(p).
In [12] it was proved that
sup
n≥1, a2
j=aj
∥X∥1
∥X∥2
= 2
p
N(p∗)p∗(1 −p∗)N(p∗)−1,
(1.10)
where N(p) = ⌊
1
1−(1−p)2 ⌋, and p∗= min(p, 1 −p). Therefore, the best δ one may
hope to get in (1.9) using the argument in Section 2.4, assuming R(p) equals to
the right hand side of (1.10), is π
2 −
R 1
1/2 2
p
N((1 −p))(1 −p)pN(1−p)−1
dp
√
p(1−p) ≈
0.149... which is still far from π/2 −
p
π/2 ≈0.31748....
2. Proofs
2.1. Proof of Theorem 1. By rescaling we can rewrite the claimed inequality in
the theorem as follows
∥X∥2 −∥X∥1 ≥C(p, q)
∥X∥2
·

∥X∥p
p −∥X∥p
2
 q−2
q−p

∥X∥q
q −∥X∥q
2
 p−2
q−p
(2.1)
which holds for all Borel random variables X with ∥X∥p < ∞. If ∥X∥q = ∞there
is nothing to prove, so in what follows we consider random variables X such that
∥X∥q < ∞. Without loss of generality, we will ﬁrst assume that ∥X∥∞< ∞since
by the Dominated Convergence Theorem, we can take XN = min{|X|, N} and note
that XN ≤|X| and
XN →|X|
a.e.
Furthermore, by homogeneity of (2.1), assume for now that 0 ≤X ≤1. Note that
since we take absolute values in all Lp norms that it suﬃces to prove (2.1) when
X ≥0. As for the assumption that X ≤1, we will brieﬂy need this assumption, but

=== PAGE 5 ===
SHARPENING THE GAP BETWEEN L1 AND L2 NORMS
5
note that we will soon replace it with a more useful stipulation. We now consider
the following optimization problem:
Φ(x, y, z) =
sup
0≤X≤1
{EX,
EX2 = x,
EXp = y,
EXq = z}.
We will note that fortunately for us the space curve γ(t) = (t2, tp, tq, −t),
t ∈[0, 1]
has totally positive torsion. We recall that this means that the leading principal
minors of the 4 × 4 matrix (γ(1)(t), γ(2)(t), γ(3)(t), γ(4)(t)) are positive for all t ∈
(0, 1). For the sake of completeness, we will check total positivity here. First of all,
the matrix is given by


γ(1)(t)
γ(2)(t)
γ(3)(t)
γ(4)(t)

=


2t
ptp−1
qtq−1
−1
2
p(p −1)tp−2
q(q −1)tq−2
0
0
p(p −1)(p −2)tp−3
q(q −1)(q −2)tq−3
0
0
p(p −1)(p −2)(p −3)tp−4
q(q −1)(q −2)(q −3)tq−4
0


and that the i-th leading principal minors Aii are each respectively given by
A11 = 2t,
A22 = tp−12p(p −1),
A33 = 2pq(p −2)(q −2)(q −p)tp+q−4,
A44 = 2p(p −1)(p −2)q(q −1)(q −2)(q −p)tp+q−7,
where positivity follows from the fact that t > 0 and that 2 < p < q. We can thus
invoke Theorem 2.3, case 1 in [6], which tells us that the supremum in Φ(x, y, z) is
achieved by a random variable X taking on at most two values. To that end, we
will assume in the sequel that
P{X = a} = 1 −P{X = b} = r.
As we no longer need it, we will drop the assumption X ≤1 at this point and
by homogeneity instead replace it with the assumption that ∥X∥2 = 1. Note that
since
∥X∥2
2 = a2r + b2(1 −r),
our new normalization assumption implies that
r = b2 −1
b2 −a2 .
Taking this into account, we obtain that
∥X∥1 = 1 + ab
a + b ,
meaning the desired inequality can now be expressed as
(1 −a)(b −1)
a + b
≥C(p, q)

ap(b2−1)+bp(1−a2)
b2−a2
−1
θ

aq(b2−1)+bq(1−a2)
b2−a2
−1
θ−1 ,
where θ = q−2
q−p, and we may furthermore assume without loss of generality that
0 < a < 1 < b < ∞, otherwise arbitrary. We next note that we can further simplify
this as

aq(b2−1)+bq(1−a2)+a2−b2
(1−a)(b−1)(b−a)
θ−1
≥C(p, q)

ap(b2−1)+bp(1−a2)+a2−b2
(1−a)(b−1)(b−a)
θ
.
(2.2)

=== PAGE 6 ===
6
PAATA IVANISVILI AND YONATHAN STONE
After a change of a variables b = 1/c, where c ∈(0, 1), the inequality (2.2) rewrites
and simpliﬁes as
cq−2(aq −1) + cq(a2 −aq) + 1 −a2
(1 −c)(1 −a)(1 −ac)
θ−1
≥
C(p, q)
cp−2(ap −1) + cp(a2 −ap) + 1 −a2
(1 −c)(1 −a)(1 −ac)
θ
.
Next, by deﬁning
B(a, c, q) := cq−2(aq −1) + cq(a2 −aq) + 1 −a2
(1 −c)(1 −a)(1 −ac)
,
we can take
C(p, q) =
inf
a,c∈(0,1)
B(a, c, q)θ−1
B(a, c, p)θ .
(2.3)
To show C(p, q) > 0 we claim that we can ﬁnd constants c1(p), c2(p) > 0 such that
c1(p) ≤B(a, c, p) ≤c2(p),
Then the equality (2.3) will imply
C(p, q) > c1(q)θ−1
c2(p)θ .
We will thus move on to ﬁnding upper and lower bounds on the function B(a, c, p),
where 2 < p < ∞and (a, c) ∈(0, 1) × (0, 1).
Lemma 2.1. We have B(a, c, p) ≥min{1, p −2} for all a, c ∈(0, 1) and all p > 2.
Proof. Let L :=
p−2
p , and for each c ∈(0, 1) consider the family of functions
f(a) = f(a; c) deﬁned as
f(a) := cp−2(ap −1) + cp(a2 −ap) + 1 −a2 −L(1 −c)(1 −a)(1 −ac).
Clearly the lemma is the same as f(a) ≥0 for all a, c ∈(0, 1). To verify f(a) ≥0
we will argue as follows: we will show that
(1) f(0) ≥0.
(2) f(1) = 0.
(3) f ′(1) ≤0.
(4) f ′′(0) < 0.
(5) f ′′ changes sign at most once from −to +.
It then follows that f(a) ≥0 on [0, 1]. Indeed, if f ′′ ≤0 on [0, 1] then we are done
because of (1) and (2). If f ′′ changes sign from −to + at a point k ∈[0, 1), then
(2) and (3) imply that f ≥0 on [k, 1]. In particular f(k) ≥0. Thus f ≥0 on [0, k]
because f ′′ ≤0 on [0, k], and this ﬁnishes the proof of Lemma 2.1.
To verify (1), we have
f(0) = −cp−2 + 1 −L(1 −c).
Since ϕ(c) = −cp−2 + 1 −L(1 −c) has the properties ϕ(0) = 1 −L ≥0, ϕ(1) = 0,
and ϕ′(c) = −(p −2)cp−3 + L changes sign at most once from + to −if p > 3, and
ϕ′ ≤0 if p ∈(2, 3] it follows that ϕ(c) ≥0 on [0, 1).

=== PAGE 7 ===
SHARPENING THE GAP BETWEEN L1 AND L2 NORMS
7
The veriﬁcation of (2) is trivial, so we move to verifying (3). We have
f ′(a) = pcp−2ap−1 + 2cpa −pcpap−1 −2a + L(1 −c)(1 −ac) + Lc(1 −c)(1 −a).
f ′(1) = pcp−2 + cp(2 −p) −2 + L(1 −c)2.
Let ψ(c) = pcp−2 + cp(2 −p) −2 + L(1 −c)2. We have ψ(1) = ψ′(1) = 0. If p > 3
then we see the coeﬃcients of the pseudo-polynomial
ψ(c) = cp(2 −p) + pcp−2 + Lc2 −2Lc + L −2
when arranged according to decreasing order of powers c have the signs −+ + −−.
Here, we do not know if p−2 ≥2 or p−2 < 2 but nevertheless there will be always
two sign changes in the coeﬃcients. Therefore by Descartes rule of signs for pseudo-
polynomials (see example #77 on page 46 in [10]) we obtain that ψ(c) has at most
two roots, therefore, ψ(c) does not have roots on (0, 1). Since ψ(0) = L −2 < 0 we
get ψ(c) ≤0 on [0, 1). If 2 < p ≤3 we have
ψ′(c) = p(2 −p)cp−1 + p(p −2)cp−3 + 2Lc −2L
= p(p −2)cp−3(1 −c2) −2L(1 −c)
= (1 −c)(p(p −2)cp−3(1 + c) −2L)
> (1 −c)(p(p −2) −2L) ≥0,
which implies that ψ(c) ≤0 on [0, 1).
Next we verify (4). We have
f ′′(a) = p(p −1)cp−2ap−2 + 2cp −p(p −1)cpap−2 −2 −2Lc(1 −c)
= ap−2p(p −1)cp−2(1 −c2) −2(1 −cp) −2Lc(1 −c).
(2.4)
Therefore f ′′(0) = −2(1 −cp) −2Lc(1 −c) < 0 since c ∈(0, 1).
Finally we verify (5). If follows from (2.4) that a 7→f ′′(a) is increasing, and
f ′′(0) < 0. Therefore f ′′ can change sign at most once from −to + on (0, 1].
□
Lemma 2.2. We have B(a, c, p) ≤p2 for all a, c ∈(0, 1) and all p ≥2.
Proof. Let M = p2, and consider
g(a) = cp−2(ap −1) + cp(a2 −ap) + 1 −a2 −M(1 −c)(1 −a)(1 −ac).
It suﬃces to show that g is concave, g(1) = 0 and g′(1) ≥0. The claim g(1) = 0 is
trivial. To verify g′(1) ≥0 we have
g′(1) = cp(2 −p) + pcp−2 + Mc2 −2Mc −2 + M.
Let h(c) = cp(2 −p) + pcp−2 + Mc2 −2Mc −2 + M. We have h(0) = M −2 ≥0,
h(1) = h′(1) = 0. Also h′′(1) = 2(M −(p2 −2p)) > 0. Since the coeﬃcient in front
of cp is negative then there must exist k > 1 such that h(k) = 0. The coeﬃcients of
h(c) have at most 3 sign changes, therefore by Descartes rule of signs for pseudo-
polynomials h cannot have roots in (0, 1), so h(c) ≥0, and hence g′(1) ≥0 on
[0, 1].
To verify concavity of g we have
g′′(a) = ap−2p(p −1)cp−2(1 −c2) −2(1 −cp) −2Mc(1 −c).
Since a 7→g′′(a) is increasing in a, it suﬃces to show that g′′(1) ≤0. We have
g′′(1) = −cp(p + 1)(p −2) + cp−2p(p −1) + 2Mc2 −2Mc −2.

=== PAGE 8 ===
8
PAATA IVANISVILI AND YONATHAN STONE
If p > 3 then the pseudo-polynomial v(c) = −cp(p + 1)(p −2) + cp−2p(p −1) +
2Mc2 −2Lc −2 has at most two sign changes in its coeﬃcients, hence at most two
roots (counting with multiplicities) on [0, ∞). On the other hand v(1) = v′(1) = 0,
and v′′(1) = −4p(p −2)2 < 0, therefore v(c) ≤0 on [0, 1]. Assume 2 < p ≤3.
Consider w(c) = v(c)/c. We have w(1) = 0. Let us show that w′(c) ≥0 on (0, 1).
Indeed
w′(c) = p(p −1)(p −3)cp−4 −(p2 −1)(p −2)cp−2 + 2M + 2
c2
≥p(p −1)(p −3)cp−4 −(p2 −1)(p −2) + 2M + 2
c2 := N(c).
The expression N(c) satisﬁes N(0) = +∞and N(1) = −2p(p −2) + 2M > 0. Also
Notice that N ′(c) = (p −4)(p −3)(p −1)pcp−5 −4
c3 = 0 if and only if
c = c0(p) =

4
(4 −p)(3 −p)p(p −1)
1/(p−2)
.
On the other hand we have
c2
0N(c0) =
4(p −3)p(p −1)
(4 −p)(3 −p)p(p −1) + c2(2M −(p2 −1)(p −2)) + 2
= −2(p −2)
4 −p
+ c2
0(2M −(p2 −1)(p −2)).
Since M = p2 ≥2(p −1)(p −2) for p ∈[2, 3], we have
−2(p −2)
4 −p
+ c2
0(2M −(p2 −1)(p −2)) ≥
−2(p −2)
4 −p
+ c2
0(4(p −1)(p −2) −(p2 −1)(p −2)) =
−2(p −2)
4 −p
+

4
(4 −p)(3 −p)p(p −1)
2/(p−2)
(p −1)(p −2)(3 −p) =
2t
2 −t
"
−1 +

24−t
(2 −t)2−t(1 −t)2−t(2 + t)2(1 + t)2−t
1/t#
,
where p = 2 + t for t ∈[0, 1]. Consider
f(t) = log

24−t
(2 −t)2−t(1 −t)2−t(2 + t)2(1 + t)2−t

.
We have f(0) = f ′(0) = 0. To show f(t) ≥0 on [0, 1] it suﬃces to verify that
f ′′(t) ≥0 on [0, 1]. A direct computation shows
f ′′(t) = −3t6 −14t5 + 8t4 + 52t3 −t2 −38t + 32
(2 −t)(1 −t)2(t + 2)2(1 + t)2
.
Notice that
−3t6 −14t5 + 8t4 + 52t3 −t2 −38t + 32 ≥
−3t4 −14t4 + 8t4 + 52t3 −t2 −38t + 32 =
−9t4 + 52t3 −t2 −38t + 32 ≥43t3 −39t + 32.

=== PAGE 9 ===
SHARPENING THE GAP BETWEEN L1 AND L2 NORMS
9
On the other hand 43t3 −39t + 32 ≥40t3 −40t + 30 = 10(4t3 −4t + 3). The map
t 7→4t3 −4t + 3 is positive at t = 0 and t = 1. It has a critical point at t = 1/
√
3
where its value is 3 −8
√
3
9
> 0. Thus f ′′ ≥0 on [0, 1].
□
2.2. Proof of C(4, 6) = 1/3. Let p = 4 and q = 6. We have
(cq−2(aq −1) + cq(a2 −aq) + 1 −a2)
p−2
q−p
(cp−2(ap −1) + cp(a2 −ap) + 1 −a2)
q−2
q−p · (1 −c)(1 −a)(1 −ac) −1
3
= c(2c −1)a2 −(c + 1)2a + 3c2 −c + 2
3(1 + c)(1 + a)(1 + ac)
.
It suﬃces to show infa,c∈(0,1) c(2c −1)a2 −(c + 1)2a + 3c2 −c + 2 = 0. Indeed,
the map r(a) = c(2c −1)a2 −(c + 1)2a + 3c2 −c + 2 is decreasing in a because
the linear function r′(a) = 2c(2c −1)a −(c + 1)2 satisﬁes r′(0) < 0 and r′(1) =
2c(2c −1) −(c + 1)2 = 3c2 −4c −1 = −3c(1 −c) −c −1 < 0. On the other hand
r(1) = (2c −1)2 ≥0 and it becomes equality at c = 1/2.
2.3. Proof of Corrollary 2. Let X = | Pn
j=1 ajξj|, where ξi, i = 1, . . . , n, are
i.i.d. random variables deﬁned as in (1.5), and Pn
j=1 a2
j = 1. We have ∥X∥2 = 1,
and
E|X|4 = 3 + (Eξ4
1 −3)
n
X
j=1
a4
j ≥min{3, Eξ4
1} = 1 + min

2, (2p −1)2
p(1 −p)

.
On the other hand if ξ′
j is independent copy of ξj, j = 1, . . . , n, and εj, j = 1, . . . , n,
are i.i.d. symmetric ±1 Rademacher random variables we have b a symmetrization
argument
E|X|6 = E|
n
X
j=1
ajξj|6 ≤E|
n
X
j=1
aj(ξj −ξ′
j)|6 = Eξ,ξ′Eε|
n
X
j=1
ajεj(ξj −ξ′
j)|6
≤15Eξ,ξ′


n
X
j=1
a2
j(ξj −ξ′
j)2


3
≤
15
p3(1 −p)3 ,
where we used Khinchin’s inequality E| P εibi|6 ≤15
 P b2
i
3 with the sharp con-
stant 15. Thus Theorem 1 applied with constant C(4, 6) = 1/3 gives
∥X∥1 ≤1 −min

4p3(1 −p)3, (2p −1)4p(1 −p)
	
45 −3(p(1 −p))3
.
2.4. Proof of Corrollary 3. We need the following identity obtained in [9]. For
any f : {−1, 1}n →R, we have
f(x) −Ef(x) =
Z ∞
0
1
√
e2t −1
Eζ
n
X
j=1
δj(t)Djf(xζ(t))dt,
(2.5)
where xζ(t) = (x1ζ1(t), ..., xnζn(t)), and the ζi(t) are i.i.d. random variables with
P{ζi(t) = ±1} = 1 ± e−t
2
,

=== PAGE 10 ===
10
PAATA IVANISVILI AND YONATHAN STONE
and δi = ζi(t)−Eζi(t)
√
Var(ζi(t)). We have
E|f(x) −Ef(x)| ≤
Z ∞
0
E

n
X
j=1
δj(t)Djf(xζ(t))

dt
√
e2t −1
=
Z ∞
0
E

n
X
j=1
δj(t)Djf(x)

dt
√
e2t −1
(p= 1+e−t
2
)
=
Z 1
1/2
E

n
X
j=1
ξjDjf(x)

dp
p
p(1 −p)
(1.8)
≤
E|∇f|
Z 1
1/2
 
1 −min

4p3(1 −p)3, (2p −1)4p(1 −p)
	
45 −3(p(1 −p))3
!
dp
p
p(1 −p)
≈
π
2 −δ

E|∇f|,
where
δ =
Z 1
1/2
min

4p3(1 −p)3, (2p −1)4p(1 −p)
	
45 −3(p(1 −p))3
dp
p
p(1 −p)
≈0.00013...
Acknowledgments. We thank Ramon van Handel for helpful comments. P.I was
supported in part by NSF CAREER-DMS-2152401.
References
[1] C. Aistleitner. On a problem of Bourgain concerning the L1-norm of exponential sums.
Math. Z, 275, 681–688 (2013)
[2] D. Bartsimas, I. Popescu. Optimal Inequalities in Probability Theory: A Convex Opti-
mization Approach. SIAM Journal on OptimizationVol. 15, Iss. 3 (2005)
[3] J. Bourgain. On the spectral type Ornstein’s class one transformations. Israel Journal of
Mathematics, vol. 84, pages 53–63 (1993)
[4] J. Bourgain. On Λp subsets of squares. Israel Journal of Mathematics, vol. 67, no. 3, pages
291–311 (1989)
[5] L. Ben Efraim, F. Lust-Piquard. Poincar´e type inequalities on the discrete cube and in
the CAR algebra. Prob. Theory Related Fields 141 (2008), no. 3-4, 569–602.
[6] J. de Dios Pont, P. Ivanisvili, J. Madrid. A new proof of the description of the convex
hull of space curves with totally positive torsion. arXiv:2201.12932 (accepted to Michigan
Mathematical Journal)
[7] R. van Handel. Private communication.
[8] P. Ivanisvili, R. van Handel, A. Volberg. Improving constant in end-point Poincar´e
inequality on Hamming cube. arXiv:1811.05584
[9] P. Ivanisvili, R. van Handel, A. Volberg. Rademacher type and Enﬂo type coincide.
Annals of Mathematics, vol. 192, Issue 2, pages 665-678.
[10] G. P´olya, G. Szeg¨o. Problems and Theorems in Analysis. Vol. II, Springer-Verlag, New
York, 1976
[11] M. Talagrand. Isoperimetry, logarithmic Sobolev inequalities on the discrete cube, and
Margulis graph connectivity theorem. Geom. Funct. Anal. 3 (1993). no. 3, 275-285
[12] Y. Stone. The L1 −L2 comparison results for biased Rademacher sums, to appear.
(P.I.) Department of Mathematics, University of California, Irvine, CA 92617, USA
Email address: pivanisv@uci.edu
(Y.S.) Department of Mathematics, University of California, Irvine, CA 92617, USA
Email address: ystone@uci.edu
