                                              SHARPENING THE GAP BETWEEN L1 AND L2 NORMS

                                                               PAATA IVANISVILI AND YONATHAN STONE
arXiv:2407.04835v1 [math.PR] 5 Jul 2024




                                                  Abstract. We refine the classical Cauchy–Schwartz inequality kXk1 ≤ kXk2
                                                  by demonstrating that for any p and q with q > p > 2, there exists a constant
                                                  C = C(p, q) such that
                                                                                                   q−2
                                                                                                     q−p
                                                                                         kXkpp − 1
                                                                        kXk1 ≤ 1 − C ·            p−2
                                                                                                     q−p
                                                                                         kXkqq − 1
                                                  holds true for all Borel measurable random variables X with kXk2 = 1 and
                                                  kXkp < ∞. 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 kXkp := (E|X|p )1/p . Assum-
                                          ing kXk2 = 1 for any X ∈ F , then the Cauchy–Schwarz inequality tells us that
                                          kXk1 ≤ 1. In this paper, our goal is to understand under what conditions on X
                                          there exists ε > 0 such that
                                                                                kXk1 ≤ 1 − ε                                      (1.1)
                                          holds for all X ∈ F .
                                             The conditions on X should be such that one can easily verify them. There
                                                                                               Pn        2
                                          are two typical examples we keep in mind: F = { j=1 e2πj iθ , n ≥ 2} with θ ∼
                                                                   Pn
                                          unif([0, 1]), and F = { j=1 aj ξj , n ≥ 1, aj ∈ R}, where ξj are i.i.d. random
                                          variables. Notice that if kXk1 ≤ 1 − ε, then it follows from Hölder’s inequality
                                                                                                p−2
                                          that 1 ≤ kXkθ1 kXk1−θ
                                                              p   for any p > 2, where θ = 2(p−1)   . Hence, we arrive at the
                                          necessary condition kXkp ≥ 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 kXkq 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
                                                                                             q−2
                                                                                             q−p
                                                                                 kXkpp − 1
                                                                kXk1 ≤ 1 − C                 p−2
                                                                                             q−p
                                                                                  kXkqq − 1
                                          holds for any Borel measurable random variable X with kXk2 = 1 and kXkp < ∞.
2                      PAATA IVANISVILI AND YONATHAN STONE


Remark 1.1. We will see from the proof that
    C(p, q) =
                                                          p−2
              (cq−2 (aq − 1) + cq (a2 − aq ) + 1 − a2 ) q−p
       inf                                                q−2    · (1 − c)(1 − a)(1 − ac),
     a,c∈(0,1) (cp−2 (ap − 1) + cp (a2 − ap ) + 1 − a2 ) q−p

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
                                                          p−2
                                       (min{1, q − 2}) q−p
                           C(p, q) ≥      2(q−2)
                                                      >0
                                         p q−p
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
                               1 X 2πijθ
                              p         e     dθ ≤ 1 − ε.                    (1.2)
                          0     |S| j∈S
                                                                                  √
It was proved in [1] that if such ε > 0 exists then it must be at most 1 − 2π . Denote
                                         1 X 2πijθ
                                 XS = p           e     ,
                                          |S| j∈S
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
                                             log(|S|)
                              kXS k1 ≤ 1 − c          .
                                                |S|
   Bourgain’s question is already nontrivial for squares of integers, i.e.,
                                Q := {j 2 | 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
                                       q
                  kXQ kqq ≤ C(q)|Q| 2 −2      for some       C(q) < ∞,                   (1.3)
                  kXQ k44 ≥ 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
                                                       q−2
                                               log q−4 (|Q|)
                                               ′
                            kXQ k1 ≤ 1 − C (q)
                                                    |Q|
                  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
                                                        logN (|Q|)
                              kXQ k1 ≤ 1 − c(N )                   .
                                                           |Q|
1.3. An application to L1 Poincaré inequality on the hypercube. Fix p ∈
(0, 1), and let ξ1 , . . . , ξn be i.i.d. Bernoulli random variables such that
                                 q
                                        1−p
                                             with probability p,
                         ξ1 =          qp                                      (1.5)
                                 −        p
                                          1−p  with probability 1 − p.

Clearly Eξ1 = 0, and E|ξ1 |2 = 1. The following theorem was proved in [8].
Theorem 2. There exists ε > 0 such that
                                                       n
                                                       X
                             sup      sup          E         aj ξj < 1
                             n≥1 a21 +...+a2n =1       j=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
                                             π
                        E|f (x) − Ef (x)| ≤ ( − δ)E|∇f |(x)                          (1.6)
                                             2
holds for any f : {−1, 1}n 7→ R, all n ≥ 1, where x = (x1 , . . . , xn ) ∼ {−1, 1}n,
                                                      1/2
                                        Xn
                          |∇f |(x) =      |Dj f (x)|2  ,
                                            j=1

and
             f (x) − f (Sj (x))
  Dj f (x) :=                   , here Sj (x) = (x1 , . . . , xj−1 , −xj , xj+1 , . . . , xn ).
                     2
   The estimate (1.6) was particularly surprising due to the variety of proofs leading
to the same bound E|f (x) − Ef (x)| ≤ π2 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 Poincaré 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
                                     s
                                            (kXk44 − 1)2
                             kXk1 ≤ 1 −
                                              32kXk66
hold for all random variables X with kXk2 = 1, and kXk4 < ∞, which allowed him
to obtain
                          n
                                    r
                         X               min {4p3 (1 − p)3 , (2p − 1)4 p(1 − p)}
   sup       sup       E     aj ξj ≤ 1 −                                         . (1.7)
   n≥1 a21 +...+a2n =1   j=1
                                                          480
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
                               n
                                                 min 4p3 (1 − p)3 , (2p − 1)4 p(1 − p)
                                                    
                               X
     sup      sup          E         aj ξj   ≤1−                                         (1.8)
     n≥1 a21 +...+a2n =1       j=1
                                                          45 − 3(p(1 − p))3

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
                                          π    
                      E|f (x) − Ef (x)| ≤    − δ E|∇f |(x)                               (1.9)
                                           2
holds with δ ≈ 0.00013...
Remark 1.3. It seems to us that the best δ one can choose in (1.9) is π2 − π2 ≈
                                                                                  p
0.31748..., at least this is the case for an analogous question in Gauss space, however,
to prove such a statement onePneeds to come up with techniques different from
Section 2.4. Indeed, let X = nj=1 aj ξj be a biased Rademacher sum. Let R =
R(p) ∈ (0, 1) be the best constant, for each p ∈ (0, 1), such that kXk1 /kXk2 ≤ R(p).
In [12] it was proved that
                                 kXk1       p                      ∗
                        sup             = 2 N (p∗ )p∗ (1 − p∗ )N (p )−1 ,         (1.10)
                    n≥1, a2j =aj kXk2

                      1              ∗
where N (p) = ⌊ 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
                                        R1 p
the right hand side of (1.10), is π2 − 1/2 2 N ((1 − p))(1 − p)pN (1−p)−1 √ dp       ≈
                                                                              p(1−p)
                                        p
0.149... which is still far from π/2 − 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
                                                         q−2
                                              p        p q−p
                                C(p, q)   kXk p  − kXk 2
                  kXk2 − kXk1 ≥         ·                p−2           (2.1)
                                kXk2
                                           kXkqq − kXkq2
                                                           q−p




which holds for all Borel random variables X with kXkp < ∞. If kXkq = ∞ there
is nothing to prove, so in what follows we consider random variables X such that
kXkq < ∞. Without loss of generality, we will first assume that kXk∞ < ∞ 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
                  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 {EX,         EX 2 = x,      EX p = y,     EX q = z}.
                          0≤X≤1

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
   (1)  
                 2t ptp−1                                qtq−1
                                                                                             
    γ (t)                                                                                −1
  γ (2) (t) 2 p(p − 1)tp−2                            q(q − 1)tq−2                    0 
             =                    p−3
                                                                                             
  γ (3) (t) 0 p(p − 1)(p − 2)t                        q(q − 1)(q − 2)tq−3             0 
    γ (4) (t)    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−1 2p(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 kXk2 = 1. Note that
since
                            kXk22 = a2 r + b2 (1 − r),
our new normalization assumption implies that
                                          b2 − 1
                                        r=        .
                                         b 2 − a2
Taking this into account, we obtain that
                                             1 + ab
                                 kXk1 =               ,
                                              a+b
meaning the desired inequality can now be expressed as
                                            p 2                     θ
                                             a (b −1)+bp (1−a2 )
                (1 − a)(b − 1)                      b2 −a2       −1
                               ≥ C(p, q)  q 2                      θ−1 ,
                    a+b                     a (b −1)+bq (1−a2 )
                                                    2
                                                  b −a  2       − 1
              q−2
where θ = 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
       q 2                        θ−1                                      θ
        a (b −1)+bq (1−a2 )+a2 −b2
                                                  p 2
                                                       −1)+bp (1−a2 )+a2 −b2
             (1−a)(b−1)(b−a)            ≥ C(p, q) a (b (1−a)(b−1)(b−a)          . (2.2)
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
                   q−2 q                                θ−1
                    c (a − 1) + cq (a2 − aq ) + 1 − a2
                                                               ≥
                           (1 − c)(1 − a)(1 − ac)
                           p−2 p                               θ
                           c (a − 1) + cp (a2 − ap ) + 1 − a2
                  C(p, q)                                          .
                                  (1 − c)(1 − a)(1 − ac)
Next, by defining
                                 cq−2 (aq − 1) + cq (a2 − aq ) + 1 − a2
                 B(a, c, q) :=                                          ,
                                        (1 − c)(1 − a)(1 − ac)
we can take
                                                B(a, c, q)θ−1
                           C(p, q) =      inf                 .                    (2.3)
                                       a,c∈(0,1) B(a, c, p)θ

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
                                                c1 (q)θ−1
                                   C(p, q) >              .
                                                 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).
                 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−2 ap−1 + 2cp a − pcp ap−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−2 ap−2 + 2cp − p(p − 1)cp ap−2 − 2 − 2Lc(1 − c)
               = ap−2 p(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 + M c2 − 2M c − 2 + M.
Let h(c) = cp (2 − p) + pcp−2 + M c2 − 2M c − 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−2 p(p − 1)cp−2 (1 − c2 ) − 2(1 − cp ) − 2M c(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−2 p(p − 1) + 2M c2 − 2M c − 2.
8                          PAATA IVANISVILI AND YONATHAN STONE


If p > 3 then the pseudo-polynomial v(c) = −cp (p + 1)(p − 2) + cp−2 p(p − 1) +
2M c2 − 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
                                                                     2
          w′ (c) = p(p − 1)(p − 3)cp−4 − (p2 − 1)(p − 2)cp−2 + 2M + 2
                                                                     c
                                    p−4     2                    2
                 ≥ p(p − 1)(p − 3)c     − (p − 1)(p − 2) + 2M + 2 := N (c).
                                                                 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 − c43 = 0 if and only if
                                                         1/(p−2)
                                              4
                    c = c0 (p) =                                   .
                                   (4 − p)(3 − p)p(p − 1)
On the other hand we have
                         4(p − 3)p(p − 1)
           c20 N (c0 ) =                     + c2 (2M − (p2 − 1)(p − 2)) + 2
                      (4 − p)(3 − p)p(p − 1)
                        2(p − 2)
                     =−          + c20 (2M − (p2 − 1)(p − 2)).
                         4−p
Since M = p2 ≥ 2(p − 1)(p − 2) for p ∈ [2, 3], we have
         2(p − 2)
       −          + c20 (2M − (p2 − 1)(p − 2)) ≥
           4−p
         2(p − 2)
       −          + c20 (4(p − 1)(p − 2) − (p2 − 1)(p − 2)) =
           4−p
                                              2/(p−2)
         2(p − 2)                 4
       −          +                                      (p − 1)(p − 2)(3 − p) =
           4−p         (4 − p)(3 − p)p(p − 1)
            "                                                  1/t #
                                       24−t
                  
        2t
             −1 +                                                     ,
       2−t           (2 − t)2−t (1 − t)2−t (2 + t)2 (1 + t)2−t

where p = 2 + t for t ∈ [0, 1]. Consider
                                               24−t
                                                                      
               f (t) = log                                               .
                             (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
                                 −3t6 − 14t5 + 8t4 + 52t3 − t2 − 38t + 32
                    f ′′ (t) =                                            .
                                      (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.
                  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 9 3 > 0. Thus f ′′ ≥ 0 on [0, 1].                          

2.2. Proof of C(4, 6) = 1/3. Let p = 4 and q = 6. We have
                                                         p−2
        (cq−2 (aq − 1) + cq (a2 − aq ) + 1 − a2 ) q−p                                               1
                                                          q−2   · (1 − c)(1 − a)(1 − ac) −
        (cp−2 (ap − 1) + cp (a2 − ap ) + 1 − a2 )         q−p                                       3
           c(2c − 1)a2 − (c + 1)2 a + 3c2 − c + 2
        =                                         .
                   3(1 + c)(1 + a)(1 + ac)
It suffices to show inf a,c∈(0,1) c(2c − 1)a2 − (c + 1)2 a + 3c2 − c + 2 = 0. Indeed,
the map r(a) = c(2c − 1)a2 − (c + 1)2 a + 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.
                                            Pn
2.3. Proof of Corrollary 2. Let X = | j=1 aj ξj |, where ξi , i = 1, . . . , n, are
                                                 Pn
i.i.d. random variables defined as in (1.5), and j=1 a2j = 1. We have kXk2 = 1,
and
                              n
                                                                (2p − 1)2
                             X                                           
           4          4           4             4
       E|X| = 3 + (Eξ1 − 3)      aj ≥ min{3, Eξ1 } = 1 + min 2,             .
                             j=1
                                                                 p(1 − p)

On the other hand if ξj′ 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
                   n
                   X                     n
                                         X                                     n
                                                                               X
           6                   6
      E|X| = E|          aj ξj | ≤ E|          aj (ξj − ξj′ )|6 = Eξ,ξ′ Eε |         aj εj (ξj − ξj′ )|6
                   j=1                   j=1                                   j=1
                                                  3
                              n
                              X                                 15
               ≤ 15Eξ,ξ′           a2j (ξj − ξj′ )2  ≤               ,
                              j=1
                                                           p3 (1 − p)3
                                                      P 2 3
where we used Khinchin’s inequality E| εi bi |6 ≤ 15
                                      P
                                                          bi with the sharp con-
stant 15. Thus Theorem 1 applied with constant C(4, 6) = 1/3 gives
                           min 4p3 (1 − p)3 , (2p − 1)4 p(1 − p)
                               
               kXk1 ≤ 1 −                                        .
                                     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
                               Z ∞             n
                                       1       X
              f (x) − Ef (x) =     √        Eζ   δj (t)Dj f (xζ(t))dt,     (2.5)
                                0    e2t − 1 j=1

where xζ(t) = (x1 ζ1 (t), ..., xn ζn (t)), and the ζi (t) are i.i.d. random variables with
                                                           1 ± e−t
                                     P{ζi (t) = ±1} =              ,
                                                              2
10                             PAATA IVANISVILI AND YONATHAN STONE


and δi = ζ√i (t)−Eζi (t) . We have
              Var(ζi (t))

                                       Z ∞       n
                                                 X                        dt
          E|f (x) − Ef (x)| ≤                E       δj (t)Dj f (xζ(t)) √
                                                                          2t
                                                                         e −1
                                       0         j=1

              Z ∞            n
                             X                    dt
          =          E           δj (t)Dj f (x) √
                                                  2t
                                                 e −1
               0             j=1

                    −t       Z 1       n              (1.8)
          (p= 1+e2       )             X      dp
               =              ξj Dj f (x) p
                                   E                    ≤
                      1/2 j=1               p(1 − p)
                 Z 1                                            !
                          min 4p3 (1 − p)3 , (2p − 1)4 p(1 − p)
                              
                                                                     dp
          E|∇f |       1−                            3
                                                                  p
                  1/2               45  − 3(p(1 − p))              p(1 − p)
            π       
          ≈      − δ E|∇f |,
              2
where
              Z 1
                min 4p3 (1 − p)3 , (2p − 1)4 p(1 − p)
                   
                                                         dp
         δ=                                3
                                                      p         ≈ 0.00013...
            1/2          45 − 3(p(1  −  p))            p(1 − p)
Acknowledgments. We thank Ramon van Handel for helpful comments. P.I was
supported in part by NSF CAREER-DMS-2152401.

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     (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
