The 1971 Fortuin–Kasteleyn–Ginibre inequality for two monotone functions on a distributive lattice is well known and has seen many applications in statistical mechanics and other fields of mathematics. In 2008, one of us (Sahi) conjectured an extended version of this inequality for all n > 2 monotone functions on a distributive lattice. Here, we prove the conjecture for two special cases: for monotone functions on the unit square in $Rk$ whose upper level sets are k-dimensional rectangles and, more significantly, for arbitrary monotone functions on the unit square in $R2$. The general case for $Rk,k>2$, remains open.

For functions f, g on a probability space (Lμ), their expectation and correlation are defined by

$E1(f)=E(f)≔∫Lfdμ and E2(f,g)=E(fg)−E(f)E(g).$
(1.1)

Now suppose further that L is a distributive lattice8 and that the probability measure μ satisfies

$μ(a∨b)μ(a∧b)≥μ(a)μ(b).$
(1.2)

In this situation, if f, g are positive monotone (decreasing) functions9 on L, then one has

$E1(f)≥0 and E2(f,g)≥0.$
(1.3)

The first inequality is obvious, while the second is the celebrated FKG inequality of Fortuin–Kasteleyn–Ginibre1 that plays an important role in several areas of mathematics/physics. We will refer to a distributive lattice L with probability measure μ satisfying (1.2) as an FKG poset.

In formulating (1.2), we have tacitly assumed that the poset L is a discrete set. However, the FKG inequality also has important continuous versions, which can be proved by a discrete approximation. For example, if Qk = [0, 1]k is the unit hypercube in $Rk$ equipped with the partial order: xy if and only if xiyi for all i, then the FKG inequality holds for the Lebesgue measure and, more generally, for any absolutely continuous measure whose density function satisfies (1.2).

In Ref. 5, Sahi introduced a sequence of multilinear functionals En(f1, …, fn), n = 1, 2, 3, …, generalizing E1 and E2 (see Definition 3.1) and proposed the following conjecture:

Conjecture 1.1
(Ref.5 , Conjecture 5). Iff1, …, fnare positive monotone functions on an FKG poset, then
$En(f1,…,fn)≥0.$
(1.4)

Sahi5 proved the conjecture for the lattice {0, 1} × {0, 1} and for a certain subclass of positive monotone functions on the general power set lattice {0, 1}k equipped with a product measure. Since the functionals En satisfy the following “branching” property (Ref. 5, Theorem 6)

$En(f1,…,fn−1,1)=(n−2)En−1(f1,…,fn−1),$
(1.5)

the inequalities (1.4) form a hierarchy in the following sense: if Cn denotes the n-function positivity conjecture, then Cn implies Cn−1 for n > 2.

The work of Sahi was inspired by that of Richards4 who first had the idea of generalizing the FKG inequality to more than two functions. A natural first candidate for such an inequality is the cumulant (Ursell function) κn, but an easy example shows that the inequality already fails for κ3. Nevertheless, Richards [Ref. 4, Conjecture 2.5] conjectured the existence of such a hierarchy of inequalities, although without an explicit formula for En.

Indeed for n = 3, 4, 5, Sahi’s functional En coincides with the “conjugate” cumulant $κn′$ introduced by Richards [Ref. 4, formula (2.2)], although for n ≥ 6 one has $En≠κn′$. We note also that Ref. 4 contains two “proofs” of the positivity of $κ3′,κ4′,κ5′$—one for a discrete lattice and the other for a continuous analog. However, it seems to us that both proofs have essential gaps. Thus, beyond the special cases treated in Ref. 5, Conjecture 1.1 remains a conjecture, even for n = 3, 4, 5.

In this paper, we provide further evidence in support of Conjecture 1.1. We consider the continuous case of the Lebesgue measure on the unit hypercube Qk = [0, 1]k in $Rk$, and we prove the inequalities (1.4) for the following two additional cases:

• for arbitrary positive monotone functions on the unit square in $R2$ and

• for monotone characteristic functions of k-dimensional rectangles in [0, 1]k and, by multlinearity of En, for functions whose level sets are (not necessarily homothetic) rectangles.

First, we treat the case of three functions on $R2$ in Sec. II. This introduces several key ideas, including a reduction to a non-linear inequality involving decreasing sequences. In Sec. III, we define En for arbitrary n and prove Conjecture 1.1, first for characteristic functions of k-dimensional rectangles and then for general monotone functions on $R2$, that is, we extend Sec. II to all n > 3. This requires additional ideas involving the symmetric group Sn and an intricate induction on n. Subsections III A and III B are written in complete generality, and we hope these ideas will help in the eventual resolution of Conjecture 1.1.

Since the FKG inequality has many applications in probability, combinatorics, statistics, and physics, it reasonable to suppose that the generalized inequality will likewise prove to be useful in one or more of these areas. Although we do not have a compelling application in mind, we feel that it is important to find such an application. Indeed, the right application might provide additional insight into Conjecture 1.1 and perhaps even suggest a line of attack.

To end this section, we tantalize the reader with an interesting reformulation of the inequalities En ≥ 0 in terms of a formal power series from Ref. 5. First, if F(x) is a positive function on a probability space L, then it is natural to define the geometric mean of F by

$GF=expE(log⁡F).$
(1.6)

Now, suppose F(x, t) is a power series of the form

$F(x,t)=1−f1xt−f2xt2−⋯.$
(1.7)

Then, $logFx,t$ is a well defined power series, and formula (1.6) gives

$GF=expE(log(F(x,t))=1−c1t−c2t2−⋯,$
(1.8)

where the constants cj are certain algebraic expressions in various $E(fi1fi2⋯fip)$.

Conjecture 1.2

(Ref.5 , Conjecture 4). If thef1(x), f2(x), … is a sequence of positive monotone functions on an FKG poset, thencn ≥ 0 for alln.

It turns out that Conjectures 1.1 and 1.2 are equivalent. One implication has already been established in Sec. III of Ref. 5, and we prove the other direction in the  Appendix. We also refer the reader to Refs. 6 and 7 for related inequalities in an algebraic setting.

For three functions, the multilinear functional En introduced in Ref. 5 is given by

$E3(f,g,h)=2E(fgh)+E(f)E(g)E(h)−E(f)E(gh)−E(g)E(fh)−E(h)E(fg).$
(2.1)

We note that E3 is different from the cumulant (Ursell function), which is given by

$κ3(f,g,h)=E(fgh)+2E(f)E(g)E(h)−E(f)E(gh)−E(g)E(fh)−E(h)E(fg).$
(2.2)

We will consider the functional E3 for functions on the unit hypercube,

$Qk=[0,1]k={x=(x1,…,xk)|0≤xi≤1}$
(2.3)

equipped with the Lebesgue measure and the usual partial order: xx′ if and only if $xi≤xi′$ for all i. We say that a real valued function f on Qk is monotone (decreasing) if xx′ implies f(x) ≥ f(x′). We note that the FKG inequality is usually stated for monotonically increasing functions, but this is a somewhat arbitrary choice. Indeed, FKG and our theorems for decreasing functions are equivalent to the corresponding results for increasing functions. For a general FKG poset, this follows by reversing the partial order, and for Qk, this follows by the change of variables xi ↦ 1 − xi. We also note that monotonicity for Q1 has the usual 1-variable meaning of a decreasing function.

Theorem 2.1.

Iff, g, hare positive monotone functions on [0, 1]2, thenE3(f, g, h) ≥ 0.

The generalization of Theorem 2.1 to n functions is given in Theorem 3.6. We now reduce Theorem 2.1 to characteristic functions χS, SQk. These are defined by χS(x) = 1 if xS and χS(x) = 0 if xS. We will say S is monotone if χS is monotone.

Lemma 2.2.

It suffices to prove Theorem 2.1 forχS, χT, χUfor all monotoneS, T, U.

Proof.

Any positive f can be written as an integral over the characteristic functions of its upper level sets. Thus, $f(x)=∫0∞ξs(x)ds$, with ξs(x) = 1 if f(x) > s and 0 otherwise (see the “layer cake principle” in Ref. 2). If f is monotone, then ξs is monotone for every s. Since E3 is multi-linear in f, g, h, this reduces Theorem 2.1 to the case of monotone characteristic functions.■

We now describe a further reduction of Theorem 2.1 to a discrete family of characteristic functions. Let $A=A(m)$ be the set of decreasing m-tuples of integers, each between 0 and m,

$A(m)≔{a∈Zm|m≥a1≥⋯≥am≥0}.$
(2.4)

For each $a∈A$, we define a monotone subset Sa of Q2 = [0, 1]2 as follows. Divide Q2 uniformly into m2 little squares, write Di,j for the square with top right vertex (i/m, j/m), and set

$Sa=⋃j≤aiDi,j,χa=χSa.$
(2.5)

Then, Sa is a monotone subset of Q2, and, conversely, any monotone union of Di,j is of this form.

Lemma 2.3.

It suffices to prove Theorem 2.1 forχa, χb, χc;$a,b,c∈A(m)$; for allm.

Proof.

By Lemma 2.2, it suffices to consider monotone characteristic functions χS, χT, χU. Divide Q2 uniformly into m2 little squares Di,j as before, and let Sm, Tm, Um be the unions of the Di,j contained in S, T, U, respectively; then, these are monotone subsets of Q2 of the form (2.5). Moreover, $χSm,χSmχTm$, etc., converge to χS, χSχT, etc., in L1 as m. Thus, if $E3(χSm,χTm,χUm)≥0$, then we get $E3(χS,χT,χU)=limm→∞E3(χSm,χTm,χUm)≥0$.■

We now prove Theorem 2.1 for χa, χb, χc, which suffices by Lemma 2.3. To simplify the notation, we work directly with a, b, c and we define the product ab, expectation $E(a)$, etc., as follows:

$(ab)i=min{ai,bi},$
(2.6)
$E1(a)=E(a)=(a1+⋯+am)/m2,$
(2.7)
$E2(a,b)=E(ab)−E(a)E(b),$
(2.8)
$E3(a,b,c)=2E(abc)+E(a)E(b)E(c)−E(a)E(bc)−E(b)E(ac)−E(c)E(ab).$
(2.9)

Then, we have χab = χaχb, $E(a)=E(χa)$, E2(a, b) = E2(χa, χb), E3(a, b, c) = E3(χa, χb, χc).

In particular, by the FKG inequality, we obtain the following lemma:

Lemma 2.4.

For alla, bin$A$, we haveE2(a, b) ≥ 0.

To study E3(a, b, c), we consider certain perturbations of a. We say that $a∈A$ has a descent at i if ai > ai+1, and in this case, we can define three new sequences a = a−,i, a+ = a+,i, a = a⋆,i, also in $A$, in which the following changes, and only these, are made to a:

$ai−=ai+1,ai+1+=ai,ai+1⋆=ai+1+1.$
(2.10)

Lemma 2.5.

Ifahas a descent ati, butbdoes not, then we have$E(a+b)+E(a−b)=2E(ab)$.

Proof.
Let bi = bi+1 = β, say, then we have
$(a+b)i=(a+b)i+1=min{ai,β}=(ab)i,$
(2.11)
$(a−b)i=(a−b)i+1=min{ai+1,β}=(ab)i+1.$
(2.12)
Since the three sequences a+b, ab, and ab coincide except at i, i + 1, the result follows.■

Proposition 2.6.
Ifahas a descent ati, butbandcdo not, then
$E3(a+,b,c)+E3(a−,b,c)=2E3(a,b,c).$
(2.13)

Proof.
Each term of (2.9) has a unique factor involving a, which is of the form $E(ad)$, where d = 1, b, c, bc is a sequence in $A$ that does not have a descent at i. By Lemma 2.5, we get
$E(a+d)+E(a−d)=2E(ad).$
(2.14)
The result now follows from formula (2.9).■

Lemma 2.7.

Ifa, bhave a descent atiandbi+1ai+1, thenab = ab.

Proof.
Evidently, $(a⋆b)j=(ab)j$ for ji + 1, and since bi+1ai+1, we also have
$(a⋆b)i+1=bi+1=(ab)i+1.$
(2.15)
Thus, we get ab = ab, as claimed.■

Proposition 2.8.
Ifaandbhave a descent atiandbi+1ai+1, then we haveab = aband
$E3(a⋆,b,c)≤E3(a,b,c)for allc.$
(2.16)

Proof.
By Lemma 2.7, we get $E(a⋆b)=E(ab),E(a⋆bc)=E(abc)$, and it follows that
$E3(a,b,c)−E3(a⋆,b,c)=E2(b,c)E(a⋆)−E(a)+E(b)E(a⋆c)−E(ac).$
(2.17)
Evidently, we have $E(a⋆)≥E(a)$ and $E(a⋆c)≥E(ac)$, and by the FKG inequality, we also have E2(b, c) ≥ 0. Thus, all terms on the right-hand side of (2.17) are positive, which proves the result.■

Theorem 2.9.

For alla, b, cin$A$, we haveE3(a, b, c) ≥ 0.

Proof.

Let $U$ be the set of triples (a, b, c) in $A$ for which E3(a, b, c) attains its minimum, and let $V$ be the subset of $U$ for which the quantity $E(a)+E(b)+E(c)$ attains its maximum.

We claim that if $(a,b,c)∈V$, then a, b, c are constant sequences. If this is not the case, then a, say, has a descent at some i. If b, c do not have a descent at i, then by Proposition 2.6 we get
$E3(a,b,c)=E3(a+,b,c)+E3(a−,b,c)/2.$
By minimality, E3(a, b, c) ≤ E3(a±, b, c), which forces E3(a, b, c) = E3(a±, b, c). Replacing a by a+, we reach a contradiction since $E(a+)>E(a)$.

If b, say, also has a descent at i, then by symmetry we may assume bi+1ai+1. Then, by Proposition 2.8, E3(a, b, c) ≤ E3(a, b, c), and we again reach a contradiction since $E(a⋆)>E(a)$.

Now, we may assume a, b, c are constant sequences and, by symmetry, further assume that
$a≡mα,b≡mβ,c≡mγ,0≤α≤β≤γ≤1,$
and it follows that E3(a, b, c) = 2α + αβγ − (αβ + αβ + αγ) = α(1 − β) (2 − γ) ≥ 0.■

This proves Theorem 2.1 for χa, χb, χc and, thus, by Lemma 2.3, in general.

In this subsection and Subsection III B, we work with arbitrary functions on a probability space. We start by recalling the definition of the multilinear functional En(f1, …, fn) from Ref. 5. This involves the decomposition of a permutation σ in the symmetric group Sn as a product of disjoint cycles,

$σ=(i1,…,ip)(j1,…,jq)⋯.$
(3.1)

For σ as in (3.1), we write Cσ for the number of cycles in σ and we set

$Eσ(f1,…,fn)=E(fi1⋯fip)E(fj1⋯fjq)⋯.$
(3.2)

Then, the following definition is due to Sahi.5

Definition 3.1.
For functions f1, …, fn on a probability space X, we define
$En(f1,…,fn)=∑σ∈Sn(−1)Cσ−1Eσ(f1,…,fn).$
(3.3)

Using (3.3), one can easily verify that E1, E2, E3 coincide with their earlier definitions. We note that the factor of 2 in the term 2E(f1f2f3) in formula (2.1) comes from the two 3-cycles (123) and (213). More generally, En will have repeated terms because Eσ is unchanged if we rearrange the indices within a cycle. For example, for n = 4, we have

$E4(f1,f2,f3,f4)=6E(f1f2f3f4)−2E(f1)E(f2f3f4)+E(f2)E(f1f3f4)+⋯+E(f1)E(f2)E(f3f4)+E(f1)E(f3)E(f2f4)+⋯−E(f1f2)E(f3f4)+E(f1f3)E(f2f4)+⋯−E(f1)E(f2)E(f3)E(f4).$

We now give an explicit formula for En in a special case.

Lemma 3.2.
LetX = [0, 1] be the unit interval equipped with Lebesgue measure, and letfibe the characteristic function$χ[0,ai],0≤ai≤1$, with 0 ≤ a1 ≤ ⋯ ≤ an ≤ 1. Then, we have
$En(f1,…,fn)=a1(1−a2)⋯(n−1−an).$

We note that the above formula implies that En is positive, i.e., Conjecture 1.1 holds for the Lebesgue measure on [0, 1]. While it is easy enough to give a direct proof the lemma, we prefer to postpone the proof to Subsection III B where we will derive it as a consequence of a more general result.

We first prove a recursive formula relating En to En−1.

Proposition 3.3.
We haveEn(f1, …, fn−1, f) = e1 + ⋯ + en−1en, where
$ei=En−1(f1,…,fif,…,fn−1)if1≤i≤n−1,En−1(f1,…,fn−1)E(f)if i=n.$
(3.4)

Proof.
We write f = fn and consider expression (3.3) for En(f1, …, fn) as a sum over the symmetric group Sn. We decompose Sn as a disjoint union,
$Sn=S(1)∪⋯∪S(n),S(i)={σ∈Sn∣σ(i)=n}.$
(3.5)
Then, S(n) is a subgroup of Sn, naturally isomorphic to Sn−1. By (3.3), we have
$En(f1,…,fn)=Σ1+⋯+Σn,Σi=∑σ∈S(i)(−1)Cσ−1Eσ(f1,…,fn).$
(3.6)
To study Σi, we consider the map $σ↦σ̄$ defined by dropping n from the cycle decomposition of σ. Thus, for n = 5, we have (13)(245) ↦ (13)(24), (12)(34)(5) ↦ (12)(34), etc. Then, $σ↦σ̄$ defines a bijection from eachS(i) to Sn−1. If σ is in S(i) and in, then i and n occur in the same cycle of σ, and dropping n does not change the cycle count. This gives
$Cσ=Cσ̄,Eσ(f1,…,fn−1,f)=Eσ̄(f1,…,fif,…,fn−1),$
which implies Σi = ei. If σ is in S(n), then (n) occurs as a separate cycle in σ and we get
$Cσ=Cσ̄+1,Eσ(f1,…,fn−1,f)=Eσ̄(f1,…,fn−1)E(f),$
which gives Σn = −en. This proves the proposition.■

Lemma 3.2 is now an easy consequence.

Proof of Lemma 3.2.
Let $fi=χ[0,ai]$. Since aian for all i, we get
$fifn=χ[0,ai]χ[0,an]=χ[0,an]=fi.$
Now applying Proposition 3.3 with f = fn, we deduce that
$En(f1,…,fn)=(n−1)−E(fn)En−1(f1,…,fn−1)=(n−1−an)En−1(f1,…,fn−1).$
The result follows by a straightforward induction on n.■

We next establish a useful formula for the partial sum Pc of En over the set of permutations containing a fixed cycle c.

Proposition 3.4.
LetScdenote the set of permutationsσSnthat contain a fixed cyclec = (i1, …, ip) and letJc = {j1, j2, …} = {1, …, n}\{i1, …, ip}, then we have
$Pc≔∑σ∈Sc(−1)Cσ−1Eσ(f1,…,fn)=−E(fi1⋯fip)En−p(fj1,fj2,…)ifp
(3.7)

Proof.
The set Sc consists of a single permutation if p = n. Otherwise, it consists of permutations of the form σ = c · τ, where τ is a permutation of Jc. Evidently, the number of cycles in σ and τ are related by Cτ = Cσ − 1. Thus, in this case, we have
$(−1)Cσ−1Eσ(f1⋯fn)=−E(fi1⋯fip)(−1)Cτ−1Eτ(fj1,fj2,…).$
(3.8)
Now, the result follows by summing (3.8) over τ.■

By a rectangle in dimension k or a k-rectangle, we mean a subset of [0, 1]k of the form

$[0,r1]×⋯×[0,rk],0≤r1,…,rk≤1.$

Theorem 3.5.

Iffiare characteristic functions ofk-rectangles, thenEn(f1, …, fn) ≥ 0.

Proof.
We proceed by induction on k ≥ 1 and for a given k by induction on n ≥ 1. The base cases k = 1 and n = 1 are straightforward and the former by Lemma 3.2. Thus, we may assume k > 1 and n > 1, and we can write
$fi=gi×χ[0,ai],$
where gi is the characteristic function of a (k − 1)-rectangle. By symmetry of En, we may assume
$0≤a1≤⋯≤an≤1.$
(3.9)
We note that the assumption (3.9) on ai means that we have
$E(fi1⋯fip)=alE(gi1⋯gip),l=min{i1,…,ip}.$
(3.10)
Moreover, it follows from (3.3) and (3.10) that if a2 = ⋯ = an = 1, then we have
$En(f1,…,fn)=a1En(g1,…,gn).$
(3.11)
We now fix an index i > 1 and let C(i) denote all set of all cycles containing i, then we have
$En(f1,…,fn)=∑c∈C(i)Pc,$
where Pc is as in Proposition 3.4. If i is not minimal in c, then Pc is independent of ai by (3.10). If i is minimal in c, then 1 ∉ c; hence, c has length p < n, and by (3.7) and (3.10), we get
$Pc=−aibc,bc=E(gi1⋯gip)En−p(fj1,fj2,…).$
By induction on n, we have bc ≥ 0 for such c. This means that En(f1, …, fn) decreases as we increase a2, …, an subject, of course, to condition (3.9). In particular, En decreases as we successively increase
$an↗1,an−1↗1,…,a2↗1.$
By (3.11), we get En(f1, …, fn) ≥ a1En(g1, …, gn), which is positive by induction on k.■

If f is the characteristic function of a rectangle, then any level set of f is either the same rectangle or empty. However, using the layer-cake principle2 and multilinearity as in the Proof of Lemma 2.2, we obtain the following immediate extension of the previous result.

Corollary 3.6.

Iff1, …, fnare positive, monotone functions whose level sets are (not necessarily homothetic) rectangles, thenEn(f1, …, fn) ≥ 0.

Our main result is as follows:

Theorem 3.6.

Iff1, …, fnare positive and monotone on [0, 1]2, thenEn(f1, …, fn) ≥ 0.

As before, we can deduce this from the special case of χa as in (2.5).

Lemma 3.7.

It suffices to prove Theorem 3.6 for$χa1,…,χan$,$ai∈A(m)$, for allm.

Proof.

This is proved along the same lines as Lemmas 2.2 and 2.3.■

In this section, we work with $A=A(m)$ and to simplify the notation for a1, …, an in $A$, we set

$Eσ(a1,…,an)=E(χai1,…,χaip)E(χaj1,…,χajq)⋯,$
(3.12)
$En(a1,…,an)=∑σ∈Sn(−1)Cσ−1Eσ(a1,…,an).$
(3.13)

Then, we have $En(χa1,…,χan)=En(a1,…,an)$.

To study the positivity of En, we first consider a special case.

Proposition 3.8.
Ifaiiare constant sequences with 1 ≤ α1 ≤ ⋯ ≤ αn ≤ 0, then
$En(a1,…,an)=α1(1−α2)⋯(n−1−αn).$
(3.14)

Proof.
Let Ln = En(a1, …, an). Since αiαn, we have aian = ai for all i. Thus, we get
$Ln=∑i=1n−1Ln−1−Ln−1E(an)=(n−1−αn)Ln−1$
by Proposition 3.3. Now, (3.14) follows by induction on n, the case n = 1 being obvious.■

We now prove the generalization of Proposition 2.6.

Proposition 3.9.
Ifahas a descent ati, buta1, …, an−1do not, then we have
$2En(a1,…,an−1,a)=En(a1,…,an−1,a+)+En(a1,…,an−1,a−).$
(3.15)

Proof.

This is proved for each term Eσ in (3.13) in exactly the same way as Proposition 2.6 by applying Lemma 2.5 to the unique factor of Eσ involving a = an in (3.12).■

We shall prove the next three theorems together by induction on n.

Theorem 3.10.
Ifa1, …, an−2, bare in$A$;Sis a subset ofQ2; andχbχS = 0, then
$En(χa1,…,χan−2,χb,χS)≤0.$
(3.16)

Theorem 3.11.
Ifa1, …, an−2, b, care in$A$;b, chave a descent ati; andbi+1ci+1, then
$En(a1,…,an−2,b,c⋆)≤En(a1,…,an−2,b,c).$
(3.17)

Theorem 3.12.
For alla1, …, anin$A$, we have
$En(a1,…,an)≥0.$
(3.18)

Proof.

Let us write A(n), B(n), and C(n) for the assertions of Theorems 3.10, 3.11, and 3.12. Then, A(1), B(1) are vacuously true, while C(1) is evident. Therefore, it suffices to prove the implications A(n − 1) ∧ C(n − 1) ⟹ A(n) and A(n) ⟹ B(n) ⟹ C(n) for all n ≥ 2.

A(n − 1) ∧ C(n − 1) ⟹ A(n): By assumption, we have χbχS = 0, and we also have $χaiχS=χSi$, where $Si=S∩Sai$. Thus, by Proposition 3.3, we get
$En(χa1,…,χan−2,χb,χS)=e1+⋯+en−2+en−1−en,where ei≔En−1(χa1,…,χSi,…,χan−2,χb),i≤n−2,en−1≔En−1(χa1,…,χan−2,0),en≔En−1(χa1,…,χan−2,χb)E(χS).$
Now, en ≥ 0 by C(n − 1) and en−1 = 0 by (3.12) and (3.13). In addition, $χbχSi=(χbχS)χai=0$, and so by symmetry, we can apply A(n − 1) to conclude ei ≤ 0 for in − 2. This implies A(n), (3.16).
A(n) ⟹ B(n): Define $Sc,Sc⋆$ as in (2.5) and put $S=Sc⋆\Sc$, then by Lemma 2.7 we have
$χSχb=(χc⋆−χc)χb=χc⋆b−χcb=0.$
Thus, by A(n), (3.16), we get $En(χa1,…,χan−2,χb,χc⋆−χc)≤0$, which implies B(n), (3.17).

B(n) ⟹ C(n): This argument is similar to the Proof of Theorem 2.9. Let $M$ be the set of n-tuples a = (a1, …, an) in $A$ for which En(a) achieves its minimum, and let $N$ be the subset of $M$ for which $λ(a)=E(a1)+⋯+E(an)$ achieves its maximum on $M$. We claim that for a in $N$ each ai is a constant sequence; by Proposition 3.8, this clearly implies C(n), En(a) ≥ 0.

If the claim is not true, then one of the sequences has a descent at some i. First suppose that only one sequence, by symmetry an = a, has a descent at i. By Proposition 3.9 and minimality of En(a), we deduce En(a) = En(a1, …, an−1, a±). Thus, replacing a by a+ preserves En(a) but increases λ(a), which is a contradiction. If two sequences have a descent at i, then by symmetry we may assume these are an−1 = b, an = c with bi+1ci+1. Now, B(n), (3.17), implies that replacing c by c does not increase En(a), but it does increase λ(a), which is a contradiction.■

This proves Theorem 3.6 for $χai$ and, thus, by Lemma 3.7, in general.

This work was partially supported by the NSF (Grant Nos. DMS-1939600 and DMS-2001537) and the Simons Foundation (Grant No. 509766). The hospitality of the Institute for Advanced Study is gratefully acknowledged.

The authors have no conflicts to disclose.

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

We start by recalling some basic facts about partitions and permutations. For more background and details involving these ideas, we refer the reader to Ref. 3.

A partition λ of n, of length l, is a weakly decreasing sequence of positive integers,

$λ1≥λ2≥⋯≥λl>0 such that λ1+⋯+λl=n,$

we say that the λj are the parts of λ, and we write $lλ=l$ and $λ=n$.

The conjugation action of Sn permutes the indices in the cycle decomposition (3.1) of an element σ. Thus, the class of σ is uniquely determined by its “cycle type,” i.e., the partition λ whose parts are the cycle lengths of σ, arranged in decreasing order. Moreover, if $mi=miλ$ denotes the number of parts of size i, then the conjugacy class of cycle type λ contains n!/zλ elements, where

$zλ=∏i≥1imimi.$
(A1)

For a function f on a probability space, we define its moments by the formula

$pd(f)=E(fd) and pλ(f)=pλ1(f)⋯pλl(f).$
(A2)

Lemma A.1.

We have$En(f,…,f)=n!∑|λ|=n(−1)l(λ)−1zλ−1pλ(f)$.

Proof.

If σ is of class λ, then the number of disjoint cycles in σ is l(λ), and by (3.2), we have Eσ(f, …, f) = pλ(f). Thus, the sum (3.3) for En(f, …, f) is constant over conjugacy classes, with class λ contributing n!/zλ identical terms. This implies the result.■

If f is as above and u is a parameter, then we can define the formal logarithm

$log(1−uf)=−∑i≥1uifi/i.$
(A3)

Proposition A.2

We have$expElog(1−uf)=1−∑n≥1unEn(f,…,f)/n!$

Proof.
Let $Z=Elog(1−uf)$, then by (A3) we have
$Z=−∑i≥1uipi(f)/i.$
(A4)
Writing pk = pk(f) and pλ = pλ(f) for simplicity, we get
$exp(Z)=∏i≥1∑mi≥0(−1)mi(uipi)mi/imimi!=∑λ(−1)l(λ)zλ−1pλu|λ|.$
(A5)
Now, the result follows from Lemma A.1.■

Proposition A.3.
Iff1, f2, … are functions on a probability space, then we have
$1−expElog1−∑ifiti=∑n≥1∑i1,…,inEn(fi1,…,fin)ti1+⋯+in/n!.$

Proof.
Let us write A = f1t + f2t2 + ⋯, then by Proposition A.2, we get
$1−expE(log(1−A))=∑n≥1En(A,…,A)/n!,$
and by multilinearity of En, we have $En(A,…,A)=∑i1,…,inEn(fi1,…,fin)ti1+⋯in$.■

Theorem A.4.

For a set of functions$I$on a probability space, the following are equivalent:

1. For alln, we haveEn(f1, …, fn) ≥ 0 if$f1,…,fn∈I$.

2. The power series$1−expE(log(1−∑ifiti))$has positive coefficients if$f1,f2,…∈I$.

Proof.
The first statement implies the second by Proposition A.3. The converse was proved in Ref. 5, but we recall it here for completeness. Let p1, p2, …, pn be the first n primes; define
$k=p1p2⋯pn,kj=k/pj,N=k1+⋯+kn,$
and consider possible solutions of the equation s1k1 + ⋯ + snkn = N, where s1, …, sn are integers $≥0$. If some sj were 0, then pj would divide the left side but not the right; thus, we must have all sj > 0 and, hence, that s1 = ⋯ = sn = 1. Now, it follows from Proposition A.3 that the coefficient of tN in the power series $1−expE(log(1−∑j=1nfjtkj))$ is precisely En(f1, …, fn). Thus, the second statement implies the first.■

The previous theorem proves the equivalence of Conjectures 1.1 and 1.2. In particular, our Theorem 3.6 implies Conjecture 1.2 for the Lebesgue measure on the unit square in $R2$.

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9.

In this paper, we use positive as a synonym for non-negative and monotone for monotone decreasing. By reversing the partial order, our results and conjectures hold equally for monotone increasing functions. We note further that the positivity requirement on functions is redundant for the second inequality but essential for the first.