Convolution estimates and the number of disjoint partitions

Let $X$ be a finite collection of sets. We count the number of ways a disjoint union of $n-1$ subsets in $X$ is a set in $X$, and estimate this number from above by $|X|^{c(n)}$ where $$ c(n)=\left(1-\frac{(n-1)\ln (n-1)}{n\ln n} \right)^{-1}. $$ This extends the recent result of Kane-Tao, corresponding to the case $n=3$ where $c(3)\approx 1.725$, to an arbitrary finite number of disjoint $n-1$ partitions.


Introduction
Our main result is the following theorem Theorem 1. For any n, m 1, and any f 1 , . . . , f n : {0, 1} m → R we have Moreover, for each fixed n exponent p n is the best possible in the sense that it cannot be replaced by any larger number.
As an immediate application we obtain the following corollary (see Section 2.3 below).
Corollary 2. Let X be a finite collection of sets. Then where denotes the disjoint union, and |X| denotes cardinality of the set.
The corollary extends a recent result of Kane-Tao [1], corresponding to the case n = 3 where 3 p 3 ≈ 1.725, to an arbitrary finite number n 3 disjoint partitions.

The proof of the theorem
Following [1] the proof goes by induction on the dimension of the cube {0, 1} m . The case m = 1, which is the most difficult, is the main contribution of the current paper.

Basis: m = 1
In this case, set f j (0) = u j and f j (1) = v j for j = 1, . . . , n. Then the inequality (1) takes the form We do encourage the reader first to try to prove (3) in the case n = 3, or visit [1], to see what is the obstacle. For example, when n = 3 equality in (3) is attained at several points. Besides, direct differentiation of (3) reveals many "bad" critical points at which finding the values of (3) would require numerical computations [1]. The number of critical points together with equality cases increases as n becomes larger, therefore one is forced the electronic journal of combinatorics 22 (2015), #P00 to come up with a different idea. We will overcome this obstacle by looking at (3) in dual coordinates.
Without loss of generality we can assume that u j and v j are nonnegative for j = 1, . . . , n. Moreover, we can assume that v j = 0 for all j otherwise the inequality (3) is trivial.
Let us divide (3) by n j=1 v j . Denoting x j := (u j /v j ) pn we see that it is enough to prove the following lemma.
Lemma 3. For any n 2 and all x 1 , . . . , x n 0 we have where p n = ln n n (n−1) n−1 ln n Proof. For n = 2 the lemma is trivial. By induction on n, monotonicity of the map and the fact that p n is decreasing, we can assume that all x i are strictly positive. For convenience we set p := p n . Introducing new variables we rewrite (4) as follows Concavity of the function ln(x) provides us with a simple representation of the logarithmic function Therefore we are left to show that for all x i > 0 and all b i ∈ R we have where x = (x 1 , . . . , x n ) and b = (b 1 , . . . , b n ). Notice that given a vector b ∈ R n , the infimum of B(x, b) in x cannot be reached at infinity because of the slow growth of the logarithmic function. Therefore, we look at critical points of B in x Notice that for all y i > 0. It is straightforward to check that f (y) 0 on the diagonal, i.e., when y 1 = y 2 = . . . = y n .
In general, we notice that critical points of f (y) satisfy the equation Equation (5) gives the identity y −r i = n − 1, and so at critical points (5) we are only left to show ln y i − ln y i 0.
Since the mapping is increasing on (0, (1 + r) 1/r ) and decreasing on the remaining part of the ray, we can assume without loss of generality that k numbers of x i equal to u (1 + r) 1/r , and the remaining n − k numbers of x i equal to v (1 + r) 1/r . Moreover, we can assume that 0 < k < n otherwise the statement is already proved. From (5), we have From the equality of the first and the third expressions in (7) it follows that In order v to be positive we assume that the numerator of (8) is non negative. If we plug the expression for v from (8) into the first equality of (7) then after some simplifications we obtain the following equation in the variable z := u r 1 + r It follows from (7) that (ku + (n − k)v) r = z z−1 r z, and so using (9) we obtain the electronic journal of combinatorics 22 (2015), #P00 Therefore at critical points (6) simplifies as follows Now it is pretty straightforward to show that (10) is non negative even under the assumption z 1 + r for r = p − 1 = (n−1) ln n n−1 ln n . Indeed, notice that z > n n−1 , and the map . Therefore it is enough to check nonnegativity of (10) when k = 1, in which case the inequality follows again using z > n n−1 , and the fact that the map z → ln 1 + Remark 4. Choice x 1 = . . . = x n = 1 n−1 gives equality in (4), and this confirms the fact that p n is the best possible in Theorem 1.

Inductive step
Inductive step is the same as in [1] without any modifications. This is a standard argument for obtaining estimates on the Hamming cube (see for example [2]). In order to make the paper self contained we decided to repeat the argument.
Suppose (1) is true on the Hamming cube of dimension m. Without loss of generality assume f j 0, and set g j := f p j for all j. Define B n (y 1 , . . . , y n ) := y 1/pn 1 · · · y 1/pn n . For x j ∈ {0, 1} m+1 , let x j = (x j , x ′ j ) wherex j is the vector consisting of the first m coordinates of x j , and number x ′ j denotes the last m + 1 coordinate of x j . Set the electronic journal of combinatorics 22 (2015), #P00 We have

The proof of Corollary 2
Without loss of generality we may assume that all the sets A in X are subsets of {1, . . . , m} with some natural m 1 (see [1]). For j = 1, . . . , n define functions as follows: