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Observation 1.6. The Lubell function of a family is the average number of sets from on a chain, taken over all n! chains. In particular, the LubellF function of a level is 1, and the LubellF function of an antichain is the number of chains containing a set from divided by n!. The Lubell function is F F n additive across a union of disjoint families of sets. Furthermore, l( ) n/2 ([19]). |F| ≤ F ⌊ ⌋ The Lubell function was derived from the celebrated YMBL inequality which was independently discov- ered by Yamamoto, Meshalkin, Bollobás and Lubell. Using the Lubell function terminology, it states that YMBL inequality (Yamamoto, Meshalkin, Bollobás, Lubell [26, 21, 3, 19]). If 2[n] is an antichain, then l( ) 1. F ⊆ F ≤ For a poset P , let l(n, P ) be the maximum of l(n, ) over all families 2[n] which are both P -free and F F ⊆ contain the empty set. Let l(P ) = lim supn l(n, P ). Griggs, Li and Lu proved that →∞ Lemma 1.7 (Griggs, Li and Lu [14]).
n La(n, ) l( )+ o(1) . Q2 ≤ Q2 n/2 ⌊ ⌋ Kramer, Martin and Young used flag algebras to prove that
Lemma 1.8 (Kramer, Martin and Young [17]). l( )=2.25 Q2 thereby proving
2 Theorem 1.9 (Kramer, Martin and Young [17]).
n La(n, ) (2.25+ o(1)) . Q2 ≤ n/2 ⌊ ⌋
The following construction shows that l¯( 2) 2.25 in Lemma 1.8. There are other constructions known as well. Q ≥
Example 1.10. Let 2[n] consist of all the sets of the following forms: , e , e, o , o , o where F ⊆ ∅ { } { } { 1 2} e denotes any even number in [n], and o, o1 and o2 denote any odd numbers in [n]. This family is diamond-free, and l( )=2.25 o(1). F ± Example 1.11. This construction is a generalization of the previous one. Let A [n] with A = an. Let 2[n] consist of all the sets of the following forms: , e , e, o , o , o where⊆ now |e denotes| F ⊆ ∅ { } { } { 1 2} any element of A, while o, o1 and o2 denote any elements of [n] A. This family is diamond-free, and l( )=2+ a a2 o(1). This family contains all size 2 sets that\ do not form a diamond with and the singletons,F so− all maximal± diamond-free families on levels 0, 1 and 2 that contain are of this form.∅ ∅ The following restriction of the problem of diamond-free families has been investigated: How big can a diamond-free family be if it can only contain sets from the middle three levels of 2[n] (denoted (n, 3))? Better bounds are known with this restriction. Axenovich, Manske and Martin showed that B
Theorem 1.12 (Axenovich, Manske and Martin [1]). If (n, 3) is diamond-free, then n F ⊆B |F| ≤ 2.20711 + o(1) n/2 . ⌊ ⌋ n Later, Manske and Shen improved it to 2.1547 n/2 in [20] and recently, Balogh, Hu, Lidický and Liu n ⌊ ⌋ gave the best known bound of 2.15121 n/2 in [2] using flag algebras. ⌊ ⌋ Definition 1.13. We call a chain maximal–non-maximal (MNM) with respect to (w.r.t.) if it contains a set from , and the biggest set contained in on the chain is not maximal in (i.e., thereF are other sets from Fcontaining it on some other chains).F F F It is easy to see that an -free family is Λ-free if and only if the family we get by adding is diamond-free; adding increases the Lubell∅ function by 1. In Section 2 of this paper, we prove the following∅ lemma: ∅ Lemma 1.14. Let 2[n] be a Λ-free family that does not contain the empty set, nor any set of size F ⊆ bigger than n n′ for some n′ N (that can be chosen independently of n). Assume that there are cn! − ∈ MNM chains w.r.t. . Then l( ) 1 min c + 1 , 1 + min c + 1 , 1 + 3 . F F ≤ − n′ 4 n′ 4 n′ q It is easy to see that in Example 1.11 the number of MNM-chains is approximately a2n! (so a √c): these ≈ are the chains whose second set is e1,e2 with e1,e2 A. Thus, this lemma is (asymptotically) sharp, and states that for a given number of{ MNM} chains, Example∈ 1.11 cannot be beaten (with some restriction on the sizes of the sets). Barring the requirement that the topmost n′ levels be empty, Lemma 1.14 is a generalization of Lemma 1.8. The proof of Lemma 1.7 in [17] actually works with the restriction of Lemma 2/3 1.14 concerning the topmost sets (that there is no set of size bigger than n n′) with n′ = n/2 n , immediately giving a new proof of Theorem 1.9. Our proof of Lemma 1.14 includes− an intricate induction− step and a (non-combinatorial) lemma about functions involving a lot of elementary algebra and calculus; but it does not require flag algebras, and it does not use details of the structure of above the second level (except inside the induction). F Section 3 of this paper uses Lemma 1.14 to prove our main theorem:
√2+3 n n Theorem 1.15. La(n, 2) 2 + o(1) n/2 < 2.20711 + o(1) n/2 . Q ≤ ⌊ ⌋ ⌊ ⌋ The proof is inspired by the proof of Theorem 1.12 (the same bound when restricted to 3 levels) as in [1], using the idea of grouping chains by the smallest set contained in on a chain (as developed in [14] and [17]). F
3 1.25
1.25 1
1 0.75
0.5 0.75
0.25
0.5
0 0 0.25 0.5 0 0.25 0.5 0.75 1 Values of f(x,c) plotted in x, for c = Values of f(x,c) plotted in c, for x = 0, 0.05, 0.01,..., 1 (top to 0, 0.0125, 0.025,..., 0.25 (bottom to top). Note that bottom). the x ≥ 0.5 part of the plots coincide.
2 Λ-free families – Proof of Lemma 1.14
2.1 Definitions and main lemma
Definition 2.1. We define the following functions:
For x [0, 1],c [0, ), • ∈ ∈ ∞ 1 1 2 2 1 x + 4(x x2) 1 c if x 2 and c< 4 x x − − − ≤ − 1 2 1 x2 2x +1 c + √c if x and 4 x x2 c f(x, c)= − − ≤ 2 − ≤ ≤ 4 x2 2x +1.25 if x 1 and 1 c 2 4 − 1 ≤ ≤ 1 x if 2 x. − ≤ For x [0, 1),c [0, ),a [0, 1 x), a˜ 0, min a, c , • ∈ ∈ ∞ ∈ − ∈ x+a h i c a˜(x + a) g(x, c, a, a˜)= a + (1 x a)f x + a, − +2˜a(1 x a). − − 1 x a − − − −
For x [0, 1),c [0, ),a [0, 1 x), a˜ 0, min a, c x , • ∈ ∈ ∞ ∈ − ∈ x+a − h i c (x +˜a)(x + a) h(x, c, a, a˜)= a + (1 x a)f x + a, − +2˜a(1 x a)+ x 3x(x + a). − − 1 x a − − − − − Lemma 2.2. The functions above satisfy the following conditions:
1. For all c [0, ), if c˜ = min c, 1 , then f(0,c)=1 c˜ + √c˜. ∈ ∞ 4 − 4 2. f(x, c) is concave and monotonously increasing in c, and monotonously decreasing in x.
3. For all x [0, 1],c [0, ),a [0, 1 x), a˜ 0, min a, c : g(x, c, a, a˜) f(x, c). ∈ ∈ ∞ ∈ − ∈ x+a ≤ h i 4. For all x [0, 1],c [0, ),a [0, 1 x), a˜ 0, min a, c x : h(x, c, a, a˜) f(x, c). ∈ ∈ ∞ ∈ − ∈ x+a − ≤ h i 5. For all c [0, ):1 x f(x, c). ∈ ∞ − ≤ We prove Lemma 2.2 in Appendix A. Rather that proving Lemma 1.14 directly, we prove a strengthening of it – Lemma 2.3. This strengthened version involves additional parameters, X and , and their functions x, α and µ, which we introduce in order to make the inductive proof possible. LemmaX 1.14 is a special case of Lemma 2.3 with X = = . In the rest of Section 2, we prove Lemma 2.3. X ∅
[n] Lemma 2.3. Let 2 be a Λ-free family which does not contain , nor any set larger than n n′ F ⊆ ∅ X − for some n′ N. Let us assume that we are given a “forbidden” set X [n], with x = | | . Also, let ∈ ⊆ n 2[n] be a “forbidden” antichain in which each set contains exactly one element of X (and may or mayX ⊂ not be a singleton). Let us assume that the sets in are disjoint from X, and unrelated to every set in . Let α = l( ), and let µn! be the number of chainsF which start with an element of X as a singleton, butX do not containX any set in . Assume, furthermore, that there are cn! MNM chains w.r.t. . Then l( ) f x, c + µ + 1 (α Xµ x)+ 3 . F F ≤ n′ − − − n′ First we verify the base case of the induction.
Proposition 2.4. Lemma 2.3 holds for n n′. ≤ Proof. = . is an antichain, so, by the YMBL inequality, α 1. By Lemma 2.2 Point 2 and Point 5, f x,F c + µ∅+X1 (α µ x)+ 3 f(x, c + µ) (α µ x)≤ 1 x (α x) 0= l( ). n′ − − − n′ ≥ − − − ≥ − − − ≥ F From now on we assume n′ n 1. ≤ − A Notation. Let A [n] X be the set of elements of [n] that appear as singletons in , and let a = |n| . Let be the family⊆ of\ those sets in which contain at least one element of A,F but which are not singletons.B Let β be the Lubell functionF of . Let be the family of those sets in which only contain elements of [n] X A. B C F \ \ ˜ A˜ Let A = e A : ( B : e B) , and let a˜ = | |. Let c0n! be the number of chains that start with ∈ ∃ ∈B ∈ n e as a singleton for some e A˜, but do not contain any set from . Let νn! be the number of chains that{ } start with e as a singleton∈ for some e A˜, continue with an elementB of [n] X A as the second element added to{ } form the chain, yet do not contain∈ any set from . \ \ B n (x+a)n 1 Let 1 = > 1 and 1 = − 1. These correction factors will account for the difference from n 1 (x+a)(n 1) ≤ the asymptotic− behavior. (They are− both typically close to 1. If x + a =0, let 1 =1; it is irrelevant as it will always be multiplied by x + a.)
Outline of the proof: In Subsection 2.2, we make some observations on the structure of and . In X B Subsection 2.4, we will finish the proof by applying induction to the Boolean lattices oi , [n] where o [n] X A. When applying Lemma 2.3 by induction, we will use X A in the place{ } of X, while i ∈ \ \ ∪ sets from and will contribute to the family we use in the place of (which we will denote by ′). X B X Xi We know little about the parameters of each i′, but we will be able to bound their sums. The relevant calculations are done in Subsection 2.3. X
2.2 On the structure of and X B Proposition 2.5. Every D is of the form d, o1,...,ok with d X, o1,...,ok [n] X A (where k may be 0). ∈ X { } ∈ ∈ \ \
Proof. D contains exactly one element of X by definition. Let e A; then e / D for otherwise D and e would be related. ∈ ∈ { }∈F
5 Summary of notation
X “Forbidden” set (sets in are disjoint from it – parameter of Lemma 2.3) x = X /n “Forbidden” antichain (setsF in are unrelated to sets in it – parameter α = |l( | ) X of Lemma 2.3) F X µ # chains containing d for d X but no set from /n! c # {MNM chains /n! (parameter{ } ∈ of Lemma 1.14 / LemmaX} 2.3) { } A e [n]: e a = A /n { ∈ { }∈F} | | B : ( B 2, A B = ) e : e A = β = l( ) B{C ∈F : C| |≥[n] X∩ A 6 ∅ } {{ } ∈ }∪B∪C F B C{ ∈F ⊆ \ \ } A˜ e A : ( B : e B) a˜ = A˜ /n ∈ ∃ ∈B ∈ ν # chains containing e and e, ofor e A,˜ o { } { } ∈ ∈ [n] X A but no set from /n! \ \ B c # chains containing e for e A˜ but no set from /n! 0 { } ∈ B Proposition 2.6. Sets in only contain one element of A. is an antichain, and the sets in are also unrelated to every set inB . B B C
Proof. If e1 B with e1 A, and B was related to another set S , then e1 , B and S would form a Λ. This∈ applies∈ B to any∈S , as well as S = e for any e ∈= Fe A. { } ∈B∪C { 2} 1 6 2 ∈ Proposition 2.7. a˜(x + a)1 a˜(x + a)1 + ν = c c, and thus a˜ c . ≤ 0 ≤ ≤ (x+a)1 Proof. Any chain on which the singleton is e and the second set is e, d with e A˜ and d X A is always an MNM chain: e, d and any set that{ } contains it is forbidden{ from} being∈ in either∈ because∪ it is not disjoint from X (when{ }d X), or because it would contain two elements of A (whenB d A). The number of such chains is an˜ (an∈ + xn 1) (n 2)! =a ˜(x + a)n!1. And out of the chains which∈ start with e , and whose second set· is e, o −with· some− o [n] X A, νn! do not contain any set from . { } { } ∈ \ \ B We have c0 c because a chain whose first set is e for some e A˜, but does not contain any set from , is an MNM≤ chain. { } ∈ B For a family of sets 2[n], let m( )n! be the number of chains which start with an element of X as a singleton and do notA ⊆ contain any setA from . (For example, m( )= µ, and therefore l( ) m( )= A X X − X α µ.) For a fixed d X, let md( )n! be the number of chains on which the singleton is d , and do not− contain any element∈ of . A { } A Proposition 2.8. For any d X, let = D : d D . We can assume without loss of generality ∈ Xd { ∈ X ∈ } that for any d1, d2 X, D d1 : D d1 = D d2 : D d2 . That is, if does not satisfy this condition, we show∈ a family{ \{ˆ which} ∈ does, X } and{ also\{ satisfies} ∈ the X conditions} of LemmaX 2.3’s statement (each set contains exactly oneX element of X, the sets are unrelated to each other and to every set in ), F and for which f x, c + m( ˆ)+ 1 l( ˆ) m( ˆ) x f x, c + µ + 1 (α µ x). X n′ − X − X − ≤ n′ − − − Proof. Let d X be such that 0 ∈ 1 f x, c + X m ( )+ X l( ) X m ( ) x | | d0 Xd0 n − | | Xd0 − | | d0 Xd0 − ′ 1 = min f x, c + X md( d)+ X l( d) X md( d) x . d X | | X n − | | X − | | X − ∈ ′ Let ˆ = D d d : d X,D . X \{ 0}∪{ } ∈ ∈ Xd0 = d X d, so α = d X l( d). It immediately follows from the definition of d that if a chain X ∈ X ∈ X X has d as a singleton, and does not contain any set from d, then it does not contain any set from { F} P ˆ X ˆ . So µ = d X md( d). Similarly, l( ) = X l( d0 ) and m( ) = X md0 ( d0 ). Since f(x, c) is X ∈ X X | | X X | | X P
6 monotonously increasing and concave in c, using Jensen’s inequality 1 f x, c + X m ( )+ X l( ) X m ( ) x | | d0 Xd0 n − | | Xd0 − | | d0 Xd0 − ′ 1 1 f x, c + X m ( )+ X l( ) X m ( ) X x ≤ X | | d Xd n − | | Xd − | | d Xd − | | | | "d X ′ d X d X # X∈ X∈ X∈ 1 f x, c + µ + (α µ x). ≤ n − − − ′ Sets in ˆ contain exactly one element of X, and form an antichain. They are also unrelated to every set S : XS cannot contain any element of X, so it could only be related to a set in ˆ by being its subset. But∈FS must also be unrelated to every D , so it cannot be a subset of D X d d either. ∈ Xd0 ⊆ X \{ 0}∪{ } In fact we will only use the following simple corollary of Proposition 2.8. In many parts of the rest of this section we will treat the two cases of the corollary below separately. Corollary 2.9. With the assumption of Proposition 2.8,
either = d : d X (we refer to it as the singletons case), • X { } ∈ or does not contain any singleton (referred to as the no singleton case). • X Proof. Let d X. (If X = , both statements trivially hold.) If d = D : d D , then 1 ∈ ∅ { 1} ∈ Xd1 { ∈ X 1 ∈ } d1 = d1 , because sets in d1 are unrelated. So either d1 = d1 or d1 does not contain any singleton,X {{ yielding}} the two casesX above by Proposition 2.8. X {{ }} X Remark. The fact that sets in contain an element of X implies that sets in do not contain sets in . Now, let us consider what restrictionsX are imposed on by the fact that setsF in are not contained Xin the sets in , beyond the other conditions of Lemma 2.3F (namely that all the setsF in are disjoint from X). X F In the singletons case, clearly there are no such additional restrictions. However, in the no singleton case, there are two additional restrictions that are not already implied by the set X:
The union of singletons in , A [n] . • F ⊆ \ X Sets in must not be contained in setsS in . Clearly this imposes a restriction only if contains • sets biggerC than 2. X X
2 xn C (n 2)! Example 2.10. Let C [n] X, and let = d, o : d X, o C . Then α = l( )= · ·| n|·! − = C xn (xn+⊆an 1)\ (n 2)! X { } ∈ ∈ X 2x | | 1, and µ = · − · − = x(x + a)1. The only restriction on that this creates is that n n! the union of singletons A [n] X C. F X ⊆ \ \ In other words, let us assume that α = l( )=2xγ1 for γ R (without assuming that is of the above A X ∈ X form). Then it is possible that a = |n| can be as big as 1 x γ with not creating any restrictions on (depending on the actual structure of , namely, if it− is− made upX of sets of size 2 as above; then α =2C x(1 x a)1 and µ = x(x + a)1). ButX if a> 1 x γ, then α =2xγ1 implies that contains sets bigger than− 2,− and thus it creates restrictions on .− So,− in the no singleton case, one wayX to understand the calculations that follow is to check them for C = d, o : d X, o C ; then check what happens if x, c and a are fixed, but is changed. X { } ∈ ∈ X 2.3 Chain calculations
Now we estimate the numbers of certain types of chains, in preparation for applying induction. Proposition 2.11. In the no singleton case, (α x + µ) n! chains start with o for some o [n] X A, and contain a set from . − { } ∈ \ \ X Proof. A total of αn! chains contain a set D . By Proposition 2.5, the singleton on such a chain is either from X or [n] X A. The number of chains∈ X which start with an element of X as their singleton and do not contain a\ set\ from is µn!, so the number of chains which contain a set from , and which start with an element of X, is X(x µ)n!. On the rest, the singleton is from [n] X A. X − \ \
7 Proposition 2.12. β a˜(1 x a)1+ ν n! chains start with o for some o [n] X A, and contain a set from . − − − { } ∈ \ \ B Proof. A total of βn! chains contain a set from . A set in is of the form e, o ,...,o with e B B { 1 k} ∈ A,˜ o1,...,ok [n] X A, k 1. A chain that contains a B , and does not start with o for some o [n] X ∈A, must\ \ start with≥ an element of A˜, and continue∈B with an element of [n] X{ } A as the second∈ element\ \ added to form the chain. There are an˜ (1 x a)n (n 2)! =a ˜(1 \x \a)1n! such chains, out of which νn! do not contain any set from .· So− a˜(1− x · a)1− ν n! chains− contain− a set B − − − from and start with an element of A˜. The rest start with o for some o [n] X A. B { } ∈ \ \ Proposition 2.13. In the no singleton case, µ x(x + a)1; and the number of chains of the form , d , d, o ,... with d X, o [n] X A, which≥ do not contain any set from , is µ x(x + a)1 n!. ∅ { } { } ∈ ∈ \ \ X − Proof. A total of µn! chains start with an element of X and do not contain any set from . The chains X of the form , d1 , d1, d2 ,... with d1 X, d2 X A never contain a set from when contains no singleton. The∅ { number} { of} these chains is∈ xn (xn∈ + ∪an 1) (n 2)! = x(x + a)1X n!. ForX the rest, the second element added to form the chain is from· [n] X− A. · − \ \ Notation. Let X′ = X A. Let = d, o : d X, o [n] X A , and let = e, o : e A A,˜ o ∪ Y { } ∈ ∈ \ \ Z { } ∈ \ ∈ [n] X A . In the singletons case, let ′ = . (Note that here and in the rest of the paper, \ \ X Y⊔B⊔Z stands for a union of sets which are pairwise disjoint.) In the no singleton case, let ′ = . ⊔ X X⊔B⊔Z Proposition 2.14. The three families which make up ′ are indeed disjoint in each case, and their union forms an antichain. X
Proof. is an antichain by Proposition 2.6; is an antichain by definition; and and are an- tichainsB because both consist of size 2 sets only.X Let D = d, o ,...,o , YY = d,Z o , { 1 k} ∈ X { } ∈ Y B = e1,p1,...,pl and Z = e2, q with d X, e1 A˜, e2 A A˜, oi,o,pi, q [n] X A, and l{ 1. B is unrelated}∈B to D by{ definition,} ∈ Z and to Y∈because∈d / B and∈ \B 2. Z is unrelated∈ \ to\ D and Y≥because d / Z and e / D, Y ; Z is unrelated to B because ∈e / Z and| |≥e / B. ∈ 2 ∈ 1 ∈ 2 ∈ Proposition 2.15. Sets in are disjoint from X′, and they are unrelated to every set in ′ (in both cases). C X
Proof. For every C , C [n] X′ and it is unrelated to every set in by definition. C is unrelated to every set in by∈C Proposition⊆ \ 2.6. It also cannot be a superset of a YX or a Z , since those contain an elementB of X or A; neither a proper subset of Y or Z because Y∈ Y= Z =2∈ ZC . | | | | ≤ | | Proposition 2.16. The number of chains that start with an element of [n] X′ and contain a set from \ ′ is X at least x(1 x a)1+ β a˜(1 x a)1+ ν + (1 x a) (a a˜) 1 n! in the singletons case, • and − − − − − − − − at least (α x + µ)+ β a˜(1 x a)1+ ν + (1 x a) (a a˜) 1 n! in the no singleton case. • − − − − − − − Proof. The number of chains on which the singleton is o with o [n] X′ = [n] X A, and the second set is o, d with d X, is xn (1 x a)n (n 2)!{ =}x(1 x ∈a)1n\!. The number\ \ of chains on which the singleton{ } ∈ is Y o , and∈ the second· set− is− o, e· − with e −A −A˜, is (1 x a)n (a a˜)n (n 2)! = (1 x a) (a {a˜)}1n!. The rest follows from{ Proposition} ∈ Z 2.11∈ and\ Proposition− − 2.12.· − · − − − − Proposition 2.17. The number of chains on which the singleton is o with o [n] X′, the second set { } ∈ \ is o, d with d X′ = X A, and which do not contain any set from ′, is { } ∈ ∪ X νn! in the singletons case, and • µ x(x + a)1 + ν n! in the no singleton case. • −
8 Proof. Let = , A1, A2,...,An 1, [n] be a chain with A1 A2 ... An 1 [n]. Let ϕ( ) A ∅ − ∅ ⊂ ⊂ ⊂ ⊂ − ⊂ A be the chain , A2 A1, A2, A3,...,An 1, [n]. (In other words, in the order in which elements of [n] are added to form∅ the\ chain, the first two are− swapped.) ϕ is a bijection.
It is easy to check that ′ does not contain singletons. ϕ is a bijection between chains of the form X , o , o, d ,... containing no set from ′, and chains of the form , d , o, d ,... containing no set ∅ { } { } X ∅ { } { } from ′, with o [n] X′ and d X A. Below we classify the chains , d , o, d ,... based on what set d Xbelongs to and∈ count\ them separately.∈ ∪ ∅ { } { }
For d X, o, d in the singletons case. In the no singleton case, µ x(x + a)1 n! chains of • the form∈ , {d , }o, ∈ d Y,... contain no set from by Proposition 2.13; these− chains also contain no set from ∅ or{ } ,{ since} sets from those do not containX any element of X . B Z For d A˜, the number of chains of this form which contain no set from is νn!; these chains also • contain∈ no set from , or , since sets from those contain no elementB of A˜. X Y Z For d A A˜, o, d . • ∈ \ { } ∈ Z Summing these cases, we get the statement of the proposition.
2.4 Inductive step
Notation. Using standard notation for intervals, let [A, [n]] denote the Boolean lattice S [n]: A S . { ⊆ ⊆ } Let [n] X′ = [n] X A = o1, o2,...,o(1 x a)n ; and for a family of sets , let oi = S oi : S . \ \ \ { − − } A A− { \{ } ∈ A} Let i′ = oi , [n] oi, and i′ = ′ oi , [n] oi. Let αi′ = l(n 1, i′). (Here the Lubell C C ∩ { } − [n] Xo X ∩ { } − − X function on the Boolean lattice 2 \{ i} of order n 1 is used.) − [n] o ′ 2 \{ i} is a Λ-free family which does not contain (since o / A, so o / ), nor any set larger Ci ⊆ ∅ i ∈ { i} ∈F than n 1 n′. Sets in ′ are disjoint from X′, and are unrelated to sets in ′ by Proposition 2.15. − − Ci Xi Moreover, every set in i′ contains exactly one element of X′. Therefore, the conditions of Lemma 2.3 X [n] oi are satisfied for the family i′ 2 \{ } where the corresponding “forbidden” set is X′ [n] oi , with X′ C ⊆ ⊆ \{ } |n 1| = (x + a)1 and the corresponding “forbidden” antichain is i′. − X [n] oi Since i′ is an antichain, αi′ (n 1)! is the number of chains in 2 \{ } that contain a set from i′. Chains [n]X oi − [n] X of 2 \{ } correspond to chains of 2 that start with oi . So by Proposition 2.16, in the singletons case { } (1 x a)n − − α′ x(1 x a)1+ β a˜(1 x a)1+ ν + (1 x a) (a a˜) 1 n, i ≥ − − − − − − − − i=1 X and in the no singleton case
(1 x a)n − − α′ (α x + µ)+ β a˜(1 x a)1+ ν + (1 x a) (a a˜) 1 n. i ≥ − − − − − − − i=1 X [n] o Let µ′ (n 1)! be the number of chains in the Boolean lattice 2 \{ i} which start with an element of X′ i − as a singleton, but do not contain any set from ′. By Proposition 2.17, in the singletons case Xi (1 x a)n − − µi′ = νn, i=1 X and in the no singleton case (1 x a)n − − µ′ = µ x(x + a)1 + ν n. i − i=1 X [n] o [n] Let c′ (n 1)! be the number of MNM chains w.r.t. ′ in 2 \{ i}. The corresponding 2 -chains, starting i − Ci with oi , are MNM chains w.r.t. . The total number of MNM chains w.r.t. is cn!, out of which c0n! start{ with} an element of A as a singleton.F By Proposition 2.7, F
(1 x a)n − − c′ = (c c )n = (c a˜(x + a)1 ν) n. i − 0 − − i=1 X
9 The following two examples are typical cases where, in the induction step for the i′’s, we will get the singletons case and the no singleton case respectively. C Example 2.18. Let X = = and = e, o : e A, o [n] A . Then A˜ = A, β = 2a(1 a)1, X ∅ B { } ∈ ∈ \ ′ − (1 a)n X and ν = 0. X′ = A, and ′ = e : e A . − α′ = a(1 a )1n, α′ = a1 = | | and µ′ = 0. Xi { } ∈ i=1 i − i n 1 i (1 x a)n c a2 − − − 2 i=1 ci′ = c a n and the average of thePci′ ’s is 1− a . − − PExample 2.19. Let X= = and ˆ := e, o1, o2 : e A, o1, o2 [n] A . Then X′ = A, and X ∅ B ⊆ B { [n] } ∈ ∈ \ ′ e, o : e A, o [n] A o . Chains on 2 of the form , e , e , o , e ,o,e ,... do not Xi ⊆ { } ∈ ∈ \ \{ i} ∅ { 1} { 1 } { 1 2} (1 a)n 2 2 1a(n 1) 1 intersect . So − µ = ν a (1 a)1 − − n (greater if ˆ), and the average of the µ ’s i=1 i′ 1a(n 2) $ i′ B ≥ − − B B ′ ′ 2 2 1a(n 1) 1 2 x (n 1) 1 X − P− − − ˆ is a 1 1a(n 2) = x′ x′((n 1) 1) where x′ = |n 1|. In the case of = , the size of the sets in is at ≥ − − − − B B C least 3, and the size of those in ′ is at least 2. Ci Proposition 2.20.
(1 x a)n (1 x a)n 1 − − 1 − − l( )= l(n 1, ′) and l( )= a + β + l( )= a + β + l(n 1, ′). C n − Ci F C n − Ci i=1 i=1 X X (Still understanding the one parameter version l( ) as l(n, ) for a family 2[n].) F F F ⊆ Proof. Every chain in the Boolean lattice 2[n] that intersects has an o as a singleton, and thus C { i} corresponds to a chain in the Boolean lattice [ o , [n]] o that intersects ′. { i} − i Ci (1 x a)n (1 x a)n 1 1 − − 1 − − l( )= = ′ = l(n 1, ′). n! n! i n i C [n] |H ∩ C| |H ∩ C | − C is a chain in 2 i=1 is a chain in [ oi ,[n]] oi i=1 H X X H X{ } − X A Let = e : e A . Then = . So l( ) = l( )+ l( )+ l( ) with l( ) = |n| = a and l( )=A β. { } ∈ F A⊔B⊔C F A B C A B We now prove Lemma 2.3 (and thus Lemma 1.14) using induction on n. According to Proposition 2.4, Lemma 2.3 holds for n n′. By induction and Lemma 2.2 Point 2, ≤ 1 3 l(n 1, ′) f (x + a)1,c′ + µ′ + α′ µ′ (x + a)1 + − Ci ≤ i i n − i − i − n ′ ′ 1 3 f x + a,c′ + µ′ + α′ µ′ (x + a)1 + . ≤ i i n − i − i − n ′ ′ So, by Proposition 2.20, we have
(1 x a)n (1 x a)n 1 − − 1 − − 1 l( )= l(n 1, ′) f x + a,c′ + µ′ + C n − Ci ≤ n i i n i=1 i=1 ′ X X (1 x a)n (1 x a)n (1 x a)n 1 − − − − − − 1 3(1 x a)n α′ µ′ (x + a)1 + − − . − n i − i − n · n i=1 i=1 i=1 ′ X X X We handle the case of 1 x a = 0 separately. If 1 x a = 0, A = [n] X and, since any non- − − − − \ singleton e1,e2,... would form a Λ with the singletons e1 , e2 , we have = and l( )= a =1{ x. This}∈F is only possible in the singletons case, since{ a} non-si{ }∈Fngleton in wouldF haveA to containF elements− of [n] X A. In the singletons case α = x and µ = 0, so l( )=1X x f(x, c) f x, c + µ + 1 (α \µ x\)+ 3 by Lemma 2.2 Point 5. From now on, we assumeF that− 1 ≤x a> 0≤. n′ − − − n′ − − Since f is concave in c, by Jensen’s inequality, and since f is monotonously decreasing in x, (1 x a)n (1 x a)n − − c + − − µ 1 l( ) (1 x a)f x + a, i=1 i′ i=1 i′ + C ≤ − − (1 x a)n n P − −P ′ ! (1 x a)n (1 x a)n 1 − − 1 − − 3(1 x a) α′ µ′ (1 x a)(x + a)1 + − − . − n i − n i − − − n i=1 i=1 ′ X X
10 Correction term calculations that we will use later (assuming n′ n 1): ≤ − 1 x a (x +˜a)(1 x a) 1 x a (1 1)(x +˜a)(x + a)+ − − = − − + − − − n n 1 n ′ − ′ (1) (1 + x +˜a)(1 x a) 1 (x + a)2 1 − − − . ≤ n′ ≤ n′ ≤ n′ 1 x a 1 x a 1 (1 1)˜a(x + a)+ − − (1 1)(x +˜a)(x + a)+ − − . (2) − n′ ≤ − n′ ≤ n′ 3(1 x a) 2˜a(1 x a)(1 1)+2(1 1)x 2(1 1)+(1 1) x(x + a)+ − − − − − − − − − n′ (3) 2a +3x 3(1 x a) 3 + − − . ≤ n 1 n ≤ n − ′ ′ 3(1 x a) 2a 3(1 x a) 3 2˜a(1 x a)(1 1)+ − − + − − . (4) − − − n ≤ n 1 n ≤ n ′ − ′ ′ In the singletons case:
(c a˜(x + a)1 ν)+ ν 1 l( ) a + β + (1 x a)f x + a, − − + F ≤ − − 1 x a n − − ′ x(1 x a)1+ β a˜(1 x a)1+ ν + (1 x a) (a a˜) 1 − − − − − − − − − 3(1 x a) ν (1 x a)(x + a)1 + − − − − − − n′ c a˜(x + a)1 1 3(1 x a) = a + (1 x a)f x + a, − + +2˜a(1 x a)1+ − − − − 1 x a n − − n ′ ′ − − 1 x a c a˜(x + a)+(1 1)˜a(x + a)+ − ′− = a + (1 x a)f x + a, − − n +2˜a(1 x a) − − 1 x a − − − − 3(1 x a) + 2˜a(1 x a)(1 1)+ − − . − − − n′ By Lemma 2.2 Point 2 and Point 3, and (2) and (4) in the Correction term calculations (note that in this case α = x and µ =0),
1 3 1 3 1 3 l( ) g x, c + , a, a˜ + f x, c + + = f x, c + µ + (α µ x)+ . F ≤ n n ≤ n n n − − − n ′ ′ ′ ′ ′ ′ 1 1 c a˜(x+a)1 c+ ′ a˜(x+a) c+ ′ c n − n (Note that a˜ (x+a)1 , so 0 −1 x a 1 x a , and a˜ x+a .) ≤ ≤ − − ≤ − − ≤ In the no singleton case:
(c a˜(x + a)1 ν)+ µ x(x + a)1 + ν 1 l( ) a + β + (1 x a)f x + a, − − − + F ≤ − − 1 x a n − − ′ (α x + µ)+ β a˜(1 x a)1+ ν + (1 x a) (a a˜) 1 − − − − − − − − 3(1 x a) [µ x(x + a)1 + ν] (1 x a)(x + a)1 + − − − − − − − n′ c + µ (x +˜a)(x+ a)1 1 = a + (1 x a)f x + a, − + (α x(x + a)1 x) − − 1 x a n − − − − − ′ 3(1 x a) +2˜a(1 x a)1+(2 1 1)x (2 1+1)x(x + a)+ − − − − · − − · n′ 1 x a c + µ (x +˜a)(x + a)+(1 1)(x +˜a)(x + a)+ − ′− = a + (1 x a)f x + a, − − n − − 1 x a − − +2˜a(1 x a)+ x 3x(x + a) (α x(x + a)1 x) − − − − − − 3(1 x a) +2˜a(1 x a)(1 1)+2(1 1)x 2(1 1)+(1 1) x(x + a)+ − − . − − − − − − − n′
11 By Lemma 2.2 Point 2 and Point 4, Proposition 2.13. and (1) and (3) in the Correction term calculations,
1 3 1 3 l( ) h x, c + µ + , a, a˜ (α µ x)+ f x, c + µ + (α µ x)+ . F ≤ n − − − n ≤ n − − − n ′ ′ ′ ′ 1 1 c a˜(x+a)1 c+µ+ ′ (x+˜a)(x+a) c+µ+ ′ c n − n (Note that a˜ (x+a)1 , so 0 −1 x a 1 x a , and a˜ x+a x.) ≤ ≤ − − ≤ − − ≤ − 3 Diamond-free families – Proof of Theorem 1.15
Let be a diamond-free family on 2[n]. F We cite Lemma 1 from [1]: Lemma 3.1 (Axenovich, Manske, Martin [1]).
n n Ω(n1/3) Ω(n1/3) n 2 − =2− . k ≤ n/2 k 0,1,...,n ∈{ X } ⌊ ⌋ k n/2 n2/3 | − |≥
2 n By this lemma, the number of sets in in the top and bottom n′ := n/2 n 3 levels is o(1) , so, F − n/2 since we are bounding the cardinality of , we may assume that those levels do not contain any⌊ set⌋ from . F F 1 √ Notation. For c [0, 1], let c˜ = min c, 4 , and let f(c)=1 c˜+ c˜. (This is equal to f(0,c) as defined in Definition 2.1.)∈ For A , recall that [A, [n]] denotes the− Boolean lattice S [n]: A S . A chain ∈F { ⊆ ⊆ } of this lattice is of the form A A A +1 A A +2 ... An 1 [n]. (When saying just “chain”, we ⊂ | | ⊂ | | ⊂ ⊂ − ⊂ continue to mean a maximal chain in the Boolean lattice 2[n].) Let
1 is a chain in [A, [n]] : c(A)= # C . (n A )! is MNM w.r.t. [A, [n]] − | | ( C F ∩ ) Further, we can assume without loss of generality that
1 is a chain: = or 1 is a chain: = or C := # C C∩F ∅ # C C∩F ∅ . n! min( ) is not minimal in ≥ n! is MNM w.r.t. ( C∩F F) ( C F ) (If this does not hold, we can replace with [n] A : A : this family is diamond-free, has the same F { \ ∈F} 1 is a chain: cardinality, and the opposite inequality holds.) Clearly C # C . ≥ n! is MNM w.r.t. ( C F) 1 1 l( )= #( )= #( ), F n! C∩F n! C∩F is a chain A is a chain C X X∈F AC=min(X ) C∩F since each chain will be counted when A = min( ) – except if = , but then #( )=0. Continuing, C C∩F C∩F ∅ C∩F
#( ) C∩F is a chain 1 is a chain containing A : AC=min(X ) l( )= # C C∩F . F n! A = min( ) A ( ) is a chain containing A : a chain X: ∈FA=min( ) C∩F # C ∃ C C∩F A = min( ) ( C∩F ) Each chain on [A, [n]] can be extended to a full 2[n]-chain in A ! ways. Furthermore, the Boolean [n]| A| lattice [A, [n]] can be made equivalent to the Boolean lattice 2 \ by subtracting A from each set; for [A, [n]], we denote A = S A : S . If A = min( ), #( ) = #( [A, [n]] ). If A⊆ A− { \ ∈ A} C∩F C∩F C ∩ ∩F
12 A is minimal in (that is, on every chain), F #( ) C∩F A ! #( [A, [n]]) is a chain | | C∩F∩ AC=min(X ) is a chain in [A,[n]] C∩F = C X is a chain containing A : A ! (n A )! # C | | − | | A = min( ) ( C∩F ) #( (( [A, [n]]) A)) C ∩ F ∩ − is a chain in 2[n]\A = C X = l(n A , ( [A, [n]]) A). (n A )! − | | F ∩ − − | |
( [A, [n]]) A is diamond-free, so (( [A, [n]]) A) is Λ-free; and the top n′ levels are assumed to F ∩ − F ∩ 1 − 1 \∅ be empty. Using Lemma 1.14 as well as that n′ = Ω(n) = o(1) and the subadditivity of the square root function,
1 1 1 1 3 l n A , ( [A, [n]]) A 1 min c(A)+ , + min c(A)+ , + − | | F ∩ − \∅ ≤ − n 4 n 4 n ′ s ′ ′ 1 3 f(c(A)) + + = f(c(A)) + o(1), ≤ rn′ n′ so l n A , ( [A, [n]]) A 1+ f(c(A)) + o(1). Whereas if A is not minimal in , i.e. S − | | F ∩ − ≤ F ∃ ∈ F such that A % S, then for any chain for which min( ) = A, we have #( ) 2 (otherwise S C C∩F #( ) C∩F ≤ is a chain C∩F PAC=min( ) C∩F and three sets in would form a diamond), so is a chain through A: 2. C∩F # C ≤ A=min( ) C∩F
1 l( ) # is a chain containing A (1 + f(c(A)) + o(1)) F ≤ n! {C } A A is minimalX∈F in F 1 is a chain containing A : + # C 2 n! A = min( ) · A ( ) A is not minimalX∈F in C∩F F 1 2+ # is a chain containing A (f (c (A)) 1+ o(1)) . ≤ n! {C } − A A is minimalX∈F in F # is a chain containing A Since f is concave, we can use Jensen’s inequality with the weights {C (1 C)n! } (where A is minimal in ). Notice that the sum of all the weights is 1 because the sum− of numerators is the total number of chainsF where min( ) is minimal in , that is, (1 C)n!. C C∩F F − # is a chain containing A c(A) {C } A A is minimalX∈F in l( ) 2+(1 C) f F 1+ o(1) . F ≤ − (1 C)n! − − c(A) is the fraction of the chains containing A which are MNM, so # is a chain containing A c(A) is the number of MNM chains through A. In the numerator, each MNM{C chain in the whole Boolean} lattice is counted once, except if the minimal element on it is not a global minimal, then it is not counted. So the numerator is less than or equal to the total number of MNM chains in the Boolean lattice, which is at most Cn!. Substituting, we get
n!C C l( ) 2+(1 C) f 1+ o(1) =1+ C + (1 C)f + o(1). F ≤ − n!(1 C) − − 1 C − − C C 1 1 C C varies between 0 and 1. 1 C is increasing in C. Above 1 C = 4 (corresponding to C = 5 ), f( 1 C ) is − − − 5 C 9 C constant 4 , so 1+ C + (1 C)f 1 C = 4 4 is decreasing in C. So it is enough to take the maximum − − − 13 1 in the interval 0, 5 : C C |F|n l( ) max 1+ C + (1 C) 1 + + o(1) ≤ F ≤ 1 − − 1 C 1 C n/2 C [0, 5 ] !! ⌊ ⌋ ∈ − r − √2+3 = + o(1) < 2.20711 + o(1). 2
Acknowledgments
We thank the anonymous referees for their detailed comments which helped improve the presentation of our paper. We would also like to thank Dömötör Pálvölgyi for helpful discussions concerning the 3-level version of this problem. The research of the second and third authors was supported by the National Research, Development and Innovation Office – NKFIH, grant K 116769.
References
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Appendix A Proof of Lemma 2.2
Points 1 and 5 are easy to see.
1 2 2 Proof of Point 2. It is also easy to check that f is continuous at the points x = 2 , c = 4 x x , c = 1 , and that the function is monotonously decreasing in x and increasing in c in each range. − 4 f(x, 0)= 1 x; x2 2x +1 c + √c is a concave and monotonously increasing expression in c. When 1 − − −1 2 0 < x 2 , c 1 x + 4(x x2) 1 c is the tangential line of the graph of c x 2x +1 c + √c ≤ 7→ − − − 7→ − − 2 at the point c =4 x x2 , since both their values, and their derivatives at this point coincide. So f is concave in c. − Since the graph of a concave function is below the tangent line at any point, we also have that for x 0, 1 ,c 0, 1 , ∈ 2 ∈ 4 f(x, c) x2 2x +1 c + √c =: f˜(x, c). (5) ≥ − − We will use this inequality in the proof of Point 3 and 4.
Proof of Point 3. If x + a =0, g(x, c, a, a˜)= f(x, c). From now on, we assume that x + a> 0. We first show that g is monotonously increasing in a˜.
1 1 2 2 4(x x2) 1 if x 2 and c< 4 x x 1 1 ∂ − − ≤ − 2 1 if x f (x, c)= 1 1 2 2 1 4(x x ) − ≤ 2 (6) 1+ 2√c if x 2 and 4 x x c 4 − 1 ∂c − ≤ − ≤ ≤ ≤ (0 if x 0 if 1 c or 1 x 2 ≤ 4 2 ≤ ≤ So, ∂ ∂ c a˜(x + a) ∂ c a˜(x + a) g (x, c, a, a˜) = 2(1 x a)+(1 x a) f x + a, − − ∂a˜ − − − − · ∂c 1 x a · ∂a˜ 1 x a − − − − 1 1 if x + a 1 4((x+a) (x+a)2) − ≤ 2 2(1 x a) (x + a) − 1 ≥ − − − (0 if x + a! 2 ≤ 1 1 1 1 4 1 (x+a) 1 if x + a 2 if x + a 2 − ≤ 2 2 2(1 x a) (x + a) · 2(1 x a) 1 ≤ 0. ≥ − − − (0 if 1 x + a! ≥ − − − (0 if x + a! ≥ 2 ≤ 2 ≤
15 Therefore, from now on we assume a˜ = min a, c since if g(x, c, a, a˜) f(x, c) holds for a˜ = x+a ≤ min a, c then it also holds for any a˜ 0, min a, c . x+a ∈ x+a h i c x+√x2+4c Case 1. First assume a x+a (so a˜ = a), which is equivalent to a(x + a) c or a − 2 . Let c a(x≤+a) ≤ ≤ x′ = x + a and c′ = −1 x a . − −
1 2 2 Case 1.1. When x′ and 4 x′ x′ c′, we bound g from above: ≤ 2 − ≤ c a(x + a) g(x,c,a,a)= a + (1 x a)f x + a, − +2a(1 x a) − − 1 x a − − − − c a + (1 x a)f x + a, +2a(1 x a)=:g ˜(x,c,a,a). ≤ − − 1 x a − − − − c 1 2 2 We now consider subcases based on the values of c and 1 x a compared to 4 . Note that 4 x′ x′ c − − − ≤ c′ 1 x a . ≤ − − c 1 Case 1.1.1. When c 1 x a 4 , using (5), ≤ − − ≤
g(x,c,a,a) f(x, c) g˜(x,c,a,a) f˜(x, c)= a + (1 x a) (x + a)2 2(x + a)+1 − ≤ − − − − c √c + +2a(1 x a) x2 2x +1 c + √c (7) − 1 x a √1 x a − − − − − − − − − = x(1 x 2a)+(1 x a)(x + a)2 + √1 x 1 √c. − − − − − − ′ − Thus, ∂ √1 x a 1 g˜(x,c,a,a) f˜(x, c) = − − − 0. ∂c − 2√c ≤ 2 2 So it is enough to check that g˜(x,c,a,a) f˜(x, c) 0 when c′ =4 x′ x′ or, equivalently, when 2 2 − ≤ − c =4 x′ x′ (1 x′)+ a(x + a); then it is also 0 for bigger c. First some auxiliary calculations: − − ≤ x 2 x √1 x 1 1 x + ′ 1= ′ 0. (8) − ′ − ≤ s − ′ 4 − − 2 ≤
2 2 2 2 2 2 1 1 4 x′ x′ 4x′ and 4 x′ x′ 4 = , so − ≤ − ≤ · 4 4 2 2 2 1 4 x′ x′ min 4x′ , x′. (9) − ≤ 4 ≤ So, 2 2 2 2 2 2 4 x′ x′ (1 x′)+ a(x + a) 4 x′ x′ (1 x′)+4 x′ x′ a − − ≥ − − − 2 2 (10) 2 2 2 = 2 x′ x′ (1 x) 2 x′ x′ (1 x) . − − ≥ − − 2 2 Putting c =4 x′ x′ (1 x′)+ a(x + a) in (7) and then using (8) and (10), − − g˜(x,c,a,a) f˜(x, c)= x(1 x 2a)+(1 x a)(x + a)2 − − − − − − 2 + √1 x 1 4 x x 2 (1 x )+ a(x + a) − ′ − ′ − ′ − ′ q 2 2 x(1 x 2a)+(1 x a)(x + a) x′ x′ x′ (1 x) ≤− − − − − − − − 1 = x (1 x 2a) (1 x a)(x + a)2 x (1 x 2a) (1 x a) − − − − − − ≤− − − − − − 4 3 3x 7a = x − − , − 4
16 3 3x 3a 3 3x′ 3 3x′ 3 which is 0 when a − 4− = −4 . Assume a> −4 . Since x′ a, x′ > 7 (and we have also ≤1 ≤ ≥ assumed x′); and 2 ≥
2 2 2 2 (3 3x′)x′ c =4 x′ x′ (1 x′)+ a(x + a) > 4 x′ x′ (1 x′)+ − − − − − 4 2 3 3 2 1 3 3 1 3 5391 1 > 4 1 + − · 2 · 7 = > · 7 − 7 − 2 4 19208 4 !
c 1 contrary to our assumption that c 1 x a 4 . ≤ − − ≤ 1 c 2 2 Case 1.1.2. When c 4 < 1 x a (and recall 4 x′ x′ (1 x′)+ a(x + a) c), using (5), ≤ − − − − ≤ g(x,c,a,a) f(x, c) g˜(x,c,a,a) f˜(x, c)= a + (1 x a) (x + a)2 2(x + a)+1.25 − ≤ − − − − +2a(1 x a) x2 2x +1 c + √c (11) − − − − − = x(1 x 2a)+(1 x a) (x + a)2 +0.25 + c √c. − − − − − −
2 2 1 Case 1.1.2.1. If 4 x′ x′ (1 x′)+ a(x + a) (1 x a) (which is 2 2 4 x x 2 (1 x )+ x 2 b 4 x x 2 (1 x )+ a(x + a) ′ − ′ − ′ ′ − − ′ − ′ − ′ q 1 q 1 (13) x(x + a) b x(x + a) b . ≤ 2 − ≤ 2 − 2 4 x x 2 (1 x )+ a(x + a) 4 x′ x′ (1 x) ′ − ′ − ′ − − q 2 2 2 2 1 (1 a) 4 x′ x′ (1 x) 4 x′ x′ (1 x′) = 4(1 x′) x′ 4 x′ = x + a. − · − − ≥ − − − ≥ · 2 So, 1 x(1 x 2a)+ 1 x(x + a) b − − − 4 x x 2 (1 x) − − ′ ′ ! (14) − − 1 a x(1 x 2 a)+ − 1 x(x + a)=0. ≤− − − x + a − 2 2 By (9), 4 x′ x′ x′, so − ≤ 2 2 2 2 2 2 2 2 2 4 x′ x′ (1 x′)+ x′ 4 x′ x′ (1 x′)+4 x′ x′ x′ = 2 x′ x′ . (15) − − ≥ − − − − 2 2 2 2 2 4 x′ x′ (1 x′)+ x′ b 2 x′ x′ , (16) − − − ≥ − 17 2 2 2 2 2 1 1 1 2 2 since if b> 0, 4 x′ x′ (1 x′)+ x′ b = 2 x′ = 2 x′ x′ (if b = − − − 4 ≥ 4 − 2 − − 0, then it holds by (15)). Now using (13) in (12), and then using (14) and (16) and thatt √t is decreasing in 0 t 1/4, we get − ≤ ≤ 2 2 2 2 g(x,c,a,a) f(x, c) (1 x a) (x + a) +0.25 +4 x′ x′ (1 x′)+ x′ b − ≤ − − − − − 2 1 4 x x 2 (1 x )+ x 2 b x(1 x 2a)+ 1 x(x + a) b − ′ − ′ − ′ ′ − − − − 4 x x 2 (1 x) − − q ′ − ′ − ! 2 2 2 2 (1 x a) (x + a) +0.25 + 2 x′ x′ 2 x′ x′ ≤ − − − − − 1 1 2 =4 x′ x′ (x′ 1) , − 4 − 2 − 1 1 which is 0 when x′ 1. When x′ < , we show that this subcase cannot hold: ≤ 4 ≤ ≤ 4 2 2 2 2 1 1 1 1 2 4 x′ x′ (1 x′)+ a(x + a) (1 x a) < 4 (1 x′)+ x′ − − − 4 − − · 4 − 4 − 4 − ! (17) 7 2 1 2 1 1 = (1 x′)+ x′ < (1 x′)+ x′ = x′ x′ + < 0. −64 − −12 − − 4 3 1 c Case 1.1.3. When 4 c (which is < 1 x a ), ≤ − − g(x,c,a,a) f(x, c) g˜(x,c,a,a) f(x, c)= a + (1 x a) (x + a)2 2(x + a)+1.25 − ≤ − − − − 1 1 +2a(1 x a) x2 2x +1.25 =˜g x, ,a,a f˜ x, 0, − − − − 4 − 4 ≤ as it falls in Case 1.1.1. or 1.1.2. above. 1 2 2 Case 1.2. When x′ and c′ 4 x′ x′ , ≤ 2 ≤ − 1 g(x,c,a,a)= a + (1 x a) 1 x′ + 1 c′ +2a(1 x a), − − − 4 x x 2 − − − ′ − ′ ! ! which is linear in c′, so also in c. f(x, c) is concave in c, so it is enough to check that g is smaller than 2 2 f in the ends of the interval c a(a + x), 4 x′ x′ (1 x′)+ a(x + a) : ∈ − − h i g(x, a(a + x),a,a)= a + (1 x a)2 +2a(1 x a) − − − − 2ax 1 x a + 1 a x a + (1 x a)2 +2a(1 x a)+ 2 − − − 2 ≤ − − − − a(x + a)+ ax + a 2 = x 2x +1 a(x + a)+ a(x + a) fpx, a(x + a) , − − ≤ 1 2 1 p since 2 x a 0 and a(x + a) x′ 4 , and using (5). Whereas the higher end of the interval was handled− above− in≥ Case 1.1. since≤f is continuous.≤ 1 Case 1.3. Finally, when x′, 2 ≤ g(x,c,a,a)= a + (1 x a)2 +2a(1 x a)= x2 2x +1 a2 + a. − − − − − − If x 1 , let c˜ = min c, 1 . Since c a(x + a) a2, and by (5) ≤ 2 4 ≥ ≥ g(x,c,a,a )= x2 2x +1 a2 + a x2 2x +1 c˜ + √c˜ f(x, c). − − ≤ − − ≤ If 1 x, then a 1 x 1 . a2 + a is monotonously increasing in a 0, 1 , so 2 ≤ ≤ − ≤ 2 − ∈ 2 g(x,c,a,a)= x2 2x +1 a2 + a x2 2x +1 (1 x)2 +1 x =1 x f(x, c), − − ≤ − − − − − ≤ by Point 5. 18 c x+√x2+4c c c x+√x2+4c Case 2. Now consider a , that is, a − . Then a˜ = = − , x+a 2 x+a −x+√x2+4c 2 ≥ ≥ ≤ x+ 2 c a˜(x+a) c a˜(x+a) and −1 x a =0, so f x + a, −1 x a =1 x a. − − − − − − c g(x, c, a, a˜)= a + (1 x a)2 +2 (1 x a) a + (1 x a)2 + x + x2 +4c (1 x a), − − x + a − − ≤ − − − − − p which is quadratic in a with a positive leading coefficient, so its maximum is at one end of the interval x+√x2+4c x+√x2+4c c − , 1 x . a = − (i.e., a = =a ˜) was handled above in Case 1. If a =1 x, the 2 − 2 x+a − hright side of the inequalityi equals 1 x which is f(x, c) by Point 5. − ≤ Proof of Point 4. If x + a =0, h(x, c, a, a˜)= f(x, c). From now on, we assume that x + a> 0. We first show that h is monotonously increasing in a˜. Using (6) and a calculation similar to the one in the proof of Point 3, we have ∂ ∂ c (x +˜a)(x + a) h(x, c, a, a˜) = 2(1 x a)+(1 x a) f x + a, − ∂a˜ − − − − · ∂c 1 x a − − 1 1 ∂ c (x +˜a)(x + a) 2 1 if x + a 4((x+a) (x+a) ) − ≤ 2 − 2(1 x a) (x + a) − 1 0. · ∂a˜ 1 x a ≥ − − − (0 if x + a! ≥ − − 2 ≤ Therefore, from now on we assume a˜ = min a, c x . x+a − Case 1. First assume a c x (so a˜ = a), which is equivalent to (x + a)2 c or a √c x. Let ≤ x+a − ≤ ≤ − c (x+a)2 x′ = x + a and c′ = −1 x a . − − 1 2 2 Case 1.1. When x′ and 4 x′ x′ c′, we bound h from above: ≤ 2 − ≤ c (x + a)2 h(x,c,a,a)= a + (1 x a)f x + a, − +2a(1 x a)+ x 3x(x + a) − − 1 x a − − − − − c a + (1 x a)f x + a, +2a(1 x a)+ x 3x(x + a) =: h˜(x,c,a,a). ≤ − − 1 x a − − − − − c 1 2 2 We now consider subcases based on the values of c and 1 x a compared to 4 . Note that 4 x′ x′ c − − − ≤ c′ 1 x a . ≤ − − c 1 Case 1.1.1. When c 1 x a 4 , using (5), ≤ − − ≤ h(x,c,a,a) f(x, c) h˜(x,c,a,a) f˜(x, c)= a + (1 x a) (x + a)2 2(x + a)+1 − ≤ − − − − c √c + +2a(1 x a)+ x 3x(x + a) x2 2x +1 c + √c (18) − 1 x a √1 x a − − − − − − − − − − = 2x2 xa + (1 x a)(x + a)2 + √1 x 1 √c. − − − − − ′ − ∂ √1 x a 1 h˜(x,c,a,a) f˜(x, c) = − − − 0. ∂c − 2√c ≤ 2 2 So it is enough to check that h˜(x,c,a,a) f˜(x, c) 0 when c′ =4 x′ x′ or, equivalently, when 2 2 2 − ≤ − c = 4 x′ x′ (1 x′)+ x′ ; then it is also 0 for bigger c. As seen in (15) and (8) in the − − ≤ 2 2 2 2 2 x′ proof of Point 3, 4 x′ x′ (1 x′)+ x′ 2 x′ x′ , and √1 x 1 0. Putting − − ≥ − − ′ − ≤− 2 ≤ 2 2 2 c =4 x′ x′ (1 x′)+ x′ in (18) and using these inequalities, we get − − h˜(x,c,a,a ) f˜(x, c)= 2x2 xa + (1 x a)(x + a)2 − − − − − 2 + √1 x 1 4 x x 2 (1 x )+ x 2 − ′ − ′ − ′ − ′ ′ 2 q 2 2 2 2x xa + (1 x a)(x+ a) x′ x′ x′ = 2x xa 0. ≤− − − − − − − − ≤ 19 2 1 c 2 2 2 2 Case 1.1.2. When c 4 < 1 x a (and recall 2 x′ x′ 4 x′ x′ (1 x′)+ x′ c), ≤ − − − ≤ − − ≤ using (5), h(x,c,a,a) f(x, c) h˜(x,c,a,a) f˜(x, c)= a + (1 x a) (x + a)2 2(x + a)+1.25 − ≤ − − − − +2a(1 x a)+ x 3x(x + a) x2 2x +1 c + √c (19) − − − − − − = 2x2 xa + (1 x a) (x + a)2 +0.25 + c √c. − − − − − 2 2 2 1 Case 1.1.2.1. If 4 x′ x′ (1 x′)+ x′ (1 x a) (which is 2 2 2 2 2 1 1 1 1 2 4 x′ x′ (1 x′)+ x′ (1 x a) < 4 (1 x′)+ x′ < 0, − − − 4 − − · 4 − 4 − 4 − ! like in (17) in the proof of Point 3. 1 c Case 1.1.3. When 4 c (which is < 1 x a ), ≤ − − h(x,c,a,a) f(x, c) h˜(x,c,a,a) f(x, c)= a + (1 x a) (x + a)2 2(x + a)+1.25 − ≤ − − − − 1 1 +2a(1 x a)+ x 3x(x + a) x2 2x +1.25 = h˜ x, ,a,a f˜ x, 0, − − − − − 4 − 4 ≤ as it falls in Case 1.1.1. or 1.1.2. above. 1 2 2 Case 1.2. When x′ and c′ 4 x′ x′ , ≤ 2 ≤ − 1 h(x,c,a,a)= a + (1 x a) 1 x′ + 1 c′ +2a(1 x a)+ x 3x(x + a), − − − 4 x x 2 − − − − ′ − ′ ! ! which is linear in c′, so also in c. f(x, c) is concave in c, so it is enough to check that h is smaller than 2 2 2 2 f in the ends of the interval c (x + a) , 4 x′ x′ (1 x′)+ x′ : ∈ − − h i h x, (x + a)2,a,a = a + (1 x a)2 +2a(1 x a)+ x 3x(x + a) − − − − − a + (1 x a)2 +2a(1 x a)+ x 3x(x + a)+ x(2x + a) ≤ − − − − − = x2 2x +1 (x + a)2 + (x + a) f x, (x + a)2 − − ≤ 2 2 1 since (x + a) = x′ 4 , and using (5). Whereas the higher end of the interval was handled above in Case 1.1. since f is continuous.≤ 1 2 1 Case 1.3. Finally, when x′, c (x + a) , so 2 ≤ ≥ ≥ 4 h(x,c,a,a)= a + (1 x a)2 +2a(1 x a)+ x 3x(x + a) x2 2x +1 (x + a)2 + (x + a). − − − − − ≤ − − 20 If x 1 , then ≤ 2 h(x,c,a,a) x2 2x +1 (x + a)2 + (x + a) x2 2x +1.25 = f(x, c). ≤ − − ≤ − If 1 x, then (x + a)2 + (x + a) is monotonously decreasing in a, so 2 ≤ − h(x,c,a,a) x2 2x +1 (x + a)2 + (x + a) x2 2x +1 x2 + x =1 x = f(x, c). ≤ − − ≤ − − − Case 2. Now we consider a c x, that is, a √c x. Then a˜ = min a, c x = c x ≥ x+a − ≥ − x+a − x+a − ≤ c (x+˜a)(x+a) c (x+˜a)(x+a) √c x, and − 1 x a =0, so f x + a, − 1 x a =1 x a. − − − − − − − c h(x, c, a, a˜)= a + (1 x a)2 +2 x (1 x a)+ x 3x(x + a) − − x + a − − − − a + (1 x a)2 +2 √c x (1 x a)+ x 3x(x + a), ≤ − − − − − − which is quadratic in a with a positive leading coefficient, so its maximum is at one end of the interval c [√c x, 1 x]. a = √c x (i.e., a = x+a x =˜a) was handled above in Case 1. If a =1 x, the right side− of the− inequality equals− 1 3x which− is f(x, c) by Point 5. − − ≤ 21