Discrepancy in graphs and hypergraphs

Transcription

1 Discrepacy i graphs ad hypergraphs B. Bollobás A.D. Scott Abstract Let G be a graph with vertices ad p ( edges, ad defie the discrepacies disc + { ( p (G) = max Y V (G) e(y ) p Y )} 2 ad disc p (G) = { ( max Y V (G) p Y ) } 2 e(y ). We prove that if p(1 p) 1/ the disc + p (G)disc p (G) p(1 p) 3 /6400. We also prove a similar iequality for k-uiform hypergraphs, ad give related results cocerig 2-colourigs of k-uiform hypergraphs. Our results exted those of Erdős, Goldberg, Pach ad Specer [6] ad Erdős ad Specer [7]. 1 Itroductio ) The discrepacy of a graph G is disc(g) = max Y V (G) e(y ) 1 2 2, where we write e(y ) = e(g[y ]) for the umber of edges of G spaed by Y. If G has edge desity 1/2 the the discrepacy ca be see as a measure of how uiformly the edges are distributed amog the vertices; see Sós [11] ad Beck ad Sós [1]) for more discussio ad a geeral accout of discrepacy. Erdős ad Specer [7] showed that for some costat c > 0 every graph G of order satisfies disc(g) c 3/2. More geerally, they showed that for every k 3 there is a costat c k > 0 such that if H is a k-uiform hypergraph( of order the disc(h) c k (k+1)/2, where disc(h) = max Y V (H) e(y ) 1 Y ) 2 k. By cosiderig radom graphs they showed that this boud is sharp up to the value of the costat. Triity College, Cambridge CB2 1TQ ad Departmet of Mathematical Scieces, Uiversity of Memphis, Memphis TN38152; bollobas@msci.memphis.edu Research supported i part by NSF grat DSM ad DARPA grat F C-1900 Departmet of Mathematics, Uiversity College Lodo, Gower Street, Lodo WC1E 6BT; scott@math.ucl.ac.uk 1 ( Y

2 Now suppose that G is a graph with e(g) = m = p ( ) G 2, where p < 1/2, so that we expect a radom subset Y V (G) to spa a subgraph with p ( ) Y 2 edges. The a more appropriate measure of edge distributio is give by the quatity disc p (G) = max Y V (G) e(y ) p ( Y. Erdős, Goldberg, Pach ad Specer [6] showed that i this case disc p (G) c m, where c is a absolute costat. A subset of vertices with large discrepacy ca clearly be either more or less dese tha the whole graph. Let us defie the positive discrepacy by disc + (G) = max Y V (G) ( {e(y ) 1 Y ) } ad the egative discrepacy by disc { ( (G) = max Y ) } Y V (G) 2 2 e(y ). The a radom graph G G(, 1/ shows that it is possible to have max{disc + (G), disc (G)} c 3/2. The oe-sided discrepacy ca be smaller: for istace, the complete bipartite graph K /2,/2 has positive discrepacy O(), although its egative discrepacy is c 2. Similarly, the graph 2K /2 has positive discrepacy O() but egative discrepacy c 2. These examples show that we ca guaratee small discrepacy o oe side provided we allow large discrepacy o the other. I this paper we shall prove that positive discrepacy substatially smaller tha 3/2 guaratees egative discrepacy substatially larger tha 3/2 ; ideed, we shall quatify the trade-off betwee positive ad egative discrepacies. Surprisigly, the correct measure turs out to be the product disc + (G)disc (G). We remark that a differet type of egative discrepacy was cosidered by Erdős, Faudree, Rousseau ad Schelp [5] with the idea of showig that graphs with small egative discrepacy cotai complete subgraphs of fixed size. For further recet results i this directio see Krivelevich [9] ad Keevash ad Sudakov [8]. We begi with some defiitios. For a k-uiform hypergraph G, a real p [0, 1] ad X V (G) let ( ) X d p (X) = e(x) p. k For disjoit sets of vertices X ad Y, let d p (X, Y ) = e(x, Y ) p X Y. The we defie disc + p (G) = max d p(x) (1) X V (G) 2

3 ad ad set disc p (G) = mi X V (G) d p (X), ( disc p (G) = max d p(x) = max{disc + p (G), disc p (G)}. X V (G) If p is ot specified we assume p = 1/2, so for istace disc(g) = disc 1/2 (G). ( Note that the cases p = 0 ad p = 1 are trivial, ad that if e(g) = p G ) 0 2 we have disc p (G) d p (V (G)) = p p 0 ( ) G 2. We will therefore usually take p with e(g) = p ( ) G 2. Note that, for ay p, disc + p (G) = disc 1 p(g) ad disc p (G) = disc + 1 p(g). We shall usually assume p 1/2, sice if p > 1/2 we may replace G by G ad p by 1 p. We remark that it does ot make much differece if we restrict the defiitios i (1) ad ( to sets X of size /2 (or some other size c): as oted by Erdős, Goldberg, Pach ad Specer [6], this would chage the resultig discrepacy by at most a costat factor. We shall frequetly refer to a radom bipartitio V = X Y. Uless otherwise stated, this meas a radom bipartitio i which each vertex is assiged idepedetly to X or Y with equal probability. Throughout the paper we shall use ɛ i ad ρ j for sequeces of idepedet Beroulli radom variables, with ɛ i {+1, 1} ad ρ j {0, 1}, each takig either value with probability 1/2. The rest of the paper is orgaized as follows. I sectio 2 we give lower bouds o disc p (G) for graphs; i sectio 3 we tur our attetio to hypergraphs. Fially, i sectio 4, we cosider some related results cocerig subgraphs of a fixed graph or hypergraph. 2 Discrepacy of graphs I this sectio we prove our results o graph discrepacy. Let G be a graph of order ad size p (. If G is very sparse, say 0 < p 1/( 1), the takig the uio of p ( ( /2 edges from G gives a subgraph with at most p vertices, so disc + p (G) p ( ( /2 p(p ) 2 /2 p 2 /5 p 3 4 /8 > p 2 /20 for sufficietly large, while sice G has average degree at most 1 it cotais a idepedet set of size at least /2, ad so disc (G) p ( ) /2 2 > p 2 /9 for sufficietly large. O the other had, max{disc + (G), disc (G)} e(g) < p 2 /2. 3

4 Thus disc + p (G) ad disc p (G) are both Θ(p 2 ). A similar argumet applies if G is very dese, with p 1 1/( 1). (More precise bouds are give by Erdős, Goldberg, Pach ad Specer [6].) We therefore restrict our attetio to graphs with p(1 p) 1/. Our mai result is the followig. Theorem 1. Let G be a graph of order ad size p (, where p(1 p) 1/. The disc + p (G)disc p (G) p(1 p) 3 /6400. (3) As a immediate corollary we get the followig result of Erdős, Goldberg, Pach ad Specer [6]. Corollary 2. Let G be a graph of order ad size p (, where p(1 p) 1/. The disc p (G) p(1 p) 3/2 /80. We remark that the result of Erdős ad Specer for graphs ca easily be deduced from Theorem 1: if 1/3 p 2/3 the disc(g) 1 2 (disc+ p (G) + disc p (G)) 3/2 /160, while otherwise disc(g) e(g) 2( 1 ( /6 3/2 /12. We also remark that, for r 2, the Turá graph T r () gives a boud o the optimal costat i (3). Defiig p by t r () = e(t r ()) = p (, we have p 1 1. A little calculatio shows that r disc + p (T r ()) = p 8 + O(r), (4) ad, for r eve, which implies disc p (T r ()) = (1 + o(1)) (1 p)2, 8 disc + p (T r ())disc p (T r ()) (1 + o(1)) p(1 p)3. 64 Before turig to the proof of Theorem 1, we make some commets about oe-sided discrepacies. Sice every graph with vertices ad t r () edges cotais a subgraph of order u ad size at least t r (u) for every 1 u, 4

5 the Turá graphs T r () have miimal positive p-discrepacy amog graphs of order ad size t r (). Thus (4) gives a optimal boud i these cases, which have desity p 1 1. To obtai a similar boud for arbitrary desities, r we defie a extesio of the Turá umbers for o-itegral r. Give a iteger 1 ad a real umber r 1, we ca write = qr + s, where q is a iteger ad 0 s < r. We defie the fractioal Turá umber t r () by ( ) t r () = t r (), 2 where ( ) ( ) q + 1 q t r () = s + (r s). 2 2 Note that this is cosistet with the defiitio of Turá umbers whe r is itegral; it is coveiet to work with the quatity t r () istead of t r (). A boud matchig (4) will follow from the followig result. Lemma 3. Suppose that 1 is a iteger ad 1 r. Let G be a graph with vertices ad at least t r () edges. The, for 2 u, G cotais a subgraph with u vertices ad at least t r (u) edges. Proof. It is eough to prove the theorem whe u = 1. Takig complemets, this is equivalet to showig that if e(g) t r () the there is a vertex v such that e(g \ v) t r ( 1). We may also assume r > 1, or else G is empty. Addig edges if ecessary, we may assume that e(g) = t r () = t r () η, (5) where 0 η < 1. Thus if = qr + s, e(g) = ( ) ( ) q + 1 q s + (r s) η 2 2 = 1 ( rq 2 + (2s r)q ) η. 2 A short calculatio shows that 2e(G) qs + s 2η (G) = q 1 +. (6) qr + s 5

6 By (5) ad (6), it is sufficiet to show that qs + s 2η q η t r () t r ( 1). (7) qr + s If q = 0 the we have a complete graph ad are doe immediately. Thus we may assume that q 1. Now if s 1, the it is easily see that t r () t r ( 1) = q, while if 0 s < 1, the a simple calculatio shows that t r () t r ( 1) = q 1 + s. (8) Now if s > η the qs + s > 2η, ad so the left side of (7) is at least q + η, ad thus (7) is satisfied. If s η, however, the 0 s < 1, so (8) holds. It is the sufficiet by (7) to show that qs + s 2η + η s, qr + s which holds provided qs + s 2η qr + s > 1. But qr + s = ad qs + s 2η 2η > 2, so this holds for 2. Calculatig as i (4), we obtai the followig result. Corollary 4. For 0 p 1, every graph G with vertices ad p ( edges satisfies disc + p (G) p 8 + O( 1 1 p ). We ow tur to the proof of Theorem 1. We shall eed two simple iequalities (these follow easily from the Littlewood-Khichi iequality, see [10], [12], [13]; however, we give short proofs at the ed of the sectio). Recall that ɛ i ad ρ i are i.i.d. Beroulli with ɛ i {+1, 1} ad ρ i {0, 1}. Lemma 5. For 1, E ɛ i /2. i=1 6

7 Lemma 6. Let a = (a i ) i=1 be a sequece of real umbers, ad A a real umber. The E ɛ i a i A a 1 / 2 ad E i=1 ρ i a i A a 1 / 8. i=1 Our mai tool i the proof of Theorem 1 is the followig lemma, which shows that i a radom bipartitio of a graph G, we do ot expect the vertex eighbourhoods to split too evely. Lemma 7. Let G be a graph of order ad size p (, where p(1 p) 1/. Let V (G) = X Y be a radom bipartitio. The E Γ(x) Y p Y p(1 p) 3/2 /20. x X Proof. We may assume p 1/2 sice we may take complemets ad replace p by 1 p. Suppose x V (G) has degree d = d(x) = p( 1) + r(x). For v x, defie e v = 1 if xv E(G) ad e v = 0 otherwise. The E Γ(x) Y p Y \ {x} = E v x ρ v (e v p) = E 1 (e v p) + 1 ɛ v (e v p) 2 2 v x v x max{ 1 2 d ( 1)p, 1 2 E v x ɛ v (e v p) }, sice v x (e v p) = d ( 1)p ad the distributio of v x ɛ v(e v p) is symmetric about 0. Now, by Lemma 5, E v x ɛ v (e v p) = E E d ɛ i (1 p) + i=1 d ɛ i (1 p) i=1 (1 p) d/2 7 1 i=d+1 ɛ i ( p)

8 ad so E Γ(x) Y p Y \ {x} 1 2 max{ r(x), (1 p) d(x)/2}. Now for x V = V (G) let I(x) = 1 if x X ad I(x) = 0 otherwise. The, sice I(x) ad Γ(x) Y are idepedet radom variables, E Γ(x) Y p Y = E I(x) Γ(x) Y p Y \ {x} x X x V = 1 E Γ(x) Y p Y \ {x} 2 x V 1 4 max{ r(x), (1 p) d(x)/2} x V x V 1 ( r(x) + (1 p) d(x)/. 8 x V Note that the first equality holds as Y = Y \ {x} if I(x) = 1. Furthermore, r(x) + (1 p) d(x)/2 is miimized whe r(x) = 0 ad so d(x) = p( 1). Thus 1 8 x V ( r(x) + (1 p) ) d(x)/2 sice p 1/2 ad we may assume (1 p) p( 1)/2 p(1 p) 3/2 /20, After this preparatio, we are ready to prove Theorem 1. Proof of Theorem 1. Sice (3) is symmetric i p ad 1 p, we may replace G by its complemet G, ad so we may assume that disc + p (G) disc p (G). If disc + p (G) p(1 p) 3/2 /80 we are doe. Otherwise, suppose disc + p (G) = p(1 p) 3/2 /80α, where α 1. We shall show that so disc + p (G)disc p (G) p(1 p) 3 /6400. disc p (G) α p(1 p) 3/2 /80, (9) 8

9 Let V (G) = X Y be a radom bipartitio. The sice p(1 p) 1/, it follows from Lemma 7 that E Γ(x) Y p Y p(1 p) 3/2 /20. (10) x X Now let X + = {x X : Γ(x) Y p Y } ad X = X \ X + ; so d p (X, Y ) = ( ) ( ) Γ(x) Y p Y + Γ(x) Y p Y. x X + Sice Ed p (X, Y ) = 0, we have E ad so by (10) x X + E d p (X +, Y ) = E Γ(x) Y p Y = E x X + Now E d p (Y ) = 0, so (11) implies x X x X Γ(x) Y p Y ( Γ(x) Y p Y ) p(1 p) 3/2 /40. (11) E ( d p (X +, Y ) + αd p (Y ) ) p(1 p) 3/2 /40. (1 Let X +, Y be a pair of sets achievig at least the expectatio i (1 ad let Z be a radom subset of X +, where each vertex of X + is chose idepedetly with probability 1/α. The it follows from (1 that E d p (Z Y ) = E ( d p (Z) + d p (Z, Y ) + d p (Y ) ) = 1 α 2 d p(x) + 1 α d p(x, Y ) + d p (Y ) 1 α 2 d p(x) + 1 α p(1 p) 3/2 /40. Sice disc + p (G) = p(1 p) 3/2 /80α, this implies d p (X)/α 2 (1/α) p(1 p) 3/2 /80 ad so d p (X) α p(1 p) 3/2 /80, which gives the desired lower boud o disc p (G). 9

10 Fially i this sectio we give the proofs of Lemmas 5 ad 6, postpoed from earlier. Proof of Lemma 5. A simple calculatio shows that for = 2k we have E i=1 ɛ i = 2 1 2k k ( ) 2k k ad for = 2k + 1 we have E i=1 ɛ i = 2 2k (2k + 1) ( ) 2k k = E +1 i=1 ɛ i. Let s = E i=1 ɛ i /. The, for k 1, s 2k+2 /s 2k = (k + 1)/ k(k + 1) > 1 ad, for k 0, s 2 2k+3 /s 2k+1 = (k + 3)(k + 1)/(k ) < 1. Thus (s 2k ) k=1 is icreasig ad (s 2k+1) k=0 is decreasig; both coverge to E N(0, 1) = 2/π. Therefore s s 2 = 1/ 2 for all. Proof of Lemma 6. We may clearly assume that all a i are oegative. Sice i=1 ɛ ia i is symmetric about 0, the expectatio is miimized for a give a whe A = 0. Now if a i a j the let a i = a j = (a i +a j )/2; it is easily checked that E B + ɛ i a i + ɛ j a j E B + ɛ i a i + ɛ j a j for every real B. It follows that E i=1 ɛ ia i E i=1 ɛ ia, where a = i=1 a i/. Thus, by Lemma 5, E ɛ i a i A ae ɛ i a /2 = a 1 / 2. i=1 i=1 The secod iequality follows directly from the first. Note that i fact proof of Lemma 5 implies the iequalities E i=1 a i 2/π a 1 if is odd ad E i=1 a i (1 + o(1)) 2/π a 1 for geeral. 3 Hypergraph discrepacy I this sectio we tur our attetio to hypergraphs. After defiig a little otatio, we begi with a result for weighted hypergraphs; we the tur to the cosideratio of uweighted hypergraphs. If G is the complete k-uiform hypergraph with edge-weightig w ad X V (G), we defie d(x) = w(k). K X (k) As i defiitios (1) ad ( we defie disc + (G) = max X V (G) d(x) ad disc (G) = mi X V (G) d(x); we also defie disc(g) = max{disc + (G), disc (G)}. 10

11 Note that this is cosistet with the defiitios for a uweighted hypergraph G by takig w(e) = 1 if e E(G) ad w(e) = 1 otherwise. For disjoit sets X 1,..., X t ad itegers k 1,..., k t such that t i=1 k i = k, we defie d k1,...,k t (X 1,..., X t ) = w(e), where the sum is over edges e with e X i = k i for every i. We ca ow state the first result of the sectio. Theorem 8. Let G be the complete k-uiform hypergraph of order with edge-weightig w such that w(e) = 0 ad w(e) = ( k). The disc + (G)disc (G) 2 14k2 k+1. We shall eed three lemmas. I the first lemma we use the fact that if P (x) is a polyomial of degree k with sup x [0,1] P (x) 1 the every coefficiet of P (x) has absolute value at most 2 k k 2k /k!. (Tamás Erdélyi [4] poited out to us that this is a elemetary cosequece of Markov s Iequality; see [3].) Lemma 9. If G is a complete k-uiform hypergraph with edge-weightig w ad disc(g) M the for disjoit subsets X, Y of V (G) ad 0 i k, d i,k i (X, Y ) 2 2k2 M. Proof. Let Z be a radom subset of X, where each vertex is chose idepedetly with probability p. The E(d(Z Y )) = k p i d i,k i (X, Y ). i=0 Sice disc(g) M, it follows that max 0 p 1 k i=0 pi d i,k i (X, Y ) M ad so max 0 i k d i,k i (X, Y ) 2 k k 2k M/k! 2 2k2 M. We also eed a aalogue of Lemma 7. Lemma 10. Let G be a complete k-uiform hypergraph of order with edgeweightig w. Let V (G) = U W be a radom bipartitio. The E d k 1,1 (K, W ) k2 k w(l) / 2. K U (k 1) L V (G) (k) 11

12 Proof. Let V = V (G) = U W be a radom bipartitio. Give K V (k 1), it follows from Lemma 6 that E d k 1,1 (K, W \ K) w(k {v}) / 8. v V \K Sice the evet {K U} ad the radom variable d k 1,1 (K, W \ K) are idepedet, ad each edge L V (k) occurs k times as K {v}, we have E d k 1,1 (K, W ) = P(K U)E d k 1,1 (K, W \ K) K U (k 1) K V (k 1) w(k {v}) / 8 K V (k 1) 2 k+1 v V \K = k2 k w(l) / 2. L V (k) The followig lemma will be useful several times. Lemma 11. Let G be a k-uiform hypergraph of order with edge-weightig w. Suppose that α 1 ad X, Y are disjoit subsets of V (G) with d 1,k 1 (X, Y ) + αd(y ) = M 0. (13) The either or disc + (G) 2 3k2 M/α disc (G) 2 3k2 Mα. Proof. If d i,k i (X, Y ) 2 k2 αm for some 0 i k the we are doe by Lemma 9. Otherwise, let Z be a radom subset of X, obtaied by choosig 12

13 each vertex of X idepedetly with probability 1/α. The k E d(z Y ) = E d i,k i (Z, Y ) = i=0 k d i,k i (X, Y )/α i i=0 d(y ) + d 1,k 1 (X, Y )/α M/α (k 1)2 k2 M/α 2 3k2 M/α. k 2 k2 αm/α i Sice some set Z must achieve this boud, we obtai the desired boud o disc + (G). i=2 We ca ow prove the mai theorem of this sectio. Proof of Theorem 8. As i the proof of Theorem 1, we may assume that disc + (G) disc (G). If disc + (G) 2 7k2 (k+1)/2 we are doe. Otherwise, suppose disc + (G) = 2 7k2 (k+1)/2 /α for some α > 1: we shall show disc (G) 2 7k2 α (k+1)/2. Note first that for disjoit sets X, Y V (G), if d 1,k 1 (X, Y ) + αd(y ) 2 4k2 (k+1)/2 (14) the we are doe by Lemma 11. It is therefore eough to fid disjoit X, Y satisfyig (14). Let V (G) = X k W k 1 be a radom bipartitio ad let W k 1 = X k 1 W k 2,..., W 2 = X 2 W 1 be radom bipartitios where, as usual, i each bipartitio each vertex is assiged idepedetly to either vertex class with probability 1/2. We defie weightigs w i o the i-sets i W i for each i by w i (K) = d i,1,...,1 (K, X i+1,..., X k ). (15) Let W k = V (G) ad defie w k = w. The for 1 i < k ad K W (i) i, w i (K) = d i,1,...,1 (K, X i+1,..., X k ) = x X i+1 d i+1,1,...,1 (K {x}, X i+2,..., X k ) = x X i+1 w i+1 (K {x}). 13

14 It therefore follows from Lemma 10 that give W i+1 ad w i+1, E w i (K) (i + 1)2 (i+1) w i+1 (L) / 2. (16) It follows that E K W (i) i x W 1 d 1,...,1 ({x}, X 2,..., X k ) = E L W (i+1) i+1 x W 1 w 1 (x) k!2 (k+1 2 ) = k!2 (k+1 2 ) K W (k) k w k (K) /( 2) k 1 ( ) /(2) (k 1)/2. k Let X 1 + = {x W 1 : d 1,...,1 (x, X 2,..., X k ) > 0}. The, as i (11), E d 1,...,1 (X 1 +, X 2,..., X k ) 1 ( ) 2 k!2 (k+1 2 ) /(2) (k 1)/2 2 2k2 (k+1)/2. k (17) We partitio the edges i V 0 = X 1 + k i=2 X i that meet X 1 + i exactly oe vertex as follows. For a oempty S {2,..., k}, let V S = i S X i ad E S = {K {x} : x X 1 +, K V (k 1) S, K X i > 0 i S}. Let d S = K E S w(k) ad ote that d 1,k 1 (X +, V S ) = =T S d T ad d {2,...,k} = d 1,...,1 (X 1 +, X 2,..., X k ). Let S 0 be miimal with d S0 (2k) k+ S d {2,...,k}. The max S {2,...,k} d 1,k 1(X + 1, S) d 1,k 1 (X + 1, V S0 ) d S0 =T S 0 d T Thus it follows from (17) that ( (2k) k+ S 0 d {2,...,k} /2(2k) k 1 2 k2 d {2,...,k}. S 0 1 i=1 E max d 1,k 1(X 1 +, V S ) 2 3k2 (k+1)/2 S {2,...,k} 14 k S 0 i (2k) k+i) d {2,...,k}

15 ad so there is some S {2,..., k} with E d 1,k 1 (X + 1, V S ) 2 3k2 (k+1)/2 /2 k. Now let Y = V S ad X + S = {x W 1 : d 1,k 1 ({x}, V S ) > 0}. The, sice E d 1,k 1 (W 1, V S ) = 0, we have E d 1,k 1 (X + S, V S) 2 3k2 (k+1)/2 /2 k+1 2 4k2 (k+1)/2. Fially, sice E d(v S ) = 0, we have E d 1,k 1 (X + S, V S) + αd(v S ) 2 4k2 (k+1)/2. It follows that there are sets X, Y satisfyig (14). We ote that Theorem 8 implies the followig boud o disc + p (G)disc p (G) for uweighted hypergraphs G. Corollary 12. Let G be a k-uiform hypergraph with vertices ad p ( ) k edges. The disc + p (G)disc p (G) 2 14k2 +2 p 2 (1 p) 2 k+1. Proof. The result is trivial if p = 0 or p = 1. Otherwise, let H be the complete k-uiform hypergraph o the same vertex set as G with edge-weightig w defied by w(e) = 1/2p if e ( E(G) ad w(e) = 1/2(1 p) otherwise. The w(e) = 0 ad w(e) = ) k, ad so, by Theorem 8, Now for Y V (G), Thus disc + (H)disc (H) 2 14k2 k+1. ( ) Y d (G) p (Y ) = e(y ) p 2 = (1 K E(G) p) K Y (k) = 2p(1 p)w(k) K Y (k) = 2p(1 p)d (H) (Y ). disc + p (G)disc p (G) = 4p 2 (1 p) 2 disc + (H)disc (H), (18) which implies the required boud. 15

16 We ca, however, improve upo the p 2 (1 p) 2 term i Corollary 12 (at the cost of a slightly worse costat) to obtai a boud similar to that i Theorem 1. First, however, we eed a versio of Lemma 7 for uweighted hypergraphs. Lemma 13. Let G be a k-uiform hypergraph of order with p ( k) edges, where p(1 p) 1/ ad 2k. Let V (G) = X Y be a radom bipartitio. The E dk 1,1 (K, Y ) p Y 2 2k 2 p(1 p) k 1 2. K X (k 1) Proof. We follow the argumet of Lemma 7. As before, we may assume p 1/2. Let V = V (G) = X Y be a radom bipartitio. For K V (k 1), let d(k) be the umber of edges of G cotaiig K ad defie r(k) by d(k) = p( k + 1) + r(k). Let d = p( k + 1). The, as i Lemma 7, E dk 1,1 (K, Y \ K) p Y \ K 1 2 max{ r(k), (1 p) d(k)/2}. For K V (k 1), we defie I(K) = 1 if K X ad I(K) = 0 otherwise. The I(K) ad d k 1,1 (K, Y \ K) are idepedet radom variables, so E dk 1,1 (K, Y ) p Y K X (k 1) K X (k 1) = E I(K) K V (k 1) = 2 k+1 K V (k 1) E dk 1,1 (K, Y \ K) p Y \ K dk 1,1 (K, Y \ K) p Y \ K 2 k max{ r(k), (1 p) d(k)/2} K V (k 1) K V (k 1) 2 (k+1) r(k) + (1 p) d(k)/2. K V (k 1) Sice r(k) + (1 p) d(x)/2 is miimized whe r(k) = 0 ad d(k) = p( k + 1), E d k 1,1 (K, Y ) p Y ( ) 2 (k+1) (1 p) p( k + 1)/2 k 1 > 2 2k2 p(1 p) k

17 Theorem 14. Let G be a k-uiform hypergraph of order with p ( k) edges, where p(1 p) 1/. The disc + p (G)disc p (G) 2 18k2 p(1 p) k+1. Proof. Let H be the complete k-uiform hypergraph o V (G) with weightig w(e) = 1 p if e E(G) ad w(e) = p otherwise. The disc + (H) = disc + p (G) ad disc (H) = disc p (G). Note that w(h) = 0. As usual we may assume p 1/2 ad disc (H) disc + (H) = 2 9k2 p(1 p) (k+1)/2 /α. If α 1 we are doe, so we may assume α 1. We will show that disc (H) 2 9k2 p(1 p)α (k+1)/2. If there are disjoit X, Y V (H) with d 1,k 1 (X, Y ) + αd(y ) 2 6k2 p(1 p) (k+1)/2 (19) the we are doe by Lemma 11. Thus it is eough to fid disjoit X, Y satisfyig (19). As i the proof of Theorem 8, we defie radom sets W k = X W k 1 W 1, where the i-sets i W i are weighted as i equatio (15). The by Lemma 13, E w k 1 (K) 2 2k2 p(1 p) k 1 2, (20) K W (k 1) k 1 while W 1,..., W k 2 satisfy (16). We have E w 1 (x) (k 1)! 2 (k x W 1 K W (k 1) k 1 w k 1 (K) /( 2) k 2, ad so, defiig X + 1 as before, we ca replace (17) by E d 1,...,1 (X + 1, X 2,..., X k ) 2 4k2 p(1 p) (k+1)/2. (21) The argumet is completed as before (with all bouds chaged by a factor 2 2k2 p(1 p)). The followig corollary is immediate. Corollary 15. Let G be a k-uiform hypergraph of order with p ( k) edges, where p(1 p) 1/. The disc p (G) 2 9k2 p(1 p) (k+1)/2. 17

18 We ote that Corollaries 2 ad 15 are best possible up to the value of the costat 2 9k2. To see this, let G G (k) (, p) be a radom k-uiform hypergraph, where each possible edge is preset idepedetly with probability p, ad let S V (G). Let N = ( k) ad h = (1+ɛ)k! 1/2 2p(1 p) l 2 (k+1)/2. The by stadard bouds o the tail of the biomial distributio (see [2], Theorem 1.3), provided p(1 p) c k 1 k, for ay subset S of V (G) we have P( d p (S) h) P( B(N, p) Np h) < 2 for sufficietly large. Thus there is some k-uiform hypergraph G of order with disc p (G) h. Let us also ote that the gai from p 2 (1 p) 2 to p(1 p) betwee Corollary 12 ad Theorem 14 comes because a typical vertex i G has degree p ( 1 k 1) : so if p is small, the the weight aroud a typical vertex is cocetrated i fairly few edges. We remark that o similar boud is possible for the larger class of k-uiform hypergraphs with w(e) = ( ) k such that max{w(e), 0} = p ( k) : cosider a radom k-uiform hypergraph H G (k) (, 1/, ad let G be the weighted hypergraph obtaied by givig each edge weight 2p ad each o-edge weight 2(1 p). The if e(h) = 2( 1 ) k (which happes with probability at least c k k/2 if ( k) is eve) we have w(e) = ( ( k) ad max{w(e), 0} = p k). O the other had, it follows from (18) that disc + p (G)disc p (G) = 4p 2 (1 p) 2 disc + (H)disc (H), while disc + (H) ad disc (H) are both O( (k+1)/2 ) with expoetially small failure probability. It is iterestig to ask about the rage i which Theorem 1 ad Theorem 14 are sharp (up to the costat). For istace, i the case of graphs the remarks above show that disc + p (G) ad disc p (G) ca both be aroud c p(1 p) 3/2. Whe p is (about) 1/2, the complete bipartite graph ad its complemet show that we ca have discrepacy O() o oe side (ad c 2 o the other). Thus Theorem 1 is sharp i i middle of the the scale from c to c 2, ad (for p = 1/ is sharp at the eds. How sharp is it at other parts of the scale, or at the eds whe p 1/2? The costat i Theorem 14 is clearly ot best possible. A more careful versio of the argumet should improve it to 2 ck l k ; it would be of iterest to kow the correct order of magitude. It would also be iterestig to kow what happes i the rage 1 k p 1/. 18

19 4 Subgraph discrepacy I previous sectios we have bee cocered with the discrepacy of subgraphs or, equivaletly, 2-colourigs of the complete graph. We begi this sectio by cosiderig 2-colourigs of a arbitrary graph: questios of this form were raised by Sós i [11]. For a k-uiform hypergraph G, a subgraph H of G ad a real umber p [0, 1], we defie ad disc + p (H, G) = max e(h[s]) pe(g[s]) S V (G) disc p (H, G) = max pe(g[s]) e(h[s]). S V (G) Note that if G is the complete k-uiform hypergraph the these two defiitios agree with (1) ad (. We set disc p (H, G) = max{disc + p (H, G), disc p (H, G)}. We begi with a fairly straightforward aalogue to Theorem 8. Note that arguig as i Corollary 12 gives a boud with p 2 (1 p) 2 i place of p(1 p). Theorem 16. Let G be a k-uiform hypergraph with vertices ad m edges, ad let H be a subgraph of G with pm edges, where p(1 p) 1/. The disc + p (H, G)disc p (H, G) 2 18k2 p(1 p)m 2 / k 1. We first eed a versio of Lemma 13. Lemma 17. Let G be a k-uiform hypergraph with vertices ad m edges, ad let H G be a subhypergraph of G with pm edges, where p(1 p) 1/. Let V (G) = X Y be a radom bipartitio. The E K X (k 1) d (H) k 1,1 (K, Y ) pd(g) k 1,1 (K, Y ) 2 (k+1) p(1 p)m/. Proof. For a partitio V (G) = X Y, let us write f(x, Y ) = (K, Y ) pd(g) K X (k 1) d (H) k 1,1 19 k 1,1 (K, Y ).

20 As i Lemma 13, we may assume that p 1/2 or else replace H by its complemet i G. For K V (k 1), let d H (K) be the umber of edges of H cotaiig K ad let d G (K) be the umber of edges of G cotaiig K. Defie r(k) by d H (K) = pd G (K) + r(k). The, as i Lemma 13, E d (H) k 1,1 (K, Y \ K) pd(g) k 1,1 (K, Y ) 1 2 max{ r(k), (1 p) d H (K)/2}. Thus Ef(X, Y ) 2 (k+1) r(k) + (1 p) dh (K)/2. K V (k 1) Now r(k) +(1 p) d H (K)/2 is miimized whe r(k) = 0 ad so d H (K) = pd G (K). Thus Ef(X, Y ) 2 (k+1) (1 p) pdg (K)/2 K V (k 1) 2 (k+1) (1 p)d G (K) p/2 K V (k 1) sice d G (K) <. Now K V (k 1) d G (K) = km, so Ef(X, Y ) 2 (k+1) km(1 p) p/2 2 (k+1) p(1 p)m/. Theorem 16 ow follows by a modificatio of the proof of Theorem 14. Proof of Theorem 16. Let V = V (G). We may assume p 1/2 or replace H by its complemet i G. We defie, as i Theorem 14, a edge-weightig w o V (k) by w(k) = 1 p if K E(H), w(k) = p if K E(G) \ E(H) ad w(k) = 0 otherwise. Note that the w(v ) = 0. We may assume disc p (H, G) disc + p (H, G) = 2 9k2 p(1 p)e(g)/ k 1 α. If α 1 we are doe, so we may assume α 1. If there are disjoit X, Y with d 1,k 1 (X, Y ) + αd(y ) 2 6k2 p(1 p)m/ (k 1)/2, the we are doe as before by Lemma 11. Oce agai, we defie radom subsets W k = X W k 1 W 1. Applyig Lemma 17 istead of 20

21 Lemma 13 to W k 1, we ca replace (20) by E w k 1 (K) 2 2k2 p(1 p)m/. (2 K W (k 1) k 1 As before, W 1,..., W k 2 satisfy (16); applyig this k 2 times to (2, we see that (istead of (21)) we obtai Ed(X + 1, X 2,..., X ) 2 4k2 p(1 p)m/ (k 1)/2, ad the argumet is completed as before. Corollary 18. Let G be a k-uiform hypergraph with vertices ad m edges, ad H a subgraph of G with pm edges, where p(1 p) 1/. The disc p (H, G) 2 9k2 p(1 p)m/ (k 1)/2. We obtai stroger results whe there is a restrictio o the maximum overlap betwee edges of positive ad egative weights. Theorem 19. Let G be a complete k-uiform hypergraph of order with edge-weightig w. Suppose i additio that, for some 1 s r, if w(e) > 0 ad w(e ) < 0 the e e < s. Let M = w(e) ad m = w(e). If m = (2p 1)M, where p(1 p) 1/, the disc + p (G)disc p (G) 2 18k2 p 2 (1 p) 2 M 2 / s 1. Proof. Suppose first that p = 1/2, ad let E = {e : w(e) 0}. As i the proof of Theorem 8, we may assume disc + (G) disc (G). Suppose disc + (G) = 2 9k2 e(h)/ (s 1)/2 α, where α 1. If there are disjoit X, Y with d 1,k 1 (X, Y ) + αd(y ) 2 6k2 e(h)/ (s 1)/2 the we are doe by Lemma 11. Otherwise, defie W i, X i ad w i as before, ad cosider W s ad w s. Sice w(e) > 0 ad w(e ) < 0 implies e e < s, we have, for K W (s) s, w s (K) = e W s=k, e X i =1 i>s w(e) 21

22 ad so E K W (s) s w s (K) = E e W s =s, e X i =1 i>s w(e). Let A e be the evet that e W s = s ad e W i = 1 for all i > s. The PA e > 2 k2 ad so E w K W s (s) s (K) 2 k2 M. Applyig Lemma 10 as i (16), we obtai that E d 1,...,1 ({x}, X 2,..., X k ) E w S (K) /( 2) s 1 x W 1 K W (s) s 2 2k2 M/ (s 1)/2. The rest of the argumet follows as i the proof of Theorem 8. Now suppose p 1/2. As i the proof of Corollary 12, we multiply all positive edge-weights by 1/2p ad all egative edge-weights by 1/2(1 p) to obtai a ew edge-weightig w. The result follows immediately. As a applicatio of Theorem 19, let us cosider the complete subgraphs of a graph ad its complemet. For t 2 ad a graph G, we write k t (G) for the umber of copies of K t of G. We write disc Kk (G) = max S V (G) k k(g[s]) k k (G[S]). For istace, disc K2 (G) is just disc(g). Clearly, complete subgraphs of G meet complete subgraphs of its complemet i at most oe vertex: applyig Theorem 19 to the k-uiform hypergraph of complete or idepedet k-sets gives the followig result. Corollary 20. For every graph G of order, For istace, i some subset S, disc Kk (G) c k k 1 2. k 3 (G[S]) k 3 (G[S]) c 5/2. Cosiderig radom graphs shows that this result is best possible up to the costat. A similar approach yields results i some cases for disc H (G) where H is ot a complete graph (ad disc H is defied i the obvious way). It would be iterestig to determie the correct order of magitude of disc H for all graphs H. Whe H is fairly dese, so that copies of H ad H caot overlap very much, we obtai a lower boud o disc H (G) usig Theorem 19. However, whe H is sparse this gives a much weaker boud; for istace, what ca we say whe H is a tree? 22

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