On the Rainbow Turán number of paths

Let F be a fixed graph. The rainbow Turán number of F is defined as the maximum number of edges in a graph on n vertices that has a proper edge-coloring with no rainbow copy of F (i.e., a copy of F all of whose edges have different colours). The systematic study of such problems was initiated by Keevash, Mubayi, Sudakov and Verstraëte. In this paper, we show that the rainbow Turán number of a path with k + 1 edges is less than (9k/7 + 2)n, improving an earlier estimate of Johnston, Palmer and Sarkar. Mathematics Subject Classifications: 05C35, 05C69


Introduction
Given a graph F , the maximum number of edges in a graph on n vertices that contains no copy of F is known as the Turán number of F , and is denoted by ex(n, F ).An edgecolored graph is called rainbow if all its edges have different colors.Given a graph F , the rainbow Turán number of F is defined as the maximum number of edges in a graph on n vertices that has a proper edge-coloring with no rainbow copy of F , and it is denoted by ex * (n, F ).
Keevash, Mubayi, Sudakov and Verstraëte also studied the rainbow Turán problem for even cycles.More precisely, they showed that ex * (n, C 2k ) = Ω(n 1+1/k ) using the construction of large B * k -sets of Bose and Chowla [2]-it is conjectured that the same lower bound holds for ex * (n, C 2k ) and is a well-known difficult open problem in extremal graph theory.They also proved a matching upper bound in the case of the six-cycle C 6 , so it known that ex * (n, C 6 ) = Θ(n 4/3 ) = ex(n, C 6 ).However, interestingly, they showed that ex * (n, C 6 ) is asymptotically larger than ex(n, C 6 ) by a multiplicative constant.Recently, Das, Lee and Sudakov [3] showed that ex * (n, C 2k ) = O(n 1+ (1+ k ) ln k k ), where k → 0 as k → ∞.
For an integer k, let P k denote a path of length k, where the length of a path is defined as the number of edges in it.Erdős and Gallai [4] proved that ex(n, P k+1 ) ≤ k 2 n; moreover, they showed that if k + 1 divides n, then the unique extremal graph is the vertex-disjoint union of n k+1 copies of K k+1 .On the other hand, Keevash, Mubayi, Sudakov and Verstraëte [6] showed that in some cases, the rainbow Turán number of P k can be strictly larger than the usual Turán number of P k : Maamoun and Meyniel [7] gave an example of a proper coloring of K 2 k containing no rainbow path with 2 k − 1 edges.By taking a vertex-disjoint union of such is not asymptotically equal to ex(n, P 2 k −1 ).They also mentioned that determining the asymptotic behavior of ex * (n, P k+1 ) is an interesting open problem, and stated the natural conjecture that the optimal construction is a disjoint union of cliques of size c(k), where c(k) is chosen as large as possible so that the cliques can be properly colored with no rainbow P k+1 .For P 4 , this conjecture was disproved by Johnston, Palmer and Sarkar [5]: Since any properly edge-colored K 5 contains a rainbow P 4 , and K 4 does not contain a P 4 , the conjecture for P 4 would be that ex * (n, P 4 ) ∼ 3n 2 .But they show that in fact, ex * (n, P 4 ) ∼ 2n by showing a proper edge-coloring of K 4,4 without no rainbow P 4 , and then taking n 8 vertex-disjoint copies of K 4,4 .For general k, they proved the following: Theorem 1 (Johnston, Palmer and Sarkar [5]).For any positive integer k, we have We improve the above bound by showing the following: Theorem 2. For any positive integer k, we have Let us remark that using the ideas introduced in this paper, it is conceivable that the upper bound can be further improved (at the cost of making the proof very involved).However, it would be very interesting (and seems to be difficult) to prove an upper bound less than kn or construct an example with kn edges.
We give a construction which shows that ex * (n, P 2 k ) > ex(n, P 2 k ) for any k ≥ 2.

Construction.
Let us first show a proper edge-coloring of K 2 k ,2 k (a complete bipartite graph with parts A and B, each of size 2 k ) with no rainbow P 2 k .The vertices of A and B are both identified with the vectors F k 2 .Each edge uv with u ∈ A and v ∈ B is assigned the color c(uv) := u − v. Clearly this gives a proper edge-coloring of K 2 k ,2 k .Moreover, if it contains a rainbow path v 0 v 1 . . .v 2 k then such a path must use all of the colors from F k 2 .Therefore But notice that since the length of the path v 0 v 1 . . .v 2 k is even, its terminal vertices v 0 and v 2 k are either both in A or they are both in B. So they could not have been identified with the same vector in F k 2 , a contradiction.Taking a vertex-disjoint union of such K 2 k ,2 k 's we obtain that ex * (n, This construction provides a counterexample to the above mentioned conjecture of Keevash, Mubayi, Sudakov and Verstraëte [6] whenever the largest clique that can be properly colored without a rainbow P 2 k has size 2 k .This is the case for k = 2, as noted before.The question of determining whether this is the case for any k ≥ 3 remains an interesting open question (see [1] for results in this direction).
Overview of the proof and organization.Let G be a graph which has a proper edge-coloring with no rainbow P k+1 .By induction on the length of the path, we assume there is a rainbow path v 0 v 1 . . .v k in G. Roughly speaking, we will show that the sum of degrees of the terminal vertices of the path, v 0 and v k is small.Our strategy is to find a set of distinct vertices whose size is as large as possible) such that for each 1 ≤ i ≤ m, there is a rainbow path P of length k with a i and b i as terminal vertices and V (P ) = {v 0 , v 1 , . . ., v k }; then we show that there are not many edges of G incident to the vertices of M , which will allow us to delete the vertices of M from G and apply induction.To this end, we define the set T ⊆ {v 0 , v 1 , . . ., v k } as the set of all vertices v ∈ {v 0 , v 1 , . . ., v k } where v is a terminal vertex of some rainbow path P with V (P ) = {v 0 , v 1 , . . ., v k }; we call T the set of terminal vertices.We will then find M as a subset of T ; moreover, it will turn out that if the size of T is large, then the size of M is also large -therefore, the heart of the proof lies in showing that T is large.
In Section 2.1, we introduce the notation and prove some basic claims.Using these claims, in Section 2.2, we will show that T is large (i.e., that there are many terminal vertices).Then in Section 2.3 we will find the desired subset M of T (which has few edges incident to it).

Proof of Theorem 2
Let G be a graph on n vertices, and suppose it has a proper edge-coloring c : E(G) → N without a rainbow path of length k + 1.Consider a longest rainbow path P * in G.We may suppose it is of length k, otherwise we are done by induction on k.For the base case k = 1, notice that any path of length 2, has to be a rainbow path.Thus G can contain at most n 2 < ( 9 7 + 2)n edges, so we are done.

Basic claims and Notation
In the rest of the paper, the degree of a vertex v ∈ V (G) be denoted by d(v).
Let L and R denote the sets of colors of edges incident to v 0 and v k respectively.(Notice that since the edges of G are colored properly, we have We define the following subsets of L, R and {c 1 , c 2 , . . ., c k } corresponding to P * . • Let L out (respectively R out ) be the set of colors of the edges incident to v 0 (respectively v k ) and to a vertex outside P * .
Note that L out ⊆ {c 1 , c 2 , . . ., c k } and R out ⊆ {c 1 , c 2 , . . ., c k }, otherwise we can extend P * to a rainbow path longer than k in G.
Notation.For convenience, we let |L| = l and |R| = r.Moreover, let Now we prove some inequalities connecting the quantities defined in Definition 3 for the path P * .
Similarly, by a symmetric argument, we have We will prove Theorem 2 by induction on the number of vertices n.For the base cases, note that for all n ≤ k, the number of edges is trivially at most so the statement of the theorem holds.If d(v) < 9k 7 + 2 for some vertex v of G, then we delete v from G to obtain a graph G on n − 1 vertices.By induction hypothesis, the number of edges in G is less than ( 9k 7 + 2)(n − 1).So the total number of edges in G is less than ( 9k 7 + 2)n, as desired.Therefore, from now on, we assume that for all v ∈ V (G), Since d(v 0 ) = l = l old + l new and l old ≤ k, we have that Similarly, Claim 6.We have On the other hand, by definition, By a symmetric argument, we get Adding the above two inequalities and rearranging, we get l + r − l nice − r nice ≤ 2k, so as required.
the electronic journal of combinatorics 26(1) (2019), #P1.17 such that v is a terminal vertex of some rainbow path P with For convenience, we will denote the size of T by t.
The next lemma yields a lower bound on the number of terminal vertices and is crucial to the proof of Theorem 2.
Lemma 8. We have The rest of this subsection is devoted to the proof of Lemma 8.

Proof of Lemma 8
Recall that P * = v 0 v 1 . . .v k and c(v j v j+1 ) = c j .First we prove a simple claim.
Claim 9. We may assume c(v Moreover, let j be an integer (with First let j ≥ i.In this case consider the path It is easy to see that the set of colors of the edges in this path is It is easy to see that the set of colors of the edges in this path is {c 1 , c 2 , . . ., c k } \ {c i+1 } ∪ {c(v j v k )}, so the path is rainbow again, with v i+1 as a terminal vertex.So v i+1 ∈ T .By symmetry, one can see that the same arguments used in the proofs of Claim 10 and Claim 11, imply the following two statements.
Notation.For any integers, 0 ≤ x ≤ y ≤ k, let and |T x,y | = t x,y .Notice that t = t 0,k = 2 + t 1,k−1 , as v 0 and v k are both terminal vertices.Now we will show that if a > b, then Lemma 8 holds.Suppose a > b.Then by the definition of a and b, we have By Claim 10, we know that whenever c(v 0 v i ) ∈ L new , we have v i−1 ∈ T .This shows that t 1,b−1 ≥ l new − 1.Similarly, by a symmetric argument (using Observation 12), we get t a+1,k−1 ≥ r new − 1.Therefore, Now using (1) and (2), we have proving Lemma 8. Therefore, from now on, we always assume a ≤ b.
Then by Claim 10, v i−1 ∈ T .We want to show that v i+1 ∈ T .
Observe that if i = a, then by Claim 10 again, we have v i+1 ∈ T because v k v i ∈ R new .So let us assume a < i and show that v i+1 ∈ T .Notice that there exists a * ∈ {a, a } (see Definition 13 for the definition of a and a ) such that c(v and it is easy to check that all the colors are different, so the path is rainbow with v i+1 as a terminal vertex.Now suppose c(v k v i ) ∈ R new .Then a similar argument (using Observation 12) shows that v i−1 ∈ T and v i+1 ∈ T again, completing the proof of the claim.

Notation. For any integers
new .Note that by definition of a and b, l new = l 0,a−1 new + l a,b new + 1 and r new = 1 + r a,b new + r b+1,k new .Using Claim 9, for any integer z, we have the following: Moreover, by definition of L new and R new , we have Informally speaking, Claim 11 and Claim 14 assert that each edge e = v 0 v i such that c(v In the next two claims, by double counting the total number of such pairs (e, x), we prove lower bounds on the number of terminal vertices in different ranges (i.e., t 0,a−1 , t b+1,k and t a,b ), in terms of l new , r new , l nice and r nice .
Claim 15.We have, and Proof of Claim.By Claim 11, and by the fact that there is only one j such that c(v k v j ) ∈ R 0,a−1 new , it is easy to see that for all but at most one i, we have the following new (equality here follows from (4)), then by Claim 10, Adding the previous two bounds, the total number of pairs (v new .Therefore, using that v 0 is also a terminal vertex, we have If c(v k v i ) ∈ R 0,a−1 nice , then by Observation 12, there is a vertex the electronic journal of combinatorics 26(1) (2019), #P1.17 and x ∈ T , is at least r 0,a−1 nice .By the pigeonhole principle, either the number of pairs In the first case, we get t 0,a−2 ≥ r 0,a−1 nice /2 and in the second case, we get t 1,a ≥ r 0,a−1 nice /2.As t 0,a−1 ≥ t 0,a−2 and t 0,a−1 ≥ t 1,a , in both cases we have, Therefore, adding up ( 5) and ( 6), we get Note that the equality follows from (3), ( 4) and the fact that r 0,a−1 nice = r 0,a nice because c(v k v a ) ∈ R new .By a symmetric argument, we have This finishes the proof of the claim.Now we prove a lower bound on t a,b .
Claim 16.Note that all the pairs (e, x) in S are such that x ∈ T a,b .Moreover, for each x ∈ T a,b , there are at most four pairs (e, x) in S. Therefore, we have Claim 17.The degree of every vertex u in H is at least 2k/7 + 2.
Proof of Claim.As u ∈ V (H) = T , u is a terminal vertex.So there is a rainbow path P = u 0 u 1 . . .u k in G such that u 0 = u and {u 0 , u 1 , . . ., u k } = {v 0 , v 1 , . . ., v k }.We define the sets L, R, L new , R new corresponding to P in the same way as we did for P * (in Definition 3).Moreover, since P * was defined as an arbitrary rainbow path of length k, (2) holds for P as well -i.e., |R new | = r new ≥ 2k/7 + 2. We claim that if u k u j is an edge in G such that c(u k u j ) ∈ R new , then uu j+1 ∈ E(H).Indeed, consider the path u 0 u 1 . . .u j u k u k−1 . . .u j+1 .This is clearly a rainbow path with terminal vertices u = u 0 and u j+1 .So u and u j+1 are adjacent in H, as required.This shows that degree of u in H is at least r new ≥ 2k/7 + 2, as desired.
Size of a matching is defined as the number of edges in it.The following proposition is folklore.
Proposition 18.Any graph G with minimum degree δ(G) has a matching of size ü´.
We know that δ(H) ≥ Let the edges of M be a 1 b 1 , a 2 b 2 , . . ., a m b m .Moreover, let the electronic journal of combinatorics 26(1) (2019), #P1.17 Proof of Claim.Note that the sum i n i counts each pair xy ∈ E(G) with x, y ∈ V (M ) exactly twice unless xy = a i b i for some i.Claim 20.The sum of degrees of a i and b i in G is at most 3k − n i 2 .Proof of Claim.Since a i b i is an edge in the auxiliary graph H, there is a rainbow path P = u 0 u 1 . . .u k in G such that u 0 = a i , u k = b i and {u 0 , u 1 , . . ., u k } = {v 0 , v 1 , . . ., v k }.We define the sets L, R, L in , R in , L out , R out , L new , R new and the numbers l, r, l in , r in , l out , r out , l new , r new corresponding to P in the same way as we did for P * (in Definition 3).Therefore, degree of a i is l ≤ l new + k.Similarly, degree of b i is at most r new + k.So the sum of degrees of a i and b i in G is at most 2k + l new + r new . (8) On the other hand, the sum of degrees of a i and b i in G is l + r = l in + l out + r in + r out .By Claim 5, this is at most (l in + r in ) + k − r new + k − l new = (l in + r in ) + 2k − l new − r new .Moreover, it is easy to see that l in + r in ≤ 2k − n i by the definition of n i .Therefore, the sum of degrees of a i and b i in G is at most the electronic journal of combinatorics 26(1) (2019), #P1.17

2. 2 Finding many terminal vertices Definition 7 (
Set of terminal vertices).Let T be the set of all vertices

t a,b ≥ 1 4 Ä l a+1,b− 1 nice+ r a+1,b− 1 nice+ 2 (. 1 nice∪ R a+1,b− 1 nice,
l a+1,b new + r a,b−1 new ) ä Proof of Claim.Let us construct a set S of pairs (e, x) such that e ∈ L in ∪ R in and x ∈ T with certain properties.For every edge e such that c(e) ∈ L a+1,b−Claim 11 (and Observation 12) ensures that there is a vertex x ∈ {v i−1 , v i+1 } such that x ∈ T (in particular, x ∈ T a,b ).Add all such pairs (e, x) to S. Therefore, the number of pairs (e, x) added to S so far, is l a+1,b−1 nice + r a+1,b−1 nice .For every edge e such that c(e) ∈ L a+1,b new ∪R a,b−1 new , we have both v i−1 , v i+1 ∈ T by Claim 14; we add both the pairs (e, v i−1 ) and (e, v i+1 ) to S. Therefore the number of pairs (e, x) added to S in this step is 2(l a+1,b new + r a,b−1 new ).Thus, |S| = l a+1,b−1 nice + r a+1,b−1 nice + 2(l a+1,b new + r a,b−1 new ).

Claim 19 .
The number of edges in the subgraph of G induced by M is

2 +
Therefore, the number of pairs xy ∈ E(G) in the subgraph of G induced by M is at most i n i 2 + m.Thus the number of edges of G in the subgraph induced by M is at least m), which implies the desired claim.
Adding up (8) and (9) and dividing by 2, we get that the sum of degrees of a i and b i in G is at most(2k + 2k − n i + 2k) i=1 (d(a i ) + d(b i )) countseach edge in the subgraph of G induced by M exactly twice (note that here d(v) denotes the degree of the vertex v in G).Therefore, the number of edges of G incident to the vertices of M is at most m i=1 (d(a i ) + d(b i )) − |E(G[M ])|.Now using Claim 19 and Claim 20, the number of edges of G incident to the vertices of M is at most the vertices of M from G to obtain a graph G on n − 2m vertices.By induction hypothesis, G contains less than ( 9k 7 + 2)(n − 2m) edges.Therefore, desired.This completes the proof of Theorem 2.
By the definition of a and b, l 0,b new = l new − 1 and r a,k new = r new − 1.So, we get 4t ≥ l nice + r nice + 2l new + 2r new + l 0,a nice + r b,k nice − 4 ≥ l nice + r nice + 2(l new + r new ) − 4.
Now we define an auxiliary graph H with the vertex set V (H) = T and edge set E(H) such that ab ∈ E(H) if and only if there is a rainbow path P in G with a and b as its terminal vertices and V 2k 7 + 2 by Claim 17. Moreover |V (H)| = |T | = t.So applying Proposition 18 to the graph H and using Lemma 8, we obtain that the graph H contains a matching M of size