Maximum Size of a Graph with Given Fractional Matching Number ∗

For three integers n, k, d , we determine the maximum size of a graph on n vertices with fractional matching number k and maximum degree at most d . As a consequence, we obtain the maximum size of a graph with given number of vertices and fractional matching number. This partially conﬁrms a conjecture proposed by Alon et al. on the maximum size of r -uniform hypergraph with a fractional matching number for the special case when r = 2.


Introduction
For an integer r 2, an r-uniform hypergraph or more simply, an r-graph, is a pair H = (V (H), E(H)) with vertex set V (H) and edge set E(H) ⊆ V (H) r . We call the number of the edges in H the size of H. A matching of H is a set of edges, no two of which are intersecting. The matching number of H, denoted by ν(H), is the size of a maximum matching in H. A matching M in H is perfect if every vertex of H is incident with an edge of M . An r-graph is called a graph if r = 2, denoted by G. We denote the maximum degree of the vertices of G by ∆(G). For a subset S of V (G), we use G[S] to denote the subgraph of G induced by S. For the terminologies and concepts not defined here, we refer the readers to [3,5,15].
Let g r (n, k) denote the maximum size of a r-graph H on n vertices with matching number k. In particular, we replace g 2 (n, k) by g(n, k). In 1959, Erdős and Gallai [8] determined g(n, k), that is, the maximum size of a graph G on n vertices with matching number k. Theorem 1. [8] For n 2k + 1, g(n, k) = max 2k + 1 2 , k(2n − k − 1) 2 .
Later in [7], Erdős further conjectured that, for n r(k + 1) − 1, This is an important conjecture in hypergraph theory and abundant literatures devote to it. For the recent progresses on Erdős matching conjecture, we refer to [10,11,12] for details.
In addition to matching number, some other restrictions were also considered in the literature. Let g(n, k, d) denote the maximum size of the graphs on n vertices with matching number k and the maximum degree at most d. In [6], Chvátal and Hanson determined g(n, k, d), and hence, generalized Theorem 1.
(ii) If d 2k and n 2k + k d 0 , then We refer to [2,4,13] for more details on this topic. Let us focus on the fractional version of Theorem 2 in the following. A fractional matching of an r-graph H is a function f assigning each edge with a real number r . It was shown that the fractional matching number ν f (G) of a graph G is either an integer or a semi-integer, that is, 2ν f (G) is integer (see [16], Theorem 2.1.5). Let F(n, k, d) denote the class of graphs on n vertices with fractional matching number k and the maximum degree at most d. Further, let f (n, k, d) denote the maximum size of the graphs in F(n, k, d), i.e., It is clear that f (n, k, d) = f (n, k, n − 1) when d n − 1, and hence we always assume d n − 1.
In this paper, we determine f (n, k, d) as follows.
Theorem 3. Let n, 2k, d be three positive integers with n 2k. If 2k is even, then otherwise.
Theorem 4. Let n, 2k be two positive integers with n 2k. If 2k is even, then If 2k is odd, then In  In the same article, they also showed that Conjecture 5 asymptotically holds for r ∈ {3, 4} and 0 s n r+1 when n tends to infinity. For r = 2, Theorem 1 implies that Conjecture 5 is asymptotically true when n goes to infinity. As far as we know, this conjecture still remains open even if r = 2 in general. In fact, by Theorem 4, we obtain the following corollary, which confirms that Conjecture 5 is true for r = 2.
Corollary 6. For any nonnegative integer s with n 2s, The rest of this article is organized as follows. In Section 2, we establish the necessary upper bounds on f (n, k, d). In Section 3, we construct the corresponding extremal graphs attaining these upper bounds and, hence, give a proof of Theorem 3.

Upper bounds
To obtain the upper bounds of f (n, k, d), we begin this section with the following known result, called fractional Tutte-Berge formula, which characterizes the fractional matching number of a graph.
Before the proof of upper bounds of f (n, k, d), we also need a result with regard to the maximum value of a function, which plays a vital role in our proof. Let n, 2k, d be three positive integers with n 2k and d n − 1. We now define a function on nonnegative real number x, and its maximum value in given intervals can be obtained.
Lemma 8. If 2k is even and 0 x k, then otherwise.
If 2k is odd and 0 x k − 3 2 , then the electronic journal of combinatorics 29(3) (2022), #P3.55 Lemma 8 can be proved trivially by the convexity of the function, and for the coherence and completeness we will prove it in Appendix. Now we present the upper bounds of f (n, k, d).
Lemma 9. Let n, 2k, d be three positive integers with n 2k. If 2k is even, then otherwise.
If 2k is odd, then Proof. Let G be a graph with n vertices satisfying ν f (G) = k and ∆(G) d. By Theorem Moreover, for the subgraph As a result, Since f (n, k, d) is an integer, it suffices to obtain the maximum integer which is no more than the maximum value of F (s) for any s ∈ D. Moreover, if 2k is odd, then the fact s ∈ D implies s k − 3 2 . By Lemma 8, we complete the proof.

Constructions of extremal graphs
In this section, we construct several extremal graphs satisfying conditions to attain f (n, k, d). We begin with some useful results. For a k-regular graph G, if we assign to each edge a number 1 k , then we obtain a fractional perfect matching of G. Therefore we have the following proposition.
Proposition 10. Let G be a k-regular graph. Then G has a fractional perfect matching.
We call that a sequence d 1 , d 2 , . . . , d n of nonnegative integers is graphic if it is a degree sequence of a simple graph G, and the graph G is said to realize the sequence d 1 , d 2 , . . . , d n .
The following characterization of graphic sequence is due to Erdős and Gallai.
A spanning subgraph of a graph G is said to k-factor if the degree of its each vertex is equal to k. The following result gives a sufficient condition with regard to degree sequences such that the graph contains a k-factor.

Theorem 12. [14]
If the sequences d 1 d 2 · · · d n and d 1 − k d 2 − k · · · d n − k are graphic, then the sequence d 1 , d 2 , . . . , d n can be realized by a graph G which contains a k-factor.
Let d and n be two odd numbers with 3 d n − 2. Theorem 11 ensures that two sequences d, d, . . . , d, d − 1 and d − 2, d − 2, . . . , d − 2, d − 3 of length n are both graphic. By Theorem 12, the sequence d, d, . . . , d − 1 can be realized by a graph G which contains a 2-factor. We know that if a graph G contains a 2-factor, then it has a fractional perfect matching. Therefore we have the following result, which is useful in our forthcoming argument.
Corollary 13. Let d and n be two odd numbers with 3 d n − 2. Then the sequence d, d, . . . , d, d − 1 of length n can be realized by a graph which has a fractional perfect matching.
Now we construct extremal graphs to attain f (n, k, d). Several notations are used in our constructions. We use K n and K n to denote a complete graph and an empty graph with n vertices, respectively. For two graphs G 1 and G 2 , we denote the disjoint union of G 1 and G 2 by G 1 ∪ G 2 . Let us first consider the case that 2k is even. Lemma 14. Let n, 2k, d be three positive integers with n 2k. If 2k is even, then otherwise.
Proof. Let us divide into two cases to complete the proof.
We can check that K 2k ∪ K n−2k ∈ F(n, k, d) and its size is 2k 2 . Thus f (n, k, d) Consider the parity of k(n − k + d). When k(n − k + d) is even, we take a (k − n + d)regular graph G 0 with k vertices. When k(n − k + d) is odd, we take a graph G 0 with k − 1 vertices of degree k − n + d and a vertex of degree k − n + d − 1. Add n − k independent vertices to G 0 such that they are adjacent to each vertex of G 0 , which contains a matching covering each vertex of G 0 between G 0 and these independent vertices, and denote the resulting graph by G. Clearly, ν f (G) = k and ∆(G) = d, and then G ∈ F(n, k, d). By the construction of G, we have Thus f (n, k, d) Case 2. d 2k − 1 or n d + k.
If d 2k −1 and n d+k, we take a d-regular graph G 0 with 2k vertices. Proposition 10 ensures that G 0 has a fractional perfect matching, and then ν f (G 0 ) = k. Clearly, G = G 0 ∪ K n−2k ∈ F(n, k, d) and |E(G)| = dk. Thus f (n, k, d) dk.
Next we give the lower bounds of f (n, k, d) when 2k is odd.
Lemma 15. Let n, 2k, d be three positive integers with n 2k. If 2k is odd, then Proof. The fact that 2k is odd implies d 2; otherwise d = 1, which implies fractional matching number k must be an integer, a contradiction. If k = 3 2 , then we have G = K 3 ∪ K n−3 , and hence f (n, 3 2 , d) = 3. So we may assume k 5 2 in the following discussion. Now let us discuss three cases.
Clearly, K 2k ∪ K n−2k ∈ F(n, k, d). Thus f (n, k, d) and h = n − k + 3 2 . Note that h − l 3 as n 2k. For i = 1, 2, . . . , l, joining the vertex x i with every vertex y j by an edge, where j = i + t (mod h) for t = 0, 1, . . . , d − 1, we call the resulting graph G 0 . Clearly, M = {x i y i ∈ E(G 0 ) : i = 1, 2, . . . , l} forms a matching covering X, and there exist at least three independent vertices y l+1 , y l+2 and y l+3 in Y which are not covered by M as h − l 3. Add three edges in {y l+1 , y l+2 , y l+3 } such that they form a K 3 to G 0 , and call the resulting graph G. Clearly, ν f (G) = l + 3 2 = k and ∆(G) = d, and hence G ∈ F(n, k, d).
Clearly, K 2k ∪ K n−2k ∈ F(n, k, d). Thus f (n, k, d) Consider the parity of (k − 3 2 ) is odd, we take a graph G 0 with k − 5 2 vertices of degree k − n + d − 3 2 and a vertex of degree k −n+d− 5 2 . Add n−k − 3 2 independent vertices to G 0 such that they are adjacent to each vertex of G 0 , which contains a matching covering each vertex of G 0 between G 0 and these independent vertices, and denote the resulting graph by G 1 . Therefore ν f (G 1 ) = k − 3 2 . Furthermore, add a K 3 formed by remainder three vertices to G 1 such that each vertex of K 3 are adjacent to each vertex of G 0 , and denote the resulting graph by G. Then we have ν f (G) = ν f (G 1 ) + 3 2 = k and ∆(G) = d. Then G ∈ F(n, k, d). By the construction of G, we have Thus f (n, k, d) When d is even, we can take a d-regular graph G 0 with 2k vertices. By Proposition 10, G 0 has a fractional perfect matching. When d is odd, Corollary 13 ensures that we can find a graph G 0 with 2k − 1 vertices of degree d and one vertex of degree d − 1 such that G 0 contains a fractional perfect matching. Therefore, ν f (G 0 ) = k. Clearly, G = G 0 ∪ K n−2k ∈ F(n, k, d) and |E(G)| = dk . Thus f (n, k, d) dk .

Appendix. Proof of Lemma 8
Let us recall that n 2k, d n − 1 and the function 2 < n − d always holds. We can rewrite the function . Note that F i (x) is convex for i = 1, 2, 3. To obtain the maximum value of F (x), it suffices to discuss maximum values of F i (x)'s in corresponding intervals. Case 1. 2k is even and 0 x k.
(i) If d 2k − 1 and n d + k, then k − d+1 2 0 < k n − d. Clearly, F (x) = F 2 (x) for any x with 0 x k. By the convexity of F 2 (x), we have (ii) If d 2k − 1 and n d + k, then k − d+1 2 0 < n − d k. Clearly, F (x) = F 2 (x) when 0 x n − d, and F (x) = F 3 (x) when n − d x k. By the convexity of F 3 (x), we have F 3 (n − d) max{F 3 (0), F 3 (k)}. Since F 2 (0) = F 3 (0), it follows that x k. Then we obtain and F (x) = F 3 (x) when n − d x k. By the convexity of F 2 (x), we have F 2 (n − d) Consequently, if 2k is even and 0 x k, then otherwise .