On Gupta's Co-density Conjecture

TL;DR

This paper investigates Gupta's co-density conjecture related to edge-colorings in multigraphs, providing partial proofs and linking it to other longstanding conjectures in graph theory.

Contribution

The authors prove bounds on the cover index based on co-density, confirming Gupta's conjecture in specific cases and connecting it to earlier conjectures.

Findings

Proved the conjecture when co-density is not an integer.

Established the conjecture with a one-unit relaxation when co-density is an integer.

Linked the co-density conjecture to Gupta's earlier cover index conjecture.

Abstract

Let $G=(V,E)$ be a multigraph. The {\em cover index} $\xi(G)$ of $G$ is the greatest integer $k$ for which there is a coloring of $E$ with $k$ colors such that each vertex of $G$ is incident with at least one edge of each color. Let $\delta(G)$ be the minimum degree of $G$ and let $\Phi(G)$ be the {\em co-density} of $G$, defined by \[\Phi(G)=\min \Big\{\frac{2|E^+(U)|}{|U|+1}:\,\, U \subseteq V, \,\, |U|\ge 3 \hskip 2mm {\rm and \hskip 2mm odd} \Big\},\] where $E^+(U)$ is the set of all edges of $G$ with at least one end in $U$. It is easy to see that $\xi(G) \le \min\{\delta(G), \lfloor \Phi(G) \rfloor\}$. In 1978 Gupta proposed the following co-density conjecture: Every multigraph $G$ satisfies $\xi(G)\ge \min\{\delta(G)-1, \, \lfloor \Phi(G) \rfloor\}$, which is the dual version of the Goldberg-Seymour conjecture on edge-colorings of multigraphs. In this note we prove that $\xi(G)\ge \min\{\delta(G)-1, \, \lfloor \Phi(G) \rfloor\}$ if $\Phi(G)$ is not integral and $\xi(G)\ge \min\{\delta(G)-2, \, \lfloor \Phi(G) \rfloor-1\}$ otherwise. We also show that this co-density conjecture implies another conjecture concerning cover index made by Gupta in 1967.

Equations

$$\Phi(G)=\min\Big{\{}\frac{2|E^{+}(U)|}{|U|+1}:\,\,U\subseteq V,\,\,|U|\geq 3\hskip 5.69054pt{\rm and\hskip 5.69054ptodd}\Big{\}},$$

$$\Gamma(G)=\max\Big{\{}\frac{2|E(U)|}{|U|-1}:\,\,U\subseteq V,\,\,|U|\geq 3\hskip 5.69054pt{\rm and\hskip 5.69054ptodd}\Big{\}},$$

$$\Phi(G)=\min\Big{\{}\frac{2|E^{+}(U)|}{|U|+1}:\,\,U\subseteq V,\,\,|U|\geq 3\hskip 5.69054pt{\rm and\hskip 5.69054ptodd}\Big{\}}.$$

$${\Omega }_{\_}p=\{2(p+1)\mu (G)-2p,2(p+1)\mu (G)-2p+1,\dots ,2p\mu (G)\},$$

Full text

[TABLE]

where $E^{+}(U)$ is the set of all edges of $G$ with at least one end in $U$. It is easy to see that $\xi(G)\leq\min\{\delta(G),\lfloor\Phi(G)\rfloor\}$. In 1978 Gupta proposed the following co-density conjecture: Every multigraph $G$ satisfies $\xi(G)\geq\min\{\delta(G)-1,\,\lfloor\Phi(G)\rfloor\}$, which is the dual version of the Goldberg-Seymour conjecture on edge-colorings of multigraphs. In this note we prove that $\xi(G)\geq\min\{\delta(G)-1,\,\lfloor\Phi(G)\rfloor\}$ if $\Phi(G)$ is not integral and $\xi(G)\geq\min\{\delta(G)-2,\,\lfloor\Phi(G)\rfloor-1\}$ otherwise. We also show that this co-density conjecture implies another conjecture concerning cover index made by Gupta in 1967.

1 Introduction

In this note we consider multigraphs, which may have parallel edges but contain no loops. Let $G=(V,E)$ be a multigraph. The chromatic index $\chi^{\prime}(G)$ of $G$ is the least integer $k$ for which there is a coloring of $E$ with $k$ colors such that each vertex of $G$ is incident with at most one edge of each color. Let $\Delta(G)$ be the maximum degree of $G$ and let $\Gamma(G)$ be the density of $G$, defined by

[TABLE]

where $E(U)$ is the set of all edges of $G$ with both ends in $U$. Clearly, $\chi^{\prime}(G)\geq\max\{\Delta(G),\,\Gamma(G)\}$; this lower bound, as shown by Seymour [10] using Edmonds’ matching polytope theorem [2], is precisely the fractional chromatic index of $G$, which is the optimal value of the fractional edge-coloring problem:

[TABLE]

where $A$ is the edge$-$matching incidence matrix of $G$. In the 1970s Goldberg [3] and Seymour [10] independently made the following conjecture.

Conjecture 1.1**.**

Every multigraph $G$ satisfies $\chi^{\prime}(G)\leq\max\{\Delta(G)+1,\,\lceil\Gamma(G)\rceil\}$.

Over the past four decades this conjecture has been a subject of extensive research, and has stimulated an important body of work, with contributions from many researchers; see McDonald [7] for a survey on this conjecture and Stiebitz et al. [11] for a comprehensive account of edge-colorings. Recently, three of the authors, Chen, Jing, and Zang, have announced a complete proof of Conjecture 1.1 [1].

The present note is devoted to the study of the dual version of the classical edge-coloring problem (ECP), which asks for a coloring of the edges of $G$ using the maximum number of colors in such a way that at each vertex all colors occur. It is easy to see that each color class induces an edge cover of $G$. (Recall that an edge cover is a subset $F$ of $E$ such that each vertex of $G$ is incident to at least one edge in $F$.) So this problem is actually the edge cover packing problem (ECPP). Let $\xi(G)$ denote the optimal value of ECPP, which we call the cover index of $G$. As it is $NP$-hard [6] in general to determine the chromatic index $\chi^{\prime}(G)$ of a simple cubic graph $G$, determining the cover index $\xi(G)$ is also $NP$-hard.

Let $\delta(G)$ be the minimum degree of $G$, let $E^{+}(U)$ be the set of all edges of $G$ with at least one end in $U$ for each $U\subseteq V$, and let $\Phi(G)$ be the co-density of $G$, defined by

[TABLE]

Obviously, $\xi(G)\leq\delta(G)$. Since each edge cover contains at least $(|U|+1)/2$ edges in $E^{+}(U)$ for any $U\subseteq V$ with $|U|\geq 3$ and odd, $\Phi(G)$ provides another upper bound for $\xi(G)$. So $\xi(G)\leq\min\{\delta(G),\Phi(G)\}$. Based on a polyhedral description of edge covers (see Theorem 27.3 in Schrijver [9]), Zhao, Chen, and Sang [12] observed that the parameter $\min\{\delta(G),\,\Phi(G)\}$ is exactly the fractional cover index of $G$, the optimal value of the fractional edge cover packing problem (FECPP):

[TABLE]

where $B$ is the edge$-$edge cover incidence matrix of $G$. They [12] also devised a combinatorial polynomial-time algorithm for finding the co-density $\Phi(G)$ of any multigraph $G$.

In 1978 Gupta [5] proposed the following co-density conjecture, which is the counterpart of Conjecture 1.1 on ECPP.

Conjecture 1.2**.**

Every multigraph $G$ satisfies $\xi(G)\geq\min\{\delta(G)-1,\,\lfloor\Phi(G)\rfloor\}$.

The reader is referred to Stiebitz et al. [11] for more information about this conjecture. Its validity would imply that, first, there are only two possible values for the cover index $\xi(G)$ of a multigraph $G$: $\min\{\delta(G)-1,\,\lfloor\Phi(G)\rfloor\}$ and $\min\{\delta(G),\,\lfloor\Phi(G)\rfloor\}$; second, any multigraph has a cover index within one of its fractional cover index, so FECPP also has a fascinating integer rounding property (see Schrijver [8, 9]); third, even if $P\neq NP$, the $NP$-hardness of ECPP does not preclude the possibility of designing an efficient algorithm for finding at least $\min\{\delta(G)-1,\,\lfloor\Phi(G)\rfloor\}$ disjoint edge covers in any multigraph $G$.

To our knowledge, the bound $\xi(G)\geq\min\{\lfloor\frac{7\delta(G)+1}{8}\rfloor,\lfloor\Phi(G)\rfloor\}$ established by Gupta [5] in 1978 remains to be the best approximate version of Conjecture 1.2.

As is well known, the inequality $\chi^{\prime}(G)\leq\Delta(G)+\mu(G)$ holds for any multigraph $G$, where $\mu(G)$ is the maximum multiplicity of an edge in $G$. This result has been successfully dualized by Gupta [4] to packing edge covers: $\xi(G)\geq\delta(G)-\mu(G)$. It is worthwhile pointing out that this dual version follows from Conjecture 1.2 as a corollary, because $\Phi(G)\geq\delta(G)-\mu(G)$. To see this, let $U$ be a subset of $V$ with $|U|\geq 3$ and odd, let $F(U)$ be the set of all edges of $G$ with precisely one end in $U$, and let $G[U]$ be the subgraph of $G$ induced by $U$. Since each vertex in $U$ is adjacent to at most $(|U|-1)\mu(G)$ edges in $G[U]$ and at most $|F(U)|$ edges outside $G[U]$, we have $\delta(G)\leq(|U|-1)\mu(G)+|F(U)|$, which implies that $\delta(G)|U|+|F(U)|\geq(\delta(G)-\mu(G))(|U|+1)$. As $2|E^{+}(U)|=2|E(U)|+2|F(U)|\geq\delta(G)|U|+|F(U)|$, we obtain $2|E^{+}(U)|\geq(\delta(G)-\mu(G))(|U|+1)$ and hence $\Phi(G)\geq\delta(G)-\mu(G)$, as desired.

Gupta [4] demonstrated that the lower bound $\delta(G)-\mu(G)$ for $\xi(G)$ is sharp when $\mu(G)\geq 1$ and $\delta(G)=2p\mu(G)-q$, where $p$ and $q$ are two integers satisfying $q\geq 0$ and $p>\mu(G)+\lfloor(q-1)/2\rfloor$. This led Gupta [4] to suggest the following conjecture, which aims to give a complete characterization of all values of $\delta(G)$ and $\mu(G)$ for which no multigraph $G$ with $\xi(G)=\delta(G)-\mu(G)$ exists.

Conjecture 1.3**.**

Let $G$ be a multigraph such that $\delta(G)$ cannot be expressed in the form $2p\mu(G)-q$, for any two integers $p$ and $q$ satisfying $q\geq 0$ and $p>\mu(G)+\lfloor(q-1)/2\rfloor$. Then $\xi(G)\geq\delta(G)-\mu(G)+1$.

As edge covers are more difficult to manipulate than matchings, it is no surprise that a direct proof of conjecture 1.2 would be more complicated and sophisticated than that of Conjecture 1.1 (see [1], which is under review). One purpose of this note is to establish a slightly weaker version of conjecture 1.2 by using Conjecture 1.1.

Theorem 1.1**.**

(Assuming Conjecture 1.1) Let $G$ be a multigraph. Then $\xi(G)\geq\min\{\delta(G)-1,\,\lfloor\Phi(G)\rfloor\}$ if $\Phi(G)$ is not integral and $\xi(G)\geq\min\{\delta(G)-2,\,\lfloor\Phi(G)\rfloor-1\}$ otherwise.

Remark. Suppose $\Phi(G)<\delta(G)$. By this theorem, we obtain $\xi(G)=\lfloor\Phi(G)\rfloor$ if $\Phi(G)$ is not integral and $\Phi(G)-1\leq\xi(G)\leq\Phi(G)$ otherwise, because $\xi(G)\leq\min\{\delta(G),\,\lfloor\Phi(G)\rfloor\}$.

In this note we also show that Conjecture 1.3 is contained in Conjecture 1.2 as a special case.

Theorem 1.2**.**

Conjecture 1.2 implies Conjecture 1.3.

Throughout this note we shall repeatedly use the following terminology and notations. Let $G=(V,E)$ be a multigraph. A subset $U$ of $V$ is called an odd set if $|U|$ is odd and $|U|\geq 3$. For each $v\in V$, let $d_{\_}G(v)$ be the degree of $v$ in $G$. For each $U\subseteq V$, let $E_{\_}G(U)$ be the set of all edges of $G$ with both ends in $U$, let $E_{\_}G^{+}(U)$ be the set of all edges of $G$ with at least one end in $U$, and let $F_{\_}G(U)$ be the set of all edges of $G$ with exactly one end in $U$. For any two subsets $X$ and $Y$ of $V$, let $E_{\_}G(X,Y)$ be the set of all edges of $G$ with one end in $X$ and the other end in $Y$. We write $E_{\_}G(x,y)$ for $E_{\_}G(X,Y)$ if $X=\{x\}$ and $Y=\{y\}$. We shall drop the subscript $G$ if there is no danger of confusion.

The proofs of the above two theorems will take up the entire remainder of this note.

2 Approximate Version

We present a proof of Theorem 1.1 in this section. Let $G=(V,E)$ be a multigraph and let $Z\subseteq V$. A set $C\subseteq E$ is called a $Z$-cover if every vertex of $Z$ is incident with at least one edge of $C$. Note that if $Z=V$, then $Z$-covers are precisely edge covers of $G$. Let $e\in E(x,y)$ and let $G^{\prime}$ be obtained from $G$ by adding a new vertex $x^{\prime}$ and making $e$ incident with $x^{\prime}$ instead of $x$ (yet still incident with $y$); we say that $G^{\prime}$ arises from $G$ by splitting off $e$ from $x$. To prove the theorem, we shall actually establish the following variant.

Theorem 2.1**.**

Let $G=(V,E)$ be a multigraph, let $Z\subseteq V$, let $k$ be a positive integer, and let $\epsilon$ be [math] or $1$. If $d(z)\geq k+1$ for all $z\in Z$ and $|E^{+}(U)|\geq\frac{|U|+1}{2}k+\epsilon$ for all odd sets $U\subseteq Z$, then $G$ contains $k-1+\epsilon$ disjoint $Z$-covers.

Proof. Splitting off edges from vertices outside $Z$ if necessary, we may assume that all vertices outside $Z$ have degree one. Suppose for a contradiction that Theorem 2.1 is false. We reserve the triple $(G,Z,k)$ for a counterexample with the minimum $\sum_{\_}{z\in Z}d(z)$. For convenience, we call an odd set $U\subseteq Z$ optimal if $|E^{+}(U)|=\frac{|U|+1}{2}k+\epsilon$.

By hypothesis, $d(z)\geq k+1$ for all $z\in Z$, which can be strengthened as follows.

Claim. $d(z)=k+1$ for all $z\in Z$.

Otherwise, $d(z)\geq k+2$ for some $z\in Z$. If $z$ is contained in no optimal odd set $U\subseteq Z$, letting $H$ be obtained from $G$ by splitting off an edge from $z$, then $(H,Z,k)$ would be a smaller counterexample than $(G,Z,k)$, a contradiction. Hence

(1) there exists an optimal odd set $U_{\_}1\subseteq Z$ containing $z$; subject to this, we assume that $|U_{\_}1|$ is minimum.

Since $(|U_{\_}1|+1)k+2\epsilon=2|E^{+}(U_{\_}1)|=2|E(U_{\_}1)|+2|F(U_{\_}1)|\geq(k+1)|U_{\_}1|+|F(U_{\_}1)|$, we have $|F(U_{\_}1)|\leq k-|U_{\_}1|+2\epsilon\leq k<d(z)$. So $z$ is adjacent to some vertex $y\in U_{\_}1$. Let $H$ be arising from $G$ by splitting off one edge $e\in E(y,z)$ from $z$. We propose to show that

(2) $(H,Z,k)$ is a smaller counterexample than $(G,Z,k)$.

Assume the contrary. Then $|E^{+}_{\_}H(U_{\_}2)|<\frac{|U_{\_}2|+1}{2}k+\epsilon$ for some odd set $U_{\_}2\subseteq Z$ by the hypothesis of this theorem. Thus

(3) $z\in U_{\_}2$, $y\notin U_{\_}2$, and $|E^{+}(U_{\_}2)|=\frac{|U_{\_}2|+1}{2}k+\epsilon$.

Let $T_{\_}1=U_{\_}1\backslash U_{\_}2$ and $T_{\_}2=U_{\_}2\backslash U_{\_}1$. By (3), we have $y\in U_{\_}1\backslash U_{\_}2$, so $T_{\_}1\neq\emptyset$. By the minimality assumption on $|U_{\_}1|$ (see (1)), $U_{\_}2$ is not a proper subset of $U_{\_}1$, which implies $T_{\_}2\neq\emptyset$. Since $z\in U_{\_}1\cap U_{\_}2$, we obtain $|U_{\_}1\cap U_{\_}2|\geq 1$. Let us consider two cases, according to the parity of $|U_{\_}1\cap U_{\_}2|$.

Case 1. $|U_{\_}1\cap U_{\_}2|$ is odd.

It is a routine matter to check that

(4) $|E^{+}(U_{\_}1\cup U_{\_}2)|+|E^{+}(U_{\_}1\cap U_{\_}2)|=|E^{+}(U_{\_}1)|+|E^{+}(U_{\_}2)|-|E(T_{\_}1,T_{\_}2)|$.

In this case, $U_{\_}1\cup U_{\_}2$ is an odd set. So $|E^{+}(U_{\_}1\cup U_{\_}2)|\geq\frac{|U_{\_}1\cup U_{\_}2|+1}{2}k+\epsilon$ by the hypothesis of this theorem.

(5) $|E^{+}(U_{\_}1\cap U_{\_}2)|\geq\frac{|U_{\_}1\cap U_{\_}2|+1}{2}k+\epsilon+1$.

To justify this, note that if $|U_{\_}1\cap U_{\_}2|=1$, then $|E^{+}(U_{\_}1\cap U_{\_}2)|=d(z)\geq k+2$. So (5) holds. If $|U_{\_}1\cap U_{\_}2|\geq 3$, then $U_{\_}1\cap U_{\_}2$ is not an optimal odd set by the minimality assumption on $|U_{\_}1|$ (see (1)). Thus (5) is also true.

From (4) and (5) we deduce that $\frac{|U_{\_}1\cup U_{\_}2|+1}{2}k+\epsilon\leq|E^{+}(U_{\_}1\cup U_{\_}2)|\leq|E^{+}(U_{\_}1)|+|E^{+}(U_{\_}2)|-|E^{+}(U_{\_}1\cap U_{\_}2)|\leq\frac{|U_{\_}1|+1}{2}k+\epsilon+\frac{|U_{\_}2|+1}{2}k+\epsilon-\frac{|U_{\_}1\cap U_{\_}2|+1}{2}k-\epsilon-1=\frac{|U_{\_}1\cup U_{\_}2|+1}{2}k+\epsilon-1$, a contradiction.

Case 2. $|U_{\_}1\cap U_{\_}2|$ is even.

It is easy to see that $|E^{+}(U_{\_}1)|+|E^{+}(U_{\_}2)|=|E^{+}(T_{\_}1)|+|E^{+}(T_{\_}2)|+2|E(U_{\_}1\cap U_{\_}2)|+|E(U_{\_}1\cap U_{\_}2,T_{\_}1\cup T_{\_}2)|+2|E(U_{\_}1\cap U_{\_}2,\overline{U_{\_}1\cup U_{\_}2})|$, where $\overline{U_{\_}1\cup U_{\_}2}=V-(U_{\_}1\cup U_{\_}2)$. Thus

(6) $|E^{+}(U_{\_}1)|+|E^{+}(U_{\_}2)|\geq|E^{+}(T_{\_}1)|+|E^{+}(T_{\_}2)|+2|E(U_{\_}1\cap U_{\_}2)|+|F(U_{\_}1\cap U_{\_}2)|$.

In this case, $|T_{\_}i|$ is odd, so $|E^{+}(T_{\_}i)|\geq\frac{|T_{\_}i|+1}{2}k+\epsilon$ for $i=1,2$ by the hypothesis of this theorem. It follows from (3) and (6) that $\frac{|U_{\_}1|+1}{2}k+\epsilon+\frac{|U_{\_}2|+1}{2}k+\epsilon\geq\frac{|T_{\_}1|+1}{2}k+\epsilon+\frac{|T_{\_}2|+1}{2}k+\epsilon+2|E(U_{\_}1\cap U_{\_}2)|+|F(U_{\_}1\cap U_{\_}2)|\geq\frac{|T_{\_}1|+1}{2}k+\epsilon+\frac{|T_{\_}2|+1}{2}k+\epsilon+|U_{\_}1\cap U_{\_}2|(k+1)=\frac{|U_{\_}1|+1}{2}k+\epsilon+\frac{|U_{\_}2|+1}{2}k+\epsilon+|U_{\_}1\cap U_{\_}2|$, a contradiction.

Combining the above two cases, we obtain (2). This contradiction justifies the claim.

For each odd set $U\subseteq Z$, by the above claim, we obtain $|U|(k+1)=2|E(U)|+|F(U)|=|E(U)|+|E^{+}(U)|\geq|E(U)|+\frac{|U|+1}{2}k+\epsilon$. Thus $|E(U)|\leq\frac{|U|-1}{2}(k+2)+1-\epsilon$. Hence $\frac{2|E(U)|}{|U|-1}\leq k+3$ if $\epsilon=0$ and $\frac{2|E(U)|}{|U|-1}\leq k+2$ if $\epsilon=1$. By Conjecture 1.1, the chromatic index of $G[Z]$ is at most $k+3-\epsilon$. Since all vertices outside $Z$ have degree one, we further obtain $\chi^{\prime}(G)\leq k+3-\epsilon$. So $E$ can be partitioned into $k+3-\epsilon$ matchings $M_{\_}1,M_{\_}2,\ldots,M_{\_}{k+3-\epsilon}$.

Let us first consider the case when $\epsilon=0$. By the above claim,

(7) each vertex $z\in Z$ is disjoint from precisely two of $M_{\_}1,M_{\_}2,\ldots,M_{\_}{k+3}$ (as $d(z)=k+1$).

Let $H$ be the subgraph of $G$ induced by edges in $M_{\_}k\uplus M_{\_}{k+1}\uplus M_{\_}{k+2}\uplus M_{\_}{k+3}$, where $\uplus$ is the multiset sum, and let $N$ be an orientation of $H$ such that $|d_{\_}N^{+}(v)-d_{\_}N^{-}(v)|\leq 1$ for each vertex $v$. (It is well known that every multigraph admits such an orientation.) From (7) and this orientation we see that

(8) if a vertex $z\in Z$ is disjoint from precisely one of $M_{\_}1,M_{\_}2,\ldots,M_{\_}{k-1}$, then $d_{\_}H(z)=3$ and $d_{\_}N^{-}(z)\geq 1$; if $z$ is disjoint from precisely two of $M_{\_}1,M_{\_}2,\ldots,M_{\_}{k-1}$, then $d_{\_}H(z)=4$ and $d_{\_}N^{-}(z)=2$.

For each $i=1,2,...,k-1$, let $C_{\_}i$ be obtained from $M_{\_}i$ as follows: for each $z\in Z$, if $z$ not covered by $M_{\_}i$, add an edge from $N$ that is directed to $z$ and has not yet been used in $C_{\_}1\uplus C_{\_}2\uplus\ldots\uplus C_{\_}{i-1}$, where $C_{\_}0=\emptyset$. From this construction and (8) we deduce that $C_{\_}1,C_{\_}2,...,C_{\_}{k-1}$ are pairwise disjoint and each of them is a $Z$-cover in $G$.

It remains to consider the case when $\epsilon=1$. Now

(9) each vertex $z\in Z$ is disjoint from precisely one of $M_{\_}1,M_{\_}2,\ldots,M_{\_}{k+2}$.

Let $H$ be the subgraph of $G$ induced by edges in $M_{\_}{k+1}\uplus M_{\_}{k+2}$, and let $N$ be an orientation of $H$ such that $|d_{\_}N^{+}(v)-d_{\_}N^{-}(v)|\leq 1$ for each vertex $v$. From (9) and this orientation we see that

(10) if a vertex $z\in Z$ is disjoint from precisely one of $M_{\_}1,M_{\_}2,\ldots,M_{\_}{k}$, then $d_{\_}H(z)=2$ and $d_{\_}N^{-}(z)=1$.

For each $i=1,2,...,k$, let $C_{\_}i$ be obtained from $M_{\_}i$ as follows: for each $z\in Z$, if $z$ not covered by $M_{\_}i$, add an edge from $N$ that is directed to $z$. From this construction and (10) we deduce that $C_{\_}1,C_{\_}2,...,C_{\_}{k}$ are pairwise disjoint and each of them is a $Z$-cover in $G$.

3 Implication

The purpose of this section is to show that Conjecture 1.3 can be deduced from Conjecture 1.2.

Proof of Theorem 1.2. We may assume that

(1) $G$ is connected.

To see this, let $G_{\_}1,G_{\_}2,\ldots,G_{\_}k$ be all the components of $G$. For each $i=1,2,\ldots,k$, we aim to establish the inequality $\xi(G_{\_}i)>\delta(G)-\mu(G)$. If $\delta(G_{\_}i)-\mu(G_{\_}i)>\delta(G)-\mu(G)$, then the desired inequality holds, because $\xi(G_{\_}i)\geq\delta(G_{\_}i)-\mu(G_{\_}i)$. So we assume that $\delta(G_{\_}i)-\mu(G_{\_}i)\leq\delta(G)-\mu(G)$. Since $\delta(G_{\_}i)\geq\delta(G)$ and $\mu(G_{\_}i)\leq\mu(G)$, from this assumption we deduce that $\delta(G_{\_}i)=\delta(G)$ and $\mu(G_{\_}i)=\mu(G)$. Thus $G_{\_}i$ satisfies the hypothesis of Conjecture 1.3. Hence we may assume that $G$ is connected, otherwise we consider its components separately.

By hypothesis, $\delta(G)$ cannot be expressed in the form $2p\mu(G)-q$, for any two integers $p$ and $q$ satisfying $q\geq 0$ and $p>\mu(G)+\lfloor(q-1)/2\rfloor$; these two inequalities are equivalent to $0\leq q\leq 2p-2\mu(G)$. Setting $q=0,1,\ldots,2p-2\mu(G)$ respectively, we see that $\delta(G)$ does not belong to the set

[TABLE]

where $p\geq\mu(G)$. Note that $2\mu(G)^{2}$ is the only member of $\Omega_{\_}{\mu(G)}$ and that the gap between $\Omega_{\_}p$ and $\Omega_{\_}{p+1}$ consists of all integers $i$ with $2p\mu(G)+1\leq i\leq 2(p+2)\mu(G)-(2p+3)$. So

(2) either $\delta(G)\leq 2\mu(G)^{2}-1$ or $2p\mu(G)+1\leq\delta(G)\leq 2(p+2)\mu(G)-(2p+3)$ for some $p\geq\mu(G)$.

We may assume that $\delta(G)\geq 1$, for otherwise, $\delta(G)=2p\mu(G)-q$ for $p=q=0$, contradicting the hypothesis of Conjecture 1.3. Thus $\mu(G)\geq 1$.

To prove the theorem, it suffices to show that for any odd set $U$ of $G$, we have $\frac{2|E^{+}(U)|}{|U|+1}\geq\delta(G)-\mu(G)+1$, or equivalently,

(3) $2|E(U)|+2|F(U)|\geq(|U|+1)(\delta(G)-\mu(G)+1)$.

Set $k=\mu(G)$ if $\delta(G)\leq 2\mu(G)^{2}-1$ and set $k=p+1$ if $2p\mu(G)+1\leq\delta(G)\leq 2(p+2)\mu(G)-(2p+3)$ for some $p\geq\mu(G)$. We consider two cases according to the size of $U$.

Case 1. $|U|\geq 2k+1$.

We divide the present case into two subcases.

Subcase 1.1. Either $U\subsetneq V$ or $U=V$ and $\delta(G)$ is odd. In this subcase,

(4) $2|E(U)|+2|F(U)|\geq|U|\delta(G)+1$.

Indeed, if $U\subsetneq V$, then $|F(U)|\geq 1$ by (1). If $U=V$ and $\delta(G)$ is odd, then $G$ contains at least one vertex of degree at least $\delta(G)+1$, because $|V|=|U|$ is odd and the total number of vertices with odd degree is even. Hence (4) is true.

(5) $|U|\delta(G)+1\geq(|U|+1)(\delta(G)-\mu(G)+1)$.

Note that (5) amounts to saying that $\delta(G)\leq(|U|+1)(\mu(G)-1)+1$. If $\delta(G)\leq 2\mu(G)^{2}-1$, then $\delta(G)\leq(2\mu(G)+2)(\mu(G)-1)+1=(2k+2)(\mu(G)-1)+1\leq(|U|+1)(\mu(G)-1)+1$. If $\delta(G)\leq 2(p+2)\mu(G)-(2p+3)$, then $\delta(G)\leq 2(k+1)\mu(G)-(2k+1)=(2k+2)(\mu(G)-1)+1\leq(|U|+1)(\mu(G)-1)+1$. So (5) is established.

The desired statement (3) follows instantly from (4) and (5).

Subcase 1.2. $U=V$ and $\delta(G)$ is even. In this subcase, we have $\delta(G)\leq 2\mu(G)^{2}-2$ if $\delta(G)\leq 2\mu(G)^{2}-1$ and $\delta(G)\leq 2(p+2)\mu(G)-(2p+4)$ if $\delta(G)\leq 2(p+2)\mu(G)-(2p+3)$. So $\delta(G)\leq(2k+2)(\mu(G)-1)$ by the definition of $k$ and hence

(6) $\delta(G)\leq(|U|+1)(\mu(G)-1)$.

From (6) we deduce that $|U|\delta(G)\geq(|U|+1)(\delta(G)-\mu(G)+1)$. Therefore (3) holds, because $2|E(U)|+2|F(U)|\geq|U|\delta(G)$.

Case 2. $|U|\leq 2k-1$. (So $k\geq 2$ as $|U|\geq 3$.)

By the Pigeonhole Principle, some vertex $v\in U$ is incident with at most $\frac{|F(U)|}{|U|}$ edges in $F(U)$. Note that $v$ is incident with at most $(|U|-1)\mu(G)$ edges in $G[U]$, so $d(v)\leq(|U|-1)\mu(G)+\frac{|F(U)|}{|U|}$. Hence

(7) $\delta(G)\leq(|U|-1)\mu(G)+\frac{|F(U)|}{|U|}$.

We proceed by considering two subcases.

Subcase 2.1. $2p\mu(G)+1\leq\delta(G)\leq 2(p+2)\mu(G)-(2p+3)$, where $p\geq\mu(G)$.

From (7) and the hypothesis of the present subcase, we deduce that $2p\mu(G)+1\leq(|U|-1)\mu(G)+\frac{|F(U)|}{|U|}$. Thus $|F(U)|\geq|U|(2p+1-|U|)\mu(G)+|U|$. So

(8) $|U|\delta(G)+|F(U)|\geq|U|\delta(G)+|U|(2p+1-|U|)\mu(G)+|U|$.

Let us show that

(9) $|U|\delta(G)+|U|(2p+1-|U|)\mu(G)+|U|\geq(|U|+1)(\delta(G)-\mu(G)+1)$.

To justify this, note that (9) is equivalent to

(10) $\delta(G)\leq\{|U|(2p+2-|U|)+1\}\mu(G)-1$.

By the hypothesis of the present subcase, $\delta(G)\leq 2(p+2)\mu(G)-(2p+3)$. To establish (10), we turn to proving that $2(p+2)\mu(G)-(2p+3)\leq\{|U|(2p+2-|U|)+1\}\mu(G)-1$, or equivalently

(11) $\{-|U|^{2}+2(p+1)|U|-(2p+3)\}\mu(G)\geq-(2p+2)$.

Let $f(x)=-x^{2}+2(p+1)x-(2p+3)$. Then $f(x)$ is a concave function on $\mathbb{R}$. So on any interval $[a,b]$, $f(x)$ achieves the minimum at $a$ or $b$. By the hypothesis of the present case, $|U|\leq 2k-1=2p+1$, so $3\leq|U|\leq 2p+1$. By direct computation, we obtain $f(3)=4p-6\geq-2$ and $f(2p+1)=-2$. Thus $f(|U|)\geq-2$ for $3\leq|U|\leq 2p+1$, which implies that the LHS of (11) $\geq-2\mu(G)\geq-(2p+2)=$ RHS of (11), because $p\geq\mu(G)$. This proves (11) and hence (10) and (9).

Since $2|E(U)|+2|F(U)|\geq|U|\delta(G)+|F(U)|$, the desired statement (3) follows instantly from (8) and (9).

Subcase 2.2. $\delta(G)\leq 2\mu(G)^{2}-1$.

We may assume that

(12) $\delta(G)\geq(|U|+1)(\mu(G)-1)+1$, for otherwise, $|U|\delta(G)\geq(|U|+1)(\delta(G)-\mu(G)+1)$. So (3) holds, because $2|E(U)|+2|F(U)|\geq|U|\delta(G)$.

By (12) and the hypothesis of the present subcase, either $2t(\mu(G)-1)+1\leq\delta(G)\leq 2(t+1)(\mu(G)-1)$ for some $t$ with $\frac{|U|+1}{2}\leq t\leq\mu(G)$ or $\delta(G)=2t(\mu(G)-1)+1$ for $t=\mu(G)+1$.

By (7), we have $2t(\mu(G)-1)+1\leq(|U|-1)\mu(G)+\frac{|F(U)|}{|U|}$. So $\frac{|F(U)|}{|U|}\geq(2t-|U|+1)\mu(G)-2t+1$, and hence

(13) $|U|\delta(G)+|F(U)|\geq|U|\{\delta(G)+(2t-|U|+1)\mu(G)-2t+1\}$.

We propose to show that

(14) $|U|\{\delta(G)+(2t-|U|+1)\mu(G)-2t+1\}\geq(|U|+1)(\delta(G)-\mu(G)+1)$.

To justify this, note that (14) is equivalent to

(15) $\delta(G)\leq\{|U|(2t+2-|U|)+1\}\mu(G)-|U|2t-1$.

Suppose $\delta(G)=2\mu(G)^{2}-1$. Then $t=\mu(G)+1$. So (15) says that $2\mu(G)^{2}-1\leq\{|U|(2\mu(G)+4-|U|)+1\}\mu(G)-|U|(2\mu(G)+2)-1$, or equivalently, $\{|U|(2\mu(G)+4-|U|)+1\}\mu(G)-|U|(2\mu(G)+2)\geq 2\mu(G)^{2}$. Let $g(x)=\{x(2\mu(G)+4-x)+1\}\mu(G)-x(2\mu(G)+2)$. Then $g(x)$ is a concave function on $\mathbb{R}$. So on any interval $[a,b]$, $g(x)$ achieves the minimum at $a$ or $b$. By direct computation, we obtain $g(3)=6\mu(G)^{2}-2\mu(G)-6$ and $g(2\mu(G)-1)=6\mu(G)^{2}-6\mu(G)+2$. It is easy to see that $\min\{g(3),g(2\mu(G)-1)\}\geq 2\mu(G)^{2}$, because $\mu(G)=k\geq 2$ (see the hypothesis of Case 2). Hence $g(|U|)\geq 2\mu(G)^{2}$ for $3\leq|U|\leq 2\mu(G)-1=2k-1$. This proves (15) and hence (14) and (13).

So we assume that $\delta(G)\leq 2(t+1)(\mu(G)-1)$ for some $t$ with $\frac{|U|+1}{2}\leq t\leq\mu(G)$. We prove (15) by showing that $2(t+1)(\mu(G)-1)\leq\{|U|(2t+2-|U|)+1\}\mu(G)-|U|2t-1$, or equivalently, $\{|U|(2t+2-|U|)-2t-1\}\mu(G)-|U|2t\geq-2t-1$. Let $h(x)=\{x(2t+2-x)-2t-1\}\mu(G)-2tx$. Then $h(x)$ is a concave function on $\mathbb{R}$. So on any interval $[a,b]$, $h(x)$ achieves the minimum at $a$ or $b$. By direct computation, we obtain $h(3)=4(t-1)\mu(G)-6t$ and $h(2t-1)=4(t-1)\mu(G)-2t(2t-1)$. It is easy to see that $\min\{h(3),h(2t-1)\}\geq-2t-1$, because $\mu(G)\geq t\geq\frac{|U|+1}{2}\geq 2$. Hence $h(|U|)\geq-2t-1$ for $3\leq|U|\leq 2t-1$. This proves (15) and hence (14) and (13).

Since $2|E(U)|+2|F(U)|\geq|U|\delta(G)+|F(U)|$, the desired statement (3) follows instantly from (13) and (14), competing the proof of Theorem 1.2.