## 1 Introduction

If GH are graphs, we say G is H-free if no induced subgraph of G is isomorphic to H; and for a graph G, we denote the number of vertices, the chromatic number, the size of the largest clique, and the size of the largest stable set by $$|G|, \chi (G), \omega (G),\alpha (G)$$ respectively.

The k-vertex path is denoted by $$P_k$$, and $$P_4$$-free graphs are well-understood; every $$P_4$$-free graph G with more than one vertex is either disconnected or disconnected in the complement , which implies that $$\chi (G)=\omega (G)$$. Here we study how $$\chi (G)$$ depends on $$\omega (G)$$ for $$P_5$$-free graphs G.

The Gyárfás-Sumner conjecture [10, 25] says:

1.1 Conjecture: For every forest H there is a function f such that $$\chi (G)\le f(\omega (G))$$ for every H-free graph G.

This is open in general, but has been proved  when H is a path, and for several other simple types of tree ([3, 11,12,13,14, 17, 19]; see  for a survey). The result is also known if all induced subdivisions of a tree are excluded .

A class of graphs is hereditary if the class is closed under taking induced subgraphs and under isomorphism, and a hereditary class is said to be $$\chi$$-bounded if there is a function f such that $$\chi (G)\le f(\omega (G))$$ for every graph G in the class (thus, the Gyárfás-Sumner conjecture says that, for every forest H, the class of H-free graphs is $$\chi$$-bounded). Louis Esperet  made the following conjecture:

1.2 (False) Conjecture: Let $$\mathcal G$$ be a $$\chi$$-bounded class. Then there is a polynomial function f such that $$\chi (G)\le f(\omega (G))$$ for every $$G\in \mathcal G$$.

Esperet’s conjecture was recently shown to be false by Briański, Davies and Walczak . However, this raises the further question: which $$\chi$$-bounded classes are polynomially $$\chi$$-bounded? In particular, the two conjectures 1.1 and 1.2 would together imply the following, which is still open:

1.3 Conjecture: For every forest H, there exists $$c>0$$ such that $$\chi (G)\le \omega (G)^c$$ for every H-free graph G.

This is a beautiful conjecture. In most cases where the Gyárfás-Sumner conjecture has been proved, the current bounds are very far from polynomial, and 1.3 has been only been proved for a much smaller collection of forests (see [5, 15, 16, 20,21,22,23]). In  we proved it for any $$P_5$$-free tree H, but it has not been settled for any tree H that contains $$P_5$$. In this paper we focus on the case $$H=P_5$$.

The best previously-known bound on the chromatic number of $$P_5$$-free graphs in terms of their clique number, due to Esperet, Lemoine, Maffray, and Morel , was exponential:

1.4   If G is $$P_5$$-free and $$\omega (G)\ge 3$$ then $$\chi (G)\le (5/27)3^{\omega (G)}$$.

Here we make a significant improvement, showing a “near-polynomial” bound:

1.5   If G is $$P_5$$-free and $$\omega (G)\ge 3$$ then $$\chi (G)\le \omega (G)^{\log _2(\omega (G))}$$.

(The cycle of length five shows that we need to assume $$\omega (G)\ge 3$$. Sumner  showed that $$\chi (G)\le 3$$ when $$\omega (G)=2$$.) Conjecture 1.3 when $$H=P_5$$ is of great interest, because of a famous conjecture due to Erdős and Hajnal [6, 7], that:

1.6 Conjecture: For every graph H there exists $$c>0$$ such that $$\alpha (G)\omega (G)\ge |G|^c$$ for every H-free graph G.

This is open in general, despite a great deal of effort; and in view of , the smallest graph H for which 1.6 is undecided is the graph $$P_5$$. Every forest H satisfying 1.3 also satisfies the Erdős-Hajnal conjecture, and so showing that $$H=P_5$$ satisfies 1.3 would be a significant result. (See  for some other recent progress on this question.)

We use standard notation throughout. When $$X\subseteq V(G)$$, G[X] denotes the subgraph induced on X. We write $$\chi (X)$$ for $$\chi (G[X])$$ when there is no ambiguity.

## 2 The Main Proof

We denote the set of nonnegative real numbers by $$\mathbb {R}_+$$, and the set of nonnegative integers by $$\mathbb {Z}_+$$. Let $$f:\mathbb {Z}_+\rightarrow \mathbb {R}_+$$ be a function. We say

• f is non-decreasing if $$f(y)\ge f(x)$$ for all integers $$x,y\ge 0$$ with $$y>x\ge 0$$;

• f is a binding function for a graph G if it is non-decreasing and $$\chi (H)\le f(\omega (H))$$ for every induced subgraph H of G; and

• f is a near-binding function for G if f is non-decreasing and $$\chi (H)\le f(\omega (H))$$ for every induced subgraph H of G different from G.

In this section we show that if a function f satisfies a certain inequality, then it is a binding function for all $$P_5$$-free graphs. Then at the end we will give a function that satisfies the inequality, and deduce 1.5.

A cutset in a graph G is a set X such that $$G\setminus X$$ is disconnected. A vertex $$v\in V(G)$$ is mixed on a set $$A\subseteq V(G)$$ or a subgraph A of a graph G if v is not in A and has a neighbour and a non-neighbour in A. It is complete to A if it is adjacent to every vertex of A. We begin with the following:

2.1   Let G be $$P_5$$-free, and let f be a near-binding function for G. Let G be connected, and let X be a cutset of G. Then

\begin{aligned} \chi (G\setminus X)\le f(\omega (G)-1)+ \omega (G) f(\lfloor \omega (G)/2\rfloor ). \end{aligned}

### Proof

We may assume (by replacing X by a subset if necessary) that X is a minimal cutset of G; and so $$G\setminus X$$ has at least two components, and every vertex in X has a neighbour in V(B), for every component B of $$G\setminus X$$. Let B be one such component; we will prove that $$\chi (B)\le f(\omega (G)-1)+ \omega (G) f(\lfloor \omega (G)/2\rfloor )$$, from which the result follows.

Choose $$v\in X$$ (this is possible since G is connected), and let N be the set of vertices in B adjacent to v. Let the components of $$B\setminus N$$ be $$R_1,\ldots ,R_k, S_1,\ldots ,S_\ell$$, where $$R_1,\ldots ,R_k$$ each have chromatic number more than $$f(\lfloor \omega (G)/2\rfloor )$$, and $$S_1,\ldots ,S_\ell$$ each have chromatic number at most $$f(\lfloor \omega (G)/2\rfloor )$$. Let S be the union of the graphs $$S_1,\ldots ,S_\ell$$; thus, $$\chi (S)\le f(\lfloor \omega (G)/2\rfloor )$$. For $$1\le i\le k$$, let $$Y_i$$ be the set of vertices in N with a neighbour in $$V(R_i)$$, and let $$Y=Y_1\cup \cdots \cup Y_k$$.

(1) For $$1\le i\le k$$, every vertex in $$Y_i$$ is complete to $$R_i$$.

Let $$y\in Y_i$$. Thus, y has a neighbour in $$V(R_i)$$; suppose that y is mixed on $$R_i$$. Since $$R_i$$ is connected, there is an edge ab of $$R_i$$ such that y is adjacent to a and not to b. Now v has a neighbour in each component of $$G\setminus X$$, and since there are at least two such components, there is a vertex $$u\in V(G)\setminus (X\cup V(B))$$ adjacent to v. But then $$u\hbox {-}v\hbox {-}y\hbox {-}a\hbox {-}b$$ is an induced copy of $$P_5$$, a contradiction. This proves (1).

(2) $$\chi (Y)\le (\omega (G)-1) f(\lfloor \omega (G)/2\rfloor )$$.

Let $$1\le i\le k$$. Since $$f(\lfloor \omega (G)/2\rfloor )<\chi (R_i)\le f(\omega (R_i))$$, and f is non-decreasing, it follows that $$\omega (R_i)>\omega (G)/2$$. By (1), $$\omega (G[Y_i])+\omega (R_i)\le \omega (G)$$, and so $$\omega (G[Y_i])<\omega (G)/2$$. Consequently $$\chi (Y_i)\le f(\lfloor \omega (G)/2\rfloor )$$, for $$1\le i\le k$$. Choose $$I\subseteq \{1,\ldots ,k\}$$ minimal such that $$\bigcup _{i\in I}Y_i=Y$$. From the minimality of I, for each $$i\in I$$ there exists $$y_i\in Y_i$$ such that for each $$j\in I\setminus \{i\}$$ we have that $$y_i\notin Y_j$$; and so the vertices $$y_i\;(i\in I)$$ are all distinct. For each $$i\in I$$ choose $$r_i\in V(R_i)$$. For all distinct $$i,j\in I$$, if $$y_i,y_j$$ are nonadjacent, then $$r_i\hbox {-}y_i\hbox {-}v\hbox {-}y_j\hbox {-}r_j$$ is isomorphic to $$P_5$$, a contradiction. Hence the vertices $$y_i\;(i \in I)$$ are all pairwise adjacent, and adjacent to v; and so $$|I|\le \omega (G)-1$$. Thus, $$\chi (Y)=\chi (\bigcup _{i\in I}Y_i)\le (\omega (G)-1)f(\lfloor \omega (G)/2\rfloor )$$. This proves (2).

All the vertices in $$N\setminus Y$$ are adjacent to v, and so $$\omega (G[N\setminus Y])\le \omega (G)-1$$. Moreover, for $$1\le i\le k$$, each vertex of $$R_i$$ is adjacent to each vertex in $$Y_i$$, and $$Y_i\ne \emptyset$$ since B is connected, and so $$\omega (R_i)\le \omega (G)-1$$. Since there are no edges between any two of the graphs $$G[N\setminus Y], R_1,\ldots ,R_k$$, their union (Z say) has clique number at most $$\omega (G)-1$$ and so has chromatic number at most $$f(\omega (G)-1)$$. But V(B) is the union of YV(S) and V(Z); and so

\begin{aligned} \chi (B)\le f(\omega (G)-1)+(\omega (G)-1)f(\lfloor \omega (G)/2\rfloor ) +f(\lfloor \omega (G)/2\rfloor ). \end{aligned}

This proves 2.1. $$\square$$

2.2   Let $$\Omega \ge 1$$, and let $$f:\mathbb {Z}_+\rightarrow \mathbb {R}_+$$ be non-decreasing, satisfying the following:

• f is a binding function for every $$P_5$$-free graph H with $$\omega (H)\le \Omega$$; and

• $$f(w-1) + (w+2) f(\lfloor w/2\rfloor ) \le f(w)$$ for each integer $$w> \Omega$$.

Then f is a binding function for every $$P_5$$-free graph G.

### Proof

We prove by induction on |G| that if G is $$P_5$$-free then f is a binding function for G. Thus, we may assume that G is $$P_5$$-free and f is near-binding for G. If G is not connected, or $$\omega (G)\le \Omega$$, it follows that f is binding for G, so we assume that G is connected and $$\omega (G)>\Omega$$. Let us write $$w=\omega (G)$$ and $$m=\lfloor w/2\rfloor$$. If $$\chi (G)\le f(w)$$ then f is a binding function for G, so we assume, for a contradiction, that:

(1) $$\chi (G)>f(w-1) + (w+2)f(m)$$.

We deduce that:

(2) Every cutset X of G satisfies $$\chi (X)> 2f(m)$$.

If some cutset X satisfies $$\chi (X)\le 2f(m)$$, then since $$\chi (G\setminus X)\le f(w-1)+ w f(m)$$ by 2.1, it follows that $$\chi (G)\le f(w-1)+(w+2)f(m)$$, contrary to (1). This proves (2).

(3) If PQ are cliques of G, both of cardinality at least w/2, then $$G[P\cup Q]$$ is connected.

Suppose not; then there is a minimal subset $$X\subseteq V(G){\setminus } (P\cup Q)$$ such that PQ are subsets of different components (AB say) of $$G\setminus X$$. From the minimality of X, every vertex $$x\in X$$ has a neighbour in V(A) and a neighbour in V(B). If x is mixed on A and mixed on B, then since A is connected, there is an edge $$a_1a_2$$ of A such that x is adjacent to $$a_1$$ and not to $$a_2$$; and similarly there is an edge $$b_1b_2$$ of B with x adjacent to $$b_1$$ and not to $$b_2$$. But then $$a_2\hbox {-}a_1\hbox {-}x\hbox {-}b_1\hbox {-}b_2$$ is an induced copy of $$P_5$$, a contradiction; so every $$x\in X$$ is complete to at least one of AB. The set of vertices in X complete to A is also complete to P, and hence has clique number at most m, and hence has chromatic number at most f(m); and the same for B. Thus, $$\chi (X)\le 2f(m)$$, contrary to (2). This proves (3).

If $$v\in V(G)$$, we denote its set of neighbours by N(v), or $$N_G(v)$$. Let $$a\in V(G)$$, and let B be a component of $$G{\setminus } (N(a)\cup \{a\})$$; we will show that $$\chi (B)\le (w-m+2)f(m)$$.

A subset Y of V(B) is a joint of B if there is a component C of $$B{\setminus } Y$$ such that $$\chi (C)> f(m)$$ and Y is complete to C. If $$\emptyset$$ is not a joint of B then $$\chi (B)<f(m)$$ and the claim holds, so we may assume that $$\emptyset$$ is a joint of B; let Y be a joint of B chosen with Y maximal, and let C be a component of $$B\setminus Y$$ such that $$\chi (C)> f(m)$$ and Y is complete to C.

(4) If $$v\in N(a)$$ has a neighbour in V(C), then $$\chi (V(C)\setminus N(v))\le f(m)$$.

Let $$N_C(v)$$ be the set of neighbours of v in V(C), and $$M=V(C)\setminus N_C(v)$$; and suppose that $$\chi (M)> f(m)$$. Let $$C'$$ be a component of G[M] with $$\chi (C')> f(m)$$, and let Z be the set of vertices in $$N_C(v)$$ that have a neighbour in $$V(C')$$. Thus, $$Z\ne \emptyset$$, since $$N_C(v),V(C')\ne \emptyset$$ and C is connected. If some $$z\in Z$$ is mixed on $$C'$$, let $$p_1p_2$$ be an edge of $$C'$$ such that z is adjacent to $$p_1$$ and not to $$p_2$$; then $$a\hbox {-}v\hbox {-}z\hbox {-}p_1\hbox {-}p_2$$ is an induced copy of $$P_5$$, a contradiction. So every vertex in Z is complete to $$V(C')$$; but also every vertex in Y is complete to V(C) and hence to $$V(C')$$, and so $$Y\cup Z$$ is a joint of B, contrary to the maximality of Y. This proves (4).

(5) $$\chi (Y)\le f(m)$$ and $$\chi (C)\le (w-m+1) f(m)$$.

Let X be the set of vertices in N(a) that have a neighbour in V(C). Since C is a component of $$B\setminus Y$$ and hence a component of $$G{\setminus } (X\cup Y)$$, and a belongs to a different component of $$G\setminus (X\cup Y)$$, it follows that $$X\cup Y$$ is a cutset of G. By (2), $$\chi (X\cup Y)>2f(m)$$. Since $$\omega (C)\ge m+1$$ (because $$\chi (C)>f(m)$$, and f is near-binding for G) and every vertex in Y is complete to V(C), it follows that $$\omega (G[Y])\le w-m-1\le m$$, and so has chromatic number at most f(m) as claimed; and so $$\chi (X)>f(m)$$. Consequently there is a clique $$P\subseteq X$$ with cardinality $$w-m$$. The subgraph induced on the set of vertices of C complete to P has clique number at most m, and so has chromatic number at most f(m); and for each $$v\in P$$, the set of vertices of C nonadjacent to v has chromatic number at most f(m) by (4). Thus, $$\chi (C)\le (|P|+1)f(m)= (w-m+1)f(m)$$. This proves (5).

(6) $$\chi (B)\le (w-m+2) f(m)$$.

By (3), every clique contained in $$V(B){\setminus } (V(C)\cup Y)$$ has cardinality less than w/2 (because it is anticomplete to the largest clique of C) and so

\begin{aligned} \chi (B\setminus (V(C)\cup Y))\le f(m); \end{aligned}

and hence $$\chi (B\setminus Y)\le (w-m+1)f(m)$$ by (5), since there are no edges between C and $$V(B)\setminus (V(C)\cup Y)$$. But $$\chi (Y)\le f(m)$$ by (5), and so $$\chi (B)\le (w-m+2) f(m)$$. This proves (6).

By (6), $$G\setminus N(a)$$ has chromatic number at most $$(w-m+2)f(m)$$. But G[N(a)] has clique number at most $$w-1$$ and so chromatic number at most $$f(w-1)$$; and so $$\chi (G)\le f(w-1)+(w-m+2)f(m)$$, contrary to (1). This proves 2.2. $$\square$$

Now we deduce 1.5, which we restate:

2.3   If G is $$P_5$$-free and $$\omega (G)\ge 3$$ then $$\chi (G)\le \omega (G)^{\log _2(\omega (G))}$$.

### Proof

Define $$f(0)=0$$, $$f(1)=1$$, $$f(2)=3$$, and $$f(x)=x^{\log _2(x)}$$ for every real number $$x\ge 3$$. Let G be $$P_5$$-free. If $$\omega (G)\le 2$$ then $$\chi (G)\le 3=f(2)$$, by a result of Sumner ; if $$\omega (G) = 3$$ then $$\chi (G)\le 5\le f(3)$$, by an application of the result 1.4 of Esperet, Lemoine, Maffray, and Morel ; and if $$\omega (G)=4$$ then $$\chi (G)\le 15\le f(4)$$, by another application of 1.4. Consequently every $$P_5$$-free graph G with clique number at most four has chromatic number at most $$f(\omega (G))$$.

We claim that

\begin{aligned} f(x-1) + (x+2)f(\lfloor x/2 \rfloor ) \le f(x) \end{aligned}

for each integer $$x>4$$. If that is true, then by 2.2 with $$\Omega =4$$, we deduce that $$\chi (G)\le f(\omega (G))$$ for every $$P_5$$-free graph G, and so 1.5 holds. Thus, it remains to show that

\begin{aligned} f(x-1) + (x+2)f(\lfloor x/2 \rfloor ) \le f(x) \end{aligned}

for each integer $$x>4$$. This can be verified by direct calculation when $$x=5$$, so we may assume that $$x\ge 6$$.

The derivative of $$f(x)/x^4$$ is

\begin{aligned} (2\log _2(x)-4)x^{\log _2(x)-5}, \end{aligned}

and so is nonnegative for $$x\ge 4$$. Consequently

\begin{aligned} \frac{f(x-1)}{(x-1)^4}\le \frac{f(x)}{x^4} \end{aligned}

for $$x\ge 5$$. Since $$x^2(x^2-2x-4)\ge (x-1)^4$$ when $$x\ge 5$$, it follows that

\begin{aligned} \frac{f(x-1)}{x^2-2x-4}\le \frac{f(x)}{x^2}, \end{aligned}

that is,

\begin{aligned} f(x-1) + \frac{2x+4}{x^2}f(x) \le f(x), \end{aligned}

when $$x\ge 5$$. But when $$x\ge 6$$ (so that f(x/2) is defined and the first equality below holds), we have

\begin{aligned} f(\lfloor x/2\rfloor )\le f(x/2)=(x/2)^{\log _2(x/2)}=(x/2)^{\log _2(x) - 1}=(2/x)(x/2)^{\log _2(x)} = (2/x^2)f(x), \end{aligned}

and so

\begin{aligned} f(x-1) + (x+2)f(\lfloor x/2\rfloor ) \le f(x) \end{aligned}

when $$x\ge 6$$. This proves 2.3. $$\square$$