Proof. Choose $\varepsilon <\frac 12$ small enough such that $(\delta -\varepsilon )^{t-1}>(t+1)\varepsilon $ and let $\delta '=\varepsilon /M$ , where $M=M(\varepsilon ,\frac 1\varepsilon )$ is the constant guaranteed by the regularity lemma (Theorem 2.7). Suppose G is a $K_t$ -free graph on N vertices with $\alpha (G)\leq \delta 'N$ .

Apply Theorem 2.7 with this value of $\varepsilon $ to G. This yields an $\varepsilon $ -regular partition $V(G)=V_1\cup \cdots \cup V_r\cup V_{r+1}$ with $\frac 1\varepsilon \leq r\leq M$ . We show that a substructure similar to a weighted t-clique is forbidden among the edge densities $d(V_i,V_j)$ .

Proof of claim

Order the elements of $S_1$ as $a_1,\ldots ,a_\ell $ with the elements of $S_1\setminus S_2$ listed first. Set $k=|S_1|-|S_2|$ . By Lemma 2.8, there exists a clique S of order $\ell $ in G such that $|S\cap V_{a_i}|=1$ for each $i\in [\ell ]$ ; as G is $K_t$ -free, it follows that $\ell <t$ . Our proof follows an $\ell $ -step process, where the ith step chooses one (if $i\leq k$ ) or two (if $i>k$ ) vertices from $V_{a_i}$ that are adjacent to all previously chosen vertices. For $0\leq i,j\leq \ell $ , write $W^{(i)}$ for the common neighborhood of those vertices chosen in the first i steps, and let $W_j^{(i)}=V_{a_j}\cap W^{(i)}$ . In particular, $W^{(0)}=V(G)$ . We will choose $2\ell -k$ vertices such that $|W_j^{(i)}|\geq (\delta -\varepsilon )|W_j^{(i-1)}|\geq (\delta -\varepsilon )^i|V_{a_j}|$ for all $0\leq i<j\leq \ell $ .

On the ith step with $1\leq i\leq k$ , choose one vertex $v_i\in W_i^{(i-1)}$ such that $W_j^{(i)}:=N(v_i)\cap W_j^{(i-1)}$ has cardinality at least $(\delta -\varepsilon )|W_j^{(i-1)}|$ for each $j>i$ . To show that such a vertex $v_i$ exists, consider the sets

for each $j>i$ . Observe that $d\left (X_j,W_j^{(i-1)}\right )<\delta -\varepsilon <d(V_{a_i},V_{a_j})-\varepsilon $ by construction, and that $|W_j^{(i-1)}|\geq (\delta -\varepsilon )^{i-1}|V_{a_j}|\geq (\delta -\varepsilon )^t|V_{a_j}|\geq \varepsilon |V_{a_j}|$ . Because the pair $(V_{a_i},V_{a_j})$ is $\varepsilon $ -regular, it follows that $|X_j|<\varepsilon |V_{a_i}|$ . Thus,

It follows that there is a vertex $v_i\in W_i^{(i-1)}$ such that $|W_j^{(i)}|=|N(v_i)\cap W_j^{(i-1)}|\geq (\delta -\varepsilon )|W_j^{(i-1)}|$ for each $j>i$ .

On the ith step with $k<i\leq \ell $ , choose two adjacent vertices $v_i,v^{\prime }_i\in W_i^{(i-1)}$ such that $W_j^{(i)}:=N(v_i)\cap N(v^{\prime }_i)\cap W_j^{(i-1)}$ has cardinality at least $2(\delta -\varepsilon )|W_j^{(i-1)}|$ for all $j>i$ . To verify that such vertices exist, set

for each $j>i$ . The argument from the prior paragraph shows that

Noting that $|V_{a_i}|=(N-|V_{r+1}|)/r>N/2r$ , we have

It follows that there are adjacent vertices $v_i,v^{\prime }_i\in W_i^{(i-1)}$ such that $N(v_i)\cap W_j^{(i-1)}$ and $N(v^{\prime }_i)\cap W_j^{(i-1)}$ have cardinality at least $(1/2+\delta -\varepsilon )|W_j^{(i-1)}|$ for each $j>i$ . By the pigeonhole principle,

for each $j>i$ , as desired.

After $\ell $ steps, this process results in $k+2(\ell -k)=|S_1|+|S_2|$ vertices $v_1,\ldots ,v_k,v_{k+1},v^{\prime }_{k+1},\ldots ,v_\ell ,v^{\prime }_\ell $ that form a clique in G. It follows that $|S_1|+|S_2|<t$ , because G is $K_t$ -free.

Let R be the weighted graph on $[r]$ with vertex weights $w(i)=\frac 1r$ for all $i\in [r]$ and edge weights

for all $i,j\in [r]$ . We observe that R is ${ \mathcal {K}}_t$ -free as a direct consequence of Claim 2.10.

To conclude the proof, we bound the $K_s$ -density of G. We have

If $a_1,\ldots ,a_s$ are distinct elements of $[r]$ and each pair $(V_{a_i},V_{a_j})$ is $\varepsilon $ -regular, then we may simplify the summand using the graph-counting lemma. Indeed, Lemma 2.8 implies that the number of copies of $K_s$ in $V_{a_1}\times \cdots \times V_{a_s}$ is at most

in this case. It remains to bound the contribution from terms where some $a_i$ is $r+1$ , the $a_i$ are not distinct, or some pair $(V_{a_i},V_{a_j})$ is not $\varepsilon $ -regular.

The terms where at least one index $a_i$ equals $r+1$ contribute at most

to the sum. The terms where $a_1,\ldots ,a_s$ are not all distinct contribute at most

because $r\geq 1/\varepsilon $ . The terms where a pair $(V_{a_i},V_{a_j})$ is not $\varepsilon $ -regular contribute at most

Combining these estimates, we have

To compare this to $d_{K_s}(R)$ , we observe the following inequality. If real numbers $x_1,\ldots ,x_k,y_1,\ldots ,y_k\in [0,1]$ satisfy $x_i\leq y_i+\delta $ for each $i\in [k]$ then

Applying this inequality with to the real numbers $d(V_{a_i},V_{a_j})$ and $w(a_i,a_j)$ , it follows that

Because $\varepsilon <(\delta -\varepsilon )^{t-1}<\delta ^2$ , it follows that $d_{K_s}(G)<d_{K_s}(R)+4s^2\delta $ , as desired.

Proof. Suppose that $V(R)=[r]$ for some integer r. Increasing each edge weight to the next multiple of $\frac 12$ preserves the ${ \mathcal {K}}_t$ -freeness of R, so we may assume that all edge weights of R are 0, $\frac 12$ , or $1$ . Set $\mu =\frac {\varepsilon }{\sqrt h}$ as in the Bollobás–Erdős construction.

Choose integers $n_i\geq \lfloor w(i)N\rfloor $ such that $N=n_1+\cdots +n_r$ . We construct an N-vertex graph G on vertex set $V_1\cup \cdots \cup V_r$ as follows. Intuitively, $G[V_i,V_j]$ will be complete, empty, or a randomly rotated Bollobás–Erdős graph, depending on whether $w(i,j)$ is 1, 0, or $\frac 12$ .

Suppose N is sufficiently large. For each i, we may choose a set $V_i$ of $n_i$ points on $\textsf {S}^{h-1}$ satisfying Lemma 2.11(4). Connect vertices in as follows. Within each part $V_i$ , add an edge between if . Let $G[V_i,V_j]$ be complete bipartite if $w(i,j)=1$ and empty if $w(i,j)=0$ . If $w(i,j)=\frac 12$ for some $i<j$ , then let $\rho _{ij}\in SO(h)$ be a rotation of $\textsf {S}^{h-1}$ chosen uniformly at random. Connect and if .

Observe that each induced subgraph $G[V_i]$ is $K_3$ -free with independence number $\alpha (G[V_i])\leq e^{-\varepsilon \sqrt h/4}n_i$ by Lemma 2.11(2) and (4). It follows that $\alpha (G)\leq e^{-\varepsilon \sqrt h/4}N$ . Additionally, if $w(i,j)=\frac 12$ , the induced subgraph $G[V_i\cup V_j]$ is the Bollobás–Erdős graph $\mathrm {BE}(\rho _{ij}(V_i),V_j)$ , and is thus $K_4$ -free by Lemma 2.11(3).

Using these properties, we verify that G is $K_t$ -free. Suppose, for contradiction, that $G[W]$ is a clique, where W is some set of t vertices. Let $S_1{\kern-1pt}={\kern-1pt}\{i\in [r]{\kern-1pt}:{\kern-1pt}|V_i{\kern-1pt}\cap{\kern-1pt} W|{\kern-1pt}\geq{\kern-1pt} 1\}$ and $S_2{\kern-1pt}={\kern-1pt}\{i{\kern-1pt}\in{\kern-1pt} [r]{\kern-1pt}:{\kern-1pt}|V_i{\kern-1pt}\cap{\kern-1pt} W|{\kern-1pt}\geq{\kern-1pt} 2\}\subseteq S_1$ . Because each induced subgraph $G[V_i]$ is $K_3$ -free, it follows that W has at most two points in $V_i$ , and thus that $|S_1|+|S_2|=|W|=t$ . For any distinct $i,j\in S_1$ , there is an edge between $V_i$ and $V_j$ , and it follows that $w(i,j)>0$ . Additionally, for any distinct $i,j\in S_2$ , there is a $K_4$ in $G[V_i\cup V_j]$ . Because the Bollobás–Erdős graph is $K_4$ -free, this implies that $w(i,j)>\frac 12$ . We conclude that $(S_1,S_2)$ form a weighted t-clique in R, which is a contradiction. It follows that G is $K_t$ -free.

Lastly, we verify that G has large $K_s$ -density in expectation, using the independence of the random rotations $\rho _{ij}$ . By Lemma 2.11(1), if $w(i,j)=\frac 12$ then the expected edge density between $V_i$ and $V_j$ is at least $\frac 12-2\sqrt 2\varepsilon $ . Thus, if N is sufficiently large, we have

We conclude that there is a choice of the rotations $\rho _{ij}$ such that $d_{K_s}(G)\geq (1-2\sqrt 2 s^2\varepsilon )d_{K_s}(R)$ .

Proof. Set $R = R(p,P;q,Q)$ and let u be a vertex in R with weight p. Let $R'$ be the graph obtained from R by changing the weights of all edges incident to u: if $(u,v)$ has edge weight $w\in \{\frac 12,1\}$ in R, then we assign it the weight $\frac 32-w$ in $R'$ . It is clear that $R^{\prime }_{>\frac 12}$ is a complete bipartite graph with parts of size $P-1$ and $Q+1$ .

Fix m with $2\leq m\leq P+Q$ . We verify that $d_{K_m}(R,u)<d_{K_m}(R',u)$ . We have that

Let . If $x\ge y$ then

with equality only if $x=y$ . This in turn implies that

for any $x,y$ , again with equality only if $x=y$ . Thus,

To conclude the proof, we observe that $d_{K_m}(R-u)=d_{K_m}(R'-u)$ , and thus

Proof. Observe that each $N_{m,r}$ is a linear combination of the $\left \lfloor \frac m2\right \rfloor +1$ polynomials

with nonzero coefficient if and only if $x\geq r$ . Thus, $\{N_{m,r}:0\leq r\leq m/2\}$ is a basis for the space of all linear combinations of these polynomials, which includes the $K_m$ -density

It follows that there exist real numbers $c_{r}$ such that

Moreover, by setting the coefficients of the rth term equal to each other, we conclude that

(3.2)

for each $r\leq m/2$ .

We show that the coefficients $c_{r}$ are positive by induction on r. Clearly, $c_0> 0$ by (3.2) when $r=0$ . Now, suppose $c_i> 0$ for all $i<r$ . By (3.2), we have

We claim that

for each $i<r$ . Indeed, it suffices to show that

which, recalling that $r\leq m/2$ , follows from the inequalities $2^{m-2r+1}\geq m-2r+2$ and $\frac {r-i}{m-r+1-i}=\frac {r-i}{m-2r+1+(r-i)}\geq \frac {1}{m-2r+2}$ . Hence,

which implies $c_r>0$ .

Proof. Set $R=R(p,P;q,Q)$ and $R'=R(p',P';q',Q')$ , and fix m with $2\leq m\leq P+Q$ . For convenience, we write $N_{m,r}(R)=N_{m,r}(p,P;q,Q)$ and $N_{m,r}(R')=N_{m,r}(p',P';q',Q')$ .

By Lemma 3.4, it suffices to compare $N_{m,r}(R)$ and $ N_{m,r}(R')$ . We begin with some preliminary inequalities. Let $p_-=\min \{p',q'\}$ and $p_+=\max \{p',q'\}$ .

Proof of claim

(1) Because $P\geq Q+2$ , we have

(2) Using the relation $(P-1)p\leq Qq$ , it follows that $p<\frac {P}{P-1}p=p'\leq \frac {PQ}{(P-1)^2}q<q$ and that $q>\frac {Qq}{Q+1}=q'\geq \frac {(P-1)p}{Q+1}\geq p$ .

(3) First, observe that

because $p\leq Qq/(P-1)<q$ . Thus,

if $r>0$ .

We now compare $N_{m,r}(R)$ and $N_{m,r}(R')$ .

Proof of claim

By Claim 3.6(1),

with equality only if $r=0$ . It remains to show that

(3.3)

Set $X = (\underbrace {p,p ,\ldots , p }_{P-r}, \underbrace {q,q ,\ldots ,q}_{Q-r})$ and $X' = (\underbrace {p',p' ,\ldots , p' }_{P-r-1}, \underbrace {q',q' ,\ldots ,q'}_{Q-r+1})$ . Letting $\sigma $ be the $(m-2r)$ th elementary symmetric function

we may rewrite (3.3) as $\sigma (X)\leq \sigma (X')$ .

Let Y be the $(P+Q-2r)$ -tuple obtained via the following process. Set $Y=X$ initially and repeat the following transformation, which will never decrease $\sigma (Y)$ .

1. (*) If there exist indices $i,j$ such that $Y_i<p_-$ and $Y_j>p_+$ , set $\varepsilon =\min \{p_--Y_i,~Y_j-p_+\}$ , and replace $Y_i$ and $Y_j$ with $Y_i+\varepsilon $ and $Y_j-\varepsilon $ , respectively.

Each iteration increases the number of coordinates equal to $p_-$ or $p_+$ , so the process will terminate in at most $P+Q-2r$ steps. Moreover, recalling that $p\leq p_-\leq p_+<q$ , we observe that the final tuple Y either takes the form

(Case 1)

(Case 2)

Let $X" = (p_-,\ldots ,p_-,p_+,\ldots ,p_+)$ be the result of sorting $X'$ in increasing order, and let $k\in \{P-r-1, Q-r+1\}$ be the number of occurrences of $p_-$ . In case 1, we have $Y_i\leq X^{\prime \prime }_i$ for each i, implying $\sigma (Y)\leq \sigma (X")$ . In case 2, let y be the average value of $\{Y_{k+1},\ldots ,Y_{P+Q-2r}\}$ and let $Y'=(p_-,\ldots ,p_-,y,\ldots ,y)$ be the result of replacing all but the first k terms of Y with y. By Lemma 3.2, $\sigma (Y)\leq \sigma (Y')$ . Moreover, $\mathrm {sum}(Y')=\mathrm {sum}(X)<\mathrm {sum}(X")$ by Claim 3.6(3). Because $X"$ is obtained from $Y'$ by replacing each y with $p_+$ , it follows that $y<p_+$ and $\sigma (Y')<\sigma (X")$ . Thus, we have $\sigma (X)\leq \sigma (Y)\leq \sigma (X")=\sigma (X')$ in both cases, completing the proof that $N_{m,r}(R)\leq N_{m,r}(R')$ .

Combining Lemma 3.4 and Claim 3.7, we have that

Proof. Set $R=R(p,P;q,P)$ and let u and v be vertices in R with weights p and q, respectively. Let $R'$ be the graph obtained from R by setting the weights of both u and v to $\frac {p+q}2=\frac 1{2P}$ .

Observe that $d_{K_m}(R-\{u,v\})=d_{K_m}(R'-\{u,v\})$ . Set

We have that

because $q-p$ and $q^{y-x}-p^{y-x}$ both have the same sign. Additionally,

Combining the preceding inequalities, we conclude that

for any integer $2\leq m\leq 2P$ .

Proof. We first prove (1). Suppose $t>s^2(s-1)/2$ is odd. Let $R_0$ be the weighted graph obtained from R by changing all edges weights of $1/2$ into 0 and let $R_1$ be the weighted graph obtained from R by changing all edge weights of $1/2$ into 1. We claim that

Indeed, if vertices $v_1,\ldots ,v_s\in V(R)$ induce $m\geq 1$ edges of weight $1/2$ in R, then the product of their edge weights is $\left (\frac 12\right )^m\leq \frac 12$ in R and is 1 in $R_1$ .

We have . Write $w(B_i)=\sum _{v\in B_i}w(v)$ for the total weight of vertices in $B_i$ . By Lemma 3.2, we have

Thus, setting , it suffices to show that $\frac {f(a)+f(b)}2\leq f(\frac {a+b}2)=d_{K_s}(K_r^w)$ .

First, observe that

and that

It follows that $f"(x)<0$ when : for $s=3$ , this can be checked manually, and for $s\geq 4$ , this follows from the inequality

Hence, if , Jensen’s inequality implies

with equality if and only if $a=b=r$ , (i.e., $R=K_r^w$ ). To see that , note that (A5) implies $b\leq (s-1)a$ , yielding as desired.

We now prove (2) using (1). Suppose $t>s^2(s-1)/2+s$ is even. Because $t-1=a+b=\sum _{i=1}^a(|B_i|+1)$ is odd, there exists some k such that $|B_k|+1$ is odd. Scaling $R-B_k$ up by a factor of $(1-w(B_k))^{-1}$ yields a weighted graph $R_0$ admitting a $(b',a')$ -partition, where $a'=a-1$ and $b'=b-|B_k|$ .

Set $t'=a'+b'+1=t-|B_k|-1$ , which is odd, and set $r=(t'-1)/2$ . Let $R'$ be the weighted graph obtained from R by replacing $V(R)-B_k$ with a set $B'$ of r vertices of weight $(1-w(B_k))/r$ each, and setting $w(v',v)=1$ for each $v'\in B'$ and $v\in V(R')\setminus \{v\}$ . That is, $R'[B']$ is a copy of $K_{r}^w$ scaled down by a factor of $1-w(B_k)$ . We note that $R'$ is ${ \mathcal {K}}_t$ -free. Indeed, if $(S_1,S_2)$ is a weighted clique in $R'$ , then $S_2$ contains at most one vertex from $B_k$ , so $|S_2|\leq r+1$ . Hence, $|S_1|+|S_2|\leq |V(R')|+r+1=|B_k|+2r+1=t-1$ .

Observe that $|B_k|\leq s-1$ by (A5), so $t'>s^2(s-1)/2$ . By part (1), we have

for all $m\leq s$ . Equality holds if and only if $R_0=K_r^w$ , (i.e., if and only if $|B_i|=1$ for each $i\neq k$ ). Hence, if $|B_j|>1$ for some $j\neq k$ , we have

contradicting the assumption that $d_{K_s}(R)=\pi _s({ \mathcal {K}}_t)$ . To complete the proof of (2), we recall that $|B_k|$ is even and $|B_k|\leq |B_i|+1$ for any $i\neq k$ by (A3). Hence, if $|B_i|=1$ for all $i\neq k$ then $|B_k|=2$ .

Proof. From Theorem 3.1(A2), R admits a $(b,a)$ partition for parameters $a\geq 1$ and $b\geq s$ satisfying $a+b=t-1=s+1$ . It is immediate that $a=1$ and $b=s$ .

Proof. By Theorem 3.1, R admits a $(b,a)$ -partition $B_1\cup \cdots \cup B_a$ such that $a+b=t-1$ . Moreover, combining (A5) with the inequality $t\geq 6$ , we conclude $a\geq 2$ and $|B_i|\leq 2$ for each i.

To show that $b=\lfloor t/2\rfloor =\max \{s,\lfloor t/2\rfloor \}$ when $t\geq 6$ , it suffices to show that all but at most one of the parts $B_i$ have cardinality 1. Equivalently, we must show that R does not contain disjoint edges $uv$ and $xy$ with $w(u,v)=w(x,y)=1/2$ .

Suppose the contrary. We may assume without loss of generality that $d_{K_3}(R,\{x,y\})\geq d_{K_3}(R,\{u,v\})$ . Let $R'$ be the graph obtained from R by setting $w(u,v)=0$ and $w(x,y)=1$ . We observe that $d_{K_3}(R',\{x,y\})=2d_{K_3}(R,\{x,y\})$ and $d_{K_3}(R',\{u,v\})=0$ , so

Moreover, $R'$ is ${ \mathcal {K}}_t$ -free, as the clique numbers of $R^{\prime }_{>\frac 12}$ and $R^{\prime }_{>0}$ satisfy

It follows that $R'$ is also an extremal ${ \mathcal {K}}_t$ -free weighted graph of minimal cardinality. However, this contradicts Theorem 3.1(A1), because $R'$ contains an edge of weight 0. We conclude that R cannot contain disjoint edges $uv$ and $xy$ of weight $1/2$ . This implies that all but at most one of the parts $B_i$ have cardinality 1, which is equivalent to showing that $b=\lfloor t/2\rfloor $ .

Proof. By Theorem 3.1, R admits a $(b,a)$ -partition $B_1\cup \cdots \cup B_a$ such that $a+b=t-1$ . Combining (A5) with the inequality $t\geq 7$ , we conclude that $a\geq 2$ and that each part has cardinality at most 3. If $t=7$ , then we must have $a=2$ and $b=4$ , because (A2) implies $b\geq 4$ . Henceforth, we assume $t\geq 8$ and show that $b=\lfloor t/2\rfloor $ . Equivalently, we must show that at most one of the parts $B_1,\ldots ,B_a$ has cardinality greater than 1.

Order the parts such that $3\geq |B_1|\geq |B_2|\geq \cdots \geq |B_a|\geq |B_1|-1$ ; the last inequality is a consequence of (A3). Suppose for the sake of contradiction that $|B_2|\geq 2$ . We split our proof into three cases, based on $|B_1|$ and $|B_2|$ . In each case, we derive a contradiction by constructing a ${ \mathcal {K}}_t$ -free weighted graph with larger $K_s$ -density than R; these constructions are given in Figure 1.

Figure 1 The optimizations in the proof of $s=4$ . Red edges have weight $1/2$ and black edges have weight 1.

Case 1: $|B_1|=|B_2|=3$ . In this case, the six vertices in $B_1$ and $B_2$ have the same weight p. Let $R_1$ be the weighted graph obtained from R by replacing $B_1\cup B_2$ with a set $C=\{c_1,c_2,c_3,c_4\}$ of four vertices with weight $3p/2$ each such that $w(c_i,c_j)=w(c_i,v)=1$ for any $i,j\in [4]$ and $v\in B_3\cup \cdots \cup B_a$ . We note that $(R_1)_{>\frac 12}$ and $(R_1)_{>0}$ have clique numbers satisfying

so $R_1$ is ${ \mathcal {K}}_t$ -free. Additionally, one computes that

We conclude that

contradicting the extremality of R.

Case 2: $|B_1|=3$ and $|B_2|=2$ . Let p and q be the weights of vertices in $B_1$ and $B_2$ respectively; by (A4), we have $p\leq q$ . Let $R_2$ be the weighted graph obtained from R by replacing $B_1$ with a set $C=\{c_1,c_2\}$ of two vertices with weight $3p/2$ each such that $w(c_1,c_2)=w(c_i,v)=1$ for any $i\in [2]$ and $v\in B_2\cup \cdots \cup B_a$ . We note that $(R_2)_{>\frac 12}$ and $(R_2)_{>0}$ have clique numbers satisfying

so $R_2$ is ${ \mathcal {K}}_t$ -free. One computes that

We now claim that

is positive for $2\leq m\leq 4$ . This is immediate for $m=2$ . For $m=3,4$ , only the $r=3$ term is negative, and one may check that the sum of the $r=2$ and $r=3$ terms is positive via the relations

and $d_{K_0}(R[B_2])\leq \frac 2p d_{K_1}(R[B_2])\leq \left (\frac 2p\right )^2d_{K_2}(R[B_2])$ . Thus, $d_{K_m}(R_2[C\cup B_2])>d_{K_m}(R[B_1\cup B_2])$ for $2\leq m \leq 4$ . It follows that

contradicting the extremality of R.

Case 3: $|B_1|=|B_2|=2$ . In this case, the four vertices in $B_1$ and $B_2$ have the same weight p. Moreover, we have $a\geq 3$ because $t\geq 8$ . Let $R_3$ be the weighted graph obtained from R by replacing $B_1\cup B_2$ with a set $C=\{c_1,c_2,c_3\}$ of three vertices with weight $4p/3$ each such that $w(c_i,c_j)=w(c_i,v)=1$ for any $i,j\in [3]$ and $v\in B_3\cup \cdots \cup B_a$ . We note that $(R_3)_{>\frac 12}$ and $(R_3)_{>0}$ have clique numbers satisfying

so $R_3$ is ${ \mathcal {K}}_t$ -free. One computes that

We now claim that

is positive for $2\leq m\leq 4$ . This is immediate for $m=2,3$ . For $m=4$ , only the $r=4$ term is negative. By (A4), we have $d_{K_1}(R[B_3])=\sum _{v\in B_3}w(v)\geq p$ , so

and thus $d_{K_m}(R_3[C\cup B_3])>d_{K_m}(R[B_1\cup B_2\cup B_3])$ for $2\leq m \leq 4$ . Therefore, setting $B=B_1\cup B_2\cup B_3$ , we have

contradicting the extremality of R.

Proof. First suppose t is odd (i.e., $t=2r+1$ for some integer $r\geq s$ ). Conjecture 1.2 hypothesizes that $\pi _s({ \mathcal {K}}_t)$ is attained by the complete balanced weighted graph $K_r^w$ , which has r vertices of weight $1/r$ each and has all edge weights equal to 1. Let $R'$ be a weighted graph on $(r+1)$ vertices with two disjoint edges of weight $1/2$ and all other edge weights equal to 1, such that the four vertices incident to weight- $1/2$ edges have weight $3/4r$ and the remaining $r-3$ vertices have weight $1/r$ . It is straightforward to check that $R'$ is ${ \mathcal {K}}_t$ -free. Moreover, because $s\leq r\leq 1.04s$ , we have

if $s\leq r$ is sufficiently large.

Next, suppose t is even (i.e., $t=2r$ for some integer $r\geq s$ ). Conjecture 1.2 hypothesizes that $\pi _s({ \mathcal {K}}_t)$ is attained by a weighted graph $R_1$ on r vertices with exactly one edge of weight $1/2$ . By Lemma 3.2, we have that

Let $R_2$ be a weighted graph on $(r+1)$ vertices with three disjoint edges of weight $1/2$ and all other edge weights equal to 1, such that the six vertices incident to weight-1/2 edges have weight $5/6r$ and the remaining $r-5$ vertices have weight $1/r$ . It is straightforward to check that $R_2$ is ${ \mathcal {K}}_t$ -free. Moreover, because $s\leq r\leq 1.04s$ , we have

if $s\leq r$ is sufficiently large.