Proof. For a $k$ -tuple $Q \in \!\left(\substack{B\\[3pt] k } \right)$ , let $\textrm{ext}(Q)$ denote the number of common neighbours of $Q$ in $A$ . We double-count the number of stars $K_{1,k}$ in $H$ whose central vertex is in $A$ to find that

\begin{equation*} \sum _{Q \in \left (\substack{ B\\[3pt] k }\right )} \textrm {ext}(Q) = \sum _{a \in A} \!\left (\begin {array}{c} \deg\!(a)\\[-3pt] k \end {array}\right )\geq \lvert A\rvert \left (\begin {array}{c} {\frac 1{\lvert A\rvert }\sum _{a \in A} \deg\!(a)}\\[-3pt] k \end {array}\right ) \geq \lvert A\rvert \left (\begin {array}{c} {d \lvert B\rvert }\\[-3pt] k \end {array}\right ), \end{equation*}

where the first inequality follows from convexity. If we split the left-hand side into a sum over tuples $Q$ which are non-edges of $\mathcal{H}$ , a sum over tuples $Q$ that are edges of $\mathcal{H}$ with fewer than $\zeta \lvert A\rvert$ extensions, and the remainder, we find that

\begin{align*} \lvert A\rvert \left (\begin{array}{c}{d \lvert B\rvert }\\[-3pt] k \end{array}\right ) &\leq \sum _{Q \notin E(\mathcal{H})} \textrm{ext}(Q) + \sum _{\substack{Q \in E(\mathcal{H})\\ \textrm{ext}(Q) \lt \zeta \lvert A\rvert }} \textrm{ext}(Q) + \sum _{\substack{Q \in E(\mathcal{H})\\ \textrm{ext}(Q) \geq \zeta \lvert A\rvert }} \textrm{ext}(Q)\\[5pt] &\leq \zeta \lvert A\rvert \left (\begin{array}{c}{\lvert B\rvert }\\[-3pt] k \end{array}\right ) + \zeta \lvert A\rvert \left (\begin{array}{c}{\lvert B\rvert }\\[-3pt] k \end{array}\right ) + \lvert A\rvert \lvert \{Q \in E(\mathcal{H})\;:\;\textrm{ext}(Q) \geq \zeta \lvert A\rvert \}\rvert . \end{align*}

Therefore, the number of edges of $\mathcal{H}$ with at least $\zeta \lvert A\rvert$ common neighbours is at least $ \left(\substack{d \lvert B\rvert \\[3pt] k } \right) - 2\zeta \!\left(\substack{\lvert B\rvert \\[2pt] k} \right)$ . We note that

\begin{equation*} \frac {\left (\substack{{d \lvert B\rvert }\\[3pt] k }\right )}{\left (\substack{{\lvert B\rvert }\\[3pt] k }\right )} = \frac {d \lvert B\rvert }{\lvert B\rvert } \cdot \frac {d \lvert B\rvert -1}{\lvert B\rvert -1}\dotsb \frac {d \lvert B\rvert - (k-1)}{\lvert B\rvert -(k-1)} \geq \!\left (\frac d2\right )^k =2^k \zeta, \end{equation*}

where we used our assumption that $\lvert B\rvert \geq 2k/d$ . Thus, the number of edges of $\mathcal{H}$ with at least $\zeta \lvert A\rvert$ common neighbours in $A$ is at least

\begin{equation*} \left (\begin {array}{c} {d \lvert B\rvert }\\[-3pt] k \end {array}\right ) - 2\zeta \!\left (\begin {array}{c} {\lvert B\rvert }\\[-3pt] k \end {array}\right ) \geq (2^k \zeta - 2\zeta ) \left (\begin {array}{c} {\lvert B\rvert }\\[-3pt] k \end {array}\right ) \geq \zeta \!\left (\begin {array}{c} {\lvert B\rvert }\\[-3pt] k \end {array}\right ). \end{equation*}

Proof of Theorem 1.2. Fix some $k \geq 2$ , $\theta \in (0,1)$ , and a red/blue colouring of $E(K_N)$ . Let $\sigma =(\theta/(12k^4))^k$ , $M_0=k$ , $\delta =\sigma ^2$ , and $\eta = \delta ^{k^2}$ . Let $M=M(\eta,M_0)$ be the parameter from Lemma 2.1 and let $c=\delta ^{k^2}/M^4$ and $\gamma =\delta ^{2k^2}/M^{2k}$ . We apply Lemma 2.1 to the red graph in our colouring with parameters $M_0$ and $\eta$ . This yields an equitable partition $V(K_N)=V_1 \sqcup \dotsb \sqcup V_m$ with $M_0 \leq m \leq M$ such that each part $V_i$ is $\eta$ -regular in red and, for each $i$ , there are at most $\eta m$ values of $j \neq i$ for which $(V_i,V_j)$ is not $\eta$ -regular in red.

First suppose that some part, say $V_1$ , has internal red density at least $\delta$ . By the counting lemma, Lemma 2.2, we see that $V_1$ contains at least $\frac{1}{(k+1)!}\Big(\delta ^{\left (\substack{k+1\\[3pt] 2} \right )}- \eta \!\left (\substack{ k+1\\[3pt] 2} \right )\Big)|V_1|^{k+1}$ red $K_{k+1}$ . Since each red $K_{k+1}$ contains exactly $k+1$ red $K_k$ , this implies that an average $k$ -tuple of vertices in $V_1$ lies in at least

\begin{equation*} \frac {\frac {k+1}{(k+1)!}\left(\delta ^{\left (\substack{ k+1\\[3pt] 2 }\right )}- \eta \!\left (\begin {array}{c} k+1\\[-2pt] 2 \end {array}\right )\right)|V_1|^{k+1}}{\left (\begin {array}{c} {\lvert V_1\rvert }\\[-3pt] k \end {array}\right ) }\geq \!\left (\delta ^{\left (\substack{k+1\\[3pt] 2 }\right )}-\eta \!\left (\begin {array}{c} k+1\\[-3pt] 2 \end {array}\right )\right ) |V_1|\;=\!:\; \xi M |V_1| \end{equation*}

red $K_{k+1}$ . That is, if we pick a uniformly random $k$ -tuple of vertices from $V_1$ , then the expected number of red $K_{k+1}$ containing it is at least $\xi M \lvert V_1\rvert$ . If we also define $\kappa = \Big(\delta ^{ \left(\substack{k\\[3pt] 2}\right)}- \eta \!\left(\substack{k\\[3pt] 2 } \right)\Big)/(k! M^k)$ , then Lemma 2.2 implies that $V_1$ contains at least $\kappa N^k$ red $K_k$ , with an average one having at least $\xi N$ extensions to a red $K_{k+1}$ , where we use the fact that $|V_1| \geq N/M$ since the partition is equitable and has $m \leq M$ parts. If we now set $\nu =\xi/2$ and apply Lemma 2.4, we conclude that $V_1$ contains at least $(\xi \kappa/2) N^k$ red $K_k$ , each with at least $(\xi/2)N$ extensions to a red $K_{k+1}$ . By our choice of parameters,

\begin{equation*} \xi = \frac 1M \!\left (\delta ^{\left (\substack{ k+1\\[3pt] 2 }\right )}-\eta \!\left (\begin {array}{c} k+1\\[-3pt] 2 \end {array}\right )\right ) \geq \frac {\delta ^{k^2}}{2M} \end{equation*}

and, therefore, $\xi/2 \geq c/k +\gamma$ . Similarly, $\kappa \geq \delta ^{k^2}/M^k$ and, therefore, $\xi \kappa/2 \geq \gamma$ . Thus, we find that in this case the colouring contains $(c,\gamma )$ -many books.

Therefore, we may assume that all $V_i$ have $d_R(V_i) \lt \delta$ . We build a reduced graph $G$ with vertex set $v_1,\ldots,v_m$ and declare $\{v_i, v_j\} \in E(G)$ if $(V_i,V_j)$ is $\eta$ -regular and $d_R(V_i,V_j) \geq \delta$ . Suppose that some vertex of $G$ , say $v_1$ , has degree less than $(1- \frac 1k - \sigma )m$ . Since at most $\eta m$ non-neighbours of $v_1$ can come from irregular pairs, we find that $d_B(V_1,V_i) \geq 1- \delta$ for at least $(\frac 1k+\sigma -\eta )m$ choices of $i \in [m]$ . Let $I \subseteq [m]$ be the set of such $i$ . Since $d_B(V_1) \geq 1-\delta$ and $\delta \leq 1/k^2$ , we see that, for $\alpha =k \delta/4$ ,

\begin{equation*} d_B(V_1)^{\left (\substack{ k\\[3pt] 2 }\right )} \geq (1- \delta )^{\left (\substack{ k\\[3pt] 2 }\right )} \geq 1 -\left (\begin {array}{c} k\\[-3pt] 2 \end {array}\right ) \delta \geq \alpha . \end{equation*}

Moreover, we have that $\eta \lt \alpha ^3/k^2$ by our choice of $\eta$ . Therefore, we may apply Lemma 2.3, which implies that if $Q$ is a randomly chosen blue $K_k$ in $V_1$ and $u$ is some vertex in $K_N$ , then $\mathbb{P}(u\text{ extends }Q) \geq d(u,V_1)^k-4 \alpha$ . In particular, if we sum this up over all $u \in \bigcup _{i \in I} V_i$ , we find that the expected number of blue extensions of $Q$ is at least

\begin{equation*} \sum _{i \in I} \sum _{u \in V_i} (d(u,V_1)^k - 4 \alpha ) \geq |I| \frac Nm \!\left ( (1- \delta )^k-4 \alpha \right ) \geq \!\left ( \frac 1k+\sigma -\eta \right )\left ( (1- \delta )^k-4 \alpha \right ) N, \end{equation*}

where the first inequality follows from the convexity of the function $x \mapsto x^k$ . Using $\eta \lt \sigma/2$ , we have that

\begin{align*} \left ( \frac 1k+\sigma -\eta \right )\left ( (1- \delta )^k-4 \alpha \right ) &\geq \!\left ( \frac 1k +\frac \sigma 2\right )(1-2k \delta ) \geq \frac 1k +\frac \sigma 2 - 2k \delta, \end{align*}

where the last step follows from the bound $1/k+\sigma \leq 1/k+1/k \leq 1$ . By our choice of $\delta =\sigma ^2 \leq \sigma/(8k)$ , we see that the expected number of blue extensions of $Q$ is at least $(\frac 1k+\frac \sigma 4)N$ . Moreover, by Lemma 2.2, the number of choices for $Q$ is at least $\frac{1}{k!} \Big((1-\delta )^{\left (\substack{k\\[3pt] 2 }\right )}-\eta \!\left (\substack{k\\[3pt] 2} \right )\!\Big)(N/M)^k \geq \kappa N^k$ . Therefore, if we apply Lemma 2.4 with parameters $\kappa$ , $\xi =\frac 1k+\frac \sigma 4$ , and $\nu =\frac 1k+\gamma$ , then we find that the colouring contains $(c,\gamma )$ -many books.

Therefore, we may assume that every vertex in $G$ has degree greater than $(1- \frac 1k -\sigma )m$ , so, by Theorem 3.2 and the fact that $\sigma \lt 1/(3k^2-k)$ , we see that either $G$ contains a $K_{k+1}$ or $G$ is $k$ -partite. Assume first that there is a $K_{k+1}$ in $G$ . By relabelling the vertices, we may assume that $v_1,\ldots,v_{k+1}$ form a clique. By the counting lemma, Lemma 2.2, we have that $V_1,\ldots,V_k$ span at least $\Big(\delta ^{ \left(\substack{k\\[3pt] 2 } \right)}- \eta \left(\substack{k\\[3pt] 2 } \right)\Big)(N/m)^k \geq \kappa N^k$ red $K_k$ and $V_1,\ldots,V_{k+1}$ span at least $\Big(\delta ^{ \left(\substack{k+1\\[3pt] 2 } \right)} - \eta \left(\substack{k+1\\[3pt] 2} \right)\Big)(N/m)^{k+1}$ red $K_{k+1}$ . Every such red $K_{k+1}$ contains exactly one red $K_k$ with one vertex in each of $V_1,\dots,V_k$ , so an average $k$ -tuple $(v_1,\dots,v_k) \in V_1 \times \dotsb \times V_k$ lies in at least

\begin{equation*} \frac {\left(\delta ^{\left (\substack{k+1\\[3pt] 2 }\right )} - \eta \!\left (\begin {array}{c} k+1\\[-3pt] 2 \end {array}\right )\right)(N/m)^{k+1}}{(N/m)^k} \geq \!\left (\delta ^{\left (\substack{k+1\\[3pt] 2 }\right )} - \eta \!\left (\begin {array}{c} k+1\\[-3pt] 2 \end {array}\right )\right ) \frac NM = \xi N. \end{equation*}

Thus, we have a set of at least $\kappa N^k$ red $K_k$ with at least $\xi N$ extensions on average and so, applying Lemma 2.4 as before, our colouring has $(c,\gamma )$ -many books.

Thus, we may assume that $G$ is $k$ -partite. Let this $k$ -partition of $V(G)$ be $A_1\sqcup \dotsb \sqcup A_k$ . Note that $\lvert A_\ell \rvert \leq \left(\frac 1k+\sigma \right)m$ for every $\ell$ , since the minimum degree of $G$ is at least $(1-\frac 1k - \sigma )m$ and each $A_\ell$ is an independent set in $G$ . This in turn implies that $\lvert A_\ell \rvert \geq \left(\frac 1k-k\sigma \right)m$ for every $\ell$ , since $\lvert A_\ell \rvert = m -\sum _{\ell ' \neq \ell }\lvert A_{\ell '}\rvert \geq (\frac 1k-k\sigma )m$ . We lift this partition to a partition of the vertices of $K_N$ into $k$ parts $X_1,\dots,X_k$ by letting $X_\ell = \bigcup _{v_i \in A_\ell }V_i$ , noting that our observations above imply that $\lvert X_\ell \rvert = (\frac 1k \pm k\sigma )N$ for all $\ell$ . We claim that each $X_\ell$ contains at most $\frac{3\delta }2 \!\left (\substack{\lvert X_\ell\rvert \\[3pt] 2} \right )$ red edges. Indeed, observe that if $v_i,v_j$ are two (not necessarily distinct) vertices of $G$ that are in the same part $A_\ell$ , then they must be non-adjacent in $G$ . This means that either $(V_i,V_j)$ is an irregular pair or $d_R(V_i,V_j)\lt \delta$ . There are at most $\eta m^2$ irregular pairs, so the irregular pairs can contribute at most $\eta N^2\leq 4k^2 \eta \lvert X_\ell \rvert ^2 \leq 10 k^2 \eta \!\left(\substack{\lvert X_\ell\rvert \\[3pt] 2 }\right)$ red edges inside $X_\ell$ , where we used that $\lvert X_\ell \rvert \geq (\frac 1k-k\sigma )N \geq N/(2k)$ . All other pairs of parts inside each $X_\ell$ have red density at most $\delta$ , so the total number of red edges inside $X_\ell$ is at most $\delta \!\left (\substack{\lvert X_\ell\rvert \\[3pt] 2 }\right ) + 10k^2 \eta \!\left (\substack{\lvert X_\ell\rvert \\[3pt] 2 }\right ) \leq \frac{3\delta }2 \!\left (\substack{\lvert X_\ell\rvert \\[3pt] 2 }\right )$ , since $\eta \leq \delta/(20k^2)$ . This implies that the number of ordered pairs of (not necessarily distinct) vertices in $X_\ell$ which do not form a blue edge is at most $2 \delta \lvert X_\ell \rvert ^2$ .

This already implies that the red graph can be made $k$ -partite by recolouring at most $2\delta N^2$ edges, so it only remains to show that by recolouring a small number of additional edges, we can make the red graph balanced complete $k$ -partite. For this, suppose that $d_B(X_1,X_2) \geq \theta/k^2$ . If we sample (with repetition) a random $k$ -tuple $Q$ of vertices from $X_2$ , then the probability that it does not form a blue clique is at most $\left(\substack{ k\\[3pt] 2 } \right)\cdot 2 \delta \leq k^2 \delta$ , since each random pair of vertices does not span a blue edge with probability at most $2\delta$ . Moreover, the expected number of common blue neighbours of $Q$ inside $X_2$ is at least $(1-2\delta )^k \lvert X_2\rvert - k\geq (1-2k\delta )\lvert X_2\rvert$ , by convexity. By applying Markov’s inequality as in the proof of Lemma 2.4, the probability that $Q$ has fewer than $(1-\sqrt \delta )\lvert X_2\rvert$ common blue neighbours in $X_2$ is at most $2k\sqrt \delta$ . Therefore, the probability that $Q$ is a blue clique with at least $(1-\sqrt \delta )\lvert X_2\rvert$ common blue neighbours in $X_2$ is at least $1-k^2 \delta -2k\sqrt \delta \geq 1-3k\sqrt \delta$ , since $\sqrt \delta \leq 1/k$ . Let $\mathcal{H}$ be the $k$ -uniform hypergraph with vertex set $X_2$ whose edges are all blue $K_k$ in $X_2$ with at least $(1-\sqrt \delta )\lvert X_2\rvert$ common blue neighbours in $X_2$ . Then $\mathcal{H}$ has at least $(1-3k\sqrt \delta ) \left(\substack{\lvert X_2\rvert \\[3pt] k } \right)$ edges.

We now apply Lemma 3.3 to the hypergraph $\mathcal{H}$ and to the bipartite graph of blue edges between $X_1$ and $X_2$ , which has edge density $d\geq \theta/k^2$ by assumption. We have that $(d/4)^k \geq (\theta/(4k^2))^k \geq 3k\sigma = 3k\sqrt \delta$ by our choice of $\sigma$ and $\delta$ , so we may indeed apply Lemma 3.3 to conclude that at least $(\theta/(4k^2))^k \left(\substack{\lvert X_2\rvert \\[3pt] k } \right)$ of the edges of $\mathcal{H}$ have at least $(\theta/(4k^2))^k\lvert X_1\rvert$ common blue neighbours in $X_1$ . This yields at least

\begin{equation*} \left (\frac \theta {4k^2}\right )^k \left (\begin {array}{c} {\lvert X_2\rvert }\\[-3pt] k \end {array}\right ) \geq \!\left (\frac {\theta \lvert X_2\rvert }{4k^3}\right )^k \geq \!\left (\frac {\theta }{8k^4}\right )^k N^k \geq \gamma N^k \end{equation*}

blue $K_k$ , each of which has at least

\begin{align*} (1-\sqrt \delta )\lvert X_2\rvert + \left (\frac \theta{4k^2}\right )^k \lvert X_1\rvert &\geq \!\left (1-\sqrt \delta + \left (\frac \theta{4k^2}\right )^k\right ) \left (\frac 1k -k\sigma \right )N\\[5pt] &\geq \!\left (\frac 1k + \left (\frac \theta{4k^3}\right )^k -\sqrt \delta - 2k\sigma \right )N\\[5pt] &\geq \!\left (\frac 1k + \left (\frac \theta{4k^3}\right )^k - 3k\sigma \right )N\\[5pt] &\geq \!\left (\frac 1k+\gamma \right )N \end{align*}

extensions to a blue $K_{k+1}$ , where in both computations we used the fact that $\lvert X_1\rvert,\lvert X_2\rvert \geq (\frac 1k - k\sigma )N\geq N/(2k)$ , as well as our choices of $\sqrt \delta = \sigma = (\theta/(12k^4))^k$ . Thus, in this case, we have again found $(c,\gamma )$ -many books, a contradiction.

Hence, we may assume that $d_B(X_1,X_2) \lt \theta/k^2$ . By the same argument, all the blue densities between different parts $X_\ell$ can be assumed to be at most $\theta/k^2$ . Since we have already argued that the red density inside each part is at most $2\delta$ , we see that, by recolouring at most $\left( \left(\substack{k\\[3pt] 2 } \right)\theta/k^2 + 2k\delta \right)N^2$ edges, we can make the red graph complete $k$ -partite. Finally, we recall that each part $X_\ell$ has size $\lvert X_\ell \rvert = (\frac 1k \pm k\sigma )N$ . Therefore, by moving at most $k^2\sigma N$ arbitrary vertices into another part, we see that we can make our partition equitable. We then recolour the edges incident with any moved vertex to obtain a balanced complete $k$ -partite red graph. Doing so entails recolouring at most $k^2 \sigma N^2$ additional edges. Thus, in total, we recolour at most

\begin{equation*} \left (\left (\begin {array}{c} k\\[-3pt] 2 \end {array}\right ) \frac \theta {k^2} + 2k\delta + k^2 \sigma \right )N^2 \leq \left (\frac \theta 2 + 3k^2 \sigma \right ) \leq \theta N^2 \end{equation*}

edges, where we used that $\delta \leq \sigma$ and $\sigma \leq (\theta/(12k^4))^k \leq \theta/(6k^2)$ .

Proof. First suppose that $x_j \leq \frac 1k$ for some $j \in [k]$ . Then we have that

\begin{equation*} \frac {(1-p)^{1-k}}{k} \sum _{i=1}^k (1-x_i)^k \geq \frac {(1-p)^{1-k}}{k}(1-x_j)^k \geq \frac {(1-p)^{1-k}}{k} \!\left ( 1- \frac 1k \right ) ^k \geq \frac {(1-p)^{1-k}}{e^2k} \;=\!:\; f(p,k), \end{equation*}

where we used the inequality $1-x \geq e^{-2x}$ for $x \in [0,\frac 12]$ . If $p \geq 1-5/(4e)$ , then $1-p \leq 5/(4e)$ , so $f(p,k) \geq (4/5)^{k-1} e^{k-3}/k$ . Once $k \geq 6$ , this last expression is at least $1$ , so in the case where $p \geq 1-5/(4e)$ , we may take $k_1(p)=6$ .

For $p\lt 1-5/(4e)$ , let $\lambda =\lambda (p) = \log \frac 1{1-p}$ and

\begin{equation*} k_1(p)=1+\frac {5 - \log \lambda + \log \log \frac 1\lambda }{\lambda }= 1+\frac {5 - \log \log \frac 1{1-p} + \log \!\left(\!-\!\log \log \frac 1{1-p}\right)}{\log \frac 1{1-p}}. \end{equation*}

We now claim that

(2) \begin{equation} f(p,k) \geq 1\qquad \text{ if }k \geq k_1(p). \end{equation}

By differentiating, we see that $f(p,k)$ is monotonically increasing in $k$ for $k \geq 1/\log \frac 1{1-p} = 1/\lambda$ . Since $p\lt 1-5/(4e)$ , we have that $5-\log \lambda +\log \log \frac 1\lambda \gt 1$ and so we are in the monotonicity regime. It therefore suffices to prove the statement for $k=k_1(p)$ . Note now that

\begin{equation*} (1-p)^{1-k_1(p)} = (1-p)^{(5-\log \lambda + \log \log \frac 1 \lambda )/\log (1-p)} = \frac {e^5 \log \frac 1 \lambda }{\lambda } \end{equation*}

and let $g(p)=1 + \lambda/\log \frac 1 \lambda + 5/\log \frac 1\lambda + \log \log \frac 1\lambda/\log \frac 1\lambda$ . Then we have

\begin{equation*} f(p,k_1(p)) = \frac {e^5 \log \frac 1 \lambda }{\lambda } \cdot \frac 1{e^2 k_1(p)} =\frac { e^3 \log \frac 1 \lambda }{\lambda +5+\log \frac 1 \lambda +\log \log \frac 1 \lambda } = \frac {e^3}{g(p)}. \end{equation*}

Thus, to prove that $f(p,k_1(p))\geq 1$ , it suffices to prove that $g(p) \leq e^3$ for all $p \lt 1-5/(4e)$ . By differentiating, one can check that $g(p)$ is monotonically increasing in $p \in [0,1-5/(4e)]$ . Thus, it suffices to check that $g(1-5/(4e)) \leq e^3$ . But $g(1-5/(4e))\approx 18.4 \lt e^3$ , so $f(p,k_1(p))\geq 1$ , as claimed. Hence, from now on, we may assume that all the $x_i$ are in $\big(\frac 1k,1\big]$ .

For the moment, let’s assume that all the $x_i$ are in $(\frac 1k,1)$ . Note that the function $\varphi \;:\;y \mapsto (1-e^y)^k$ is strictly convex on the interval $(\!\log \frac 1k,0)$ . By the multiplicative Jensen inequality, Lemma 2.5, this implies that, subject to the constraint $\prod _{i=1}^k x_i=z$ , the term $\frac 1k \sum _{i=1}^k \!(1-x_i)^k$ is minimized when all the $x_i$ are equal to $z^{1/k}$ . Therefore,

\begin{equation*} p^{1-k} \prod _{i=1}^k x_i+\frac {(1-p)^{1-k}}{k} \sum _{i=1}^k (1-x_i)^k \geq p^{1-k}z+(1-p)^{1-k}(1-z^{1/k})^k. \end{equation*}

So it suffices to minimize this expression as a function of $z$ . Changing variables to $w=z^{1/k}$ , it suffices to minimize

\begin{equation*} \psi (w)=p^{1-k} w^k+(1-p)^{1-k} (1-w)^k \end{equation*}

as a function of $w$ . By differentiating, we find that $\psi$ is minimized at $w=p$ , where $\psi (p)=1$ . This proves the desired result as long as all the $x_i$ are in $[0,1)$ . By continuity, the result then extends to all $x_i \in [0,1]$ .

Proof. We tackle the two cases separately: suppose first that the colouring has a $(k,\eta,\delta )$ -red-blocked configuration, say $C_1,\ldots,C_k$ . By the counting lemma, Lemma 2.2, we know that the number of blue $K_k$ with one vertex in each $C_i$ is at least

\begin{align*} \left ( \prod _{1 \leq i\lt j \leq k} d_B(C_i,C_j)-\eta \!\left (\begin{array}{c} k\\[-3pt] 2 \end{array}\right ) \right ) \prod _{i=1}^k |C_i|&\geq \!\left ( \delta ^{\left (\substack{ k\\[3pt] 2 }\right )}- \eta \left (\begin{array}{c} k\\[-3pt] 2 \end{array}\right ) \right ) \prod _{i=1}^k |C_i| \gt 0, \end{align*}

so there is at least one blue $K_k$ with one vertex in each of $C_1,\ldots,C_k$ . By a similar computation, we see that each $C_i$ contains at least one red $K_k$ .

For a vertex $v$ and $i \in [k]$ , let $x_i(v)\;:\!=\; d_B(v,C_i) \in [0,1]$ . We observe that from the definition in Lemma 4.1, we have that $k_1(1-p) \geq k_1(p)$ for all $p \geq \frac 12$ . Therefore, Lemma 4.1 implies that since $k \geq k_2 \geq k_1(p)$ , we have that

\begin{equation*} p \!\left ( p^{-k}\prod _{i=1}^k x_i(v) \right ) +(1-p) \left ( \frac {(1-p)^{-k}}{k}\sum _{i=1}^k (1-x_i(v))^k \right ) \geq 1 \end{equation*}

for all $v \in V$ . Summing this fact up over all $v$ , we find that

(3) \begin{equation} p \!\left ( p^{-k}\sum _{v \in V}\prod _{i=1}^k x_i(v) \right ) +(1-p) \left ( \frac{(1-p)^{-k}}{k}\sum _{i=1}^k \sum _{v \in V} (1-x_i(v))^k \right ) \geq N. \end{equation}

This says that a $(p,1-p)$ -weighted average of two numbers is at least $N$ , which means that at least one of them is at least $N$ . Suppose first that the first term is at least $N$ , that is, that

\begin{equation*} \sum _{v \in V} \prod _{i=1}^k x_i(v) \geq p^k N. \end{equation*}

Let $Q$ be a uniformly random blue $K_k$ spanning $C_1,\ldots,C_k$ , which must exist by our computations above. Let $\alpha =\delta ^{k^2} \leq \prod _{i\lt j} d_B(C_i,C_j)$ and observe that $\eta \leq \delta ^{4k^2} = \alpha ^4 \leq \alpha ^3/k^2$ . Thus, for any $v$ , we can apply Lemma 2.3 to conclude that the probability $v$ extends $Q$ to a blue $K_{k+1}$ is at least $\prod _i x_i(v)-4 \alpha$ . Therefore, the expected number of extensions of $Q$ to a blue $K_{k+1}$ is at least

(4) \begin{align} \sum _{v \in V} \!\left ( \prod _{i=1}^k x_i(v)-4 \alpha \right ) &\geq (p^k-4 \alpha )N\notag \\[5pt] &\geq (p^k-4 \alpha )(p^{-k}+\varepsilon )n\notag \\[5pt] &\geq (1+p^k \varepsilon -8 \alpha p^{-k}) n\notag \\[5pt] &\geq n, \end{align}

where (4) uses that $\alpha =\delta ^{k^2} \leq ((1-p)\varepsilon )^{k^2}\leq (p \varepsilon )^{k^2} \leq p^{2k} \varepsilon/8$ . Therefore, $Q$ has at least $n$ extensions in expectation, so there must exist some blue $K_k$ with at least $n$ extensions, that is, a blue $B_n ^{(k)}$ .

Now assume that the other term in the weighted average in (3) is at least $N$ , that is, that

\begin{equation*} \frac 1k \sum _{i=1}^k \sum _{v \in V} (1-x_i(v))^k \geq (1-p)^{k}N. \end{equation*}

Then there must exist some $i$ for which

\begin{equation*} \sum _{v \in V} (1-x_i(v))^k \geq (1-p)^{k} N. \end{equation*}

Therefore, if $Q$ is a random red $K_k$ inside this $C_i$ , then, by Lemma 2.3, the expected number of extensions of $Q$ is at leastFootnote ³

(5) \begin{align} \sum _{v \in V} \!\left [ (1-x_i(v))^k-4 \alpha \right ] &\geq \left [(1-p)^k-4 \alpha \right ](p^{-k}+\varepsilon )n\notag \\[5pt] &\geq \!\left (\left ( \frac{1-p}{p} \right ) ^k+(1-p)^k \varepsilon -8 \alpha p^{-k}\right )n\notag \\[5pt] &\geq cn, \end{align}

where we use the fact that $c=((1-p)/p)^k$ and that

\begin{equation*} \alpha =\delta ^{k^2}=((1-p) \varepsilon )^{k^2} \leq p^k (1-p)^k \varepsilon/8, \end{equation*}

since $1-p \leq p$ . Thus, the expected number of red extensions of a red $K_k$ in $C_i$ is at least $cn$ , so there must exist a red $B_{cn}^{(k)}$ . This concludes the proof under the assumption that the colouring contains a $(k,\eta,\delta )$ -red-blocked configuration.

Now, we instead assume that the colouring contains a $(k,\eta,\delta )$ -blue-blocked configuration and aim to conclude the same result; the proof is more or less identical, but with the role of $p$ now played by $q=1-p$ . As before, we find that there is at least one red $K_k$ spanning $C_1,\ldots,C_k$ and that each $C_i$ contains at least one blue $K_k$ . For a vertex $v$ and $i \in [k]$ , let $y_i(v)=d_R(v,C_i) \in [0,1]$ and write $q=1-p$ . Since $k \geq k_2 = k_1(q)$ , we can sum the result of applying Lemma 4.1 over all $v \in V$ to find that

\begin{equation*} q \!\left ( q^{-k} \sum _{v \in V} \prod _{i=1}^k y_i(v) \right ) +(1-q) \left ( \frac {(1-q)^{-k}}{k}\sum _{i=1}^k \sum _{v \in V} (1-y_i(v))^k \right ) \geq N. \end{equation*}

As before, this is a $(q,1-q)$ -weighted average of two terms, which means that one of the terms must be at least $N$ . Suppose first that the first term is at least $N$ , that is, that

\begin{equation*} \sum _{v \in V} \prod _{i=1}^k y_i(v) \geq q^k N. \end{equation*}

If $Q$ is a uniformly random red $K_k$ spanning $C_1,\ldots,C_k$ and $\alpha =\delta ^{k^2}$ , then, as before, we find that the expected number of extensions of $Q$ to a red $K_{k+1}$ is at least

\begin{equation*} \sum _{v \in V} \!\left ( \prod _{i=1}^k y_i(v)-4 \alpha \right ) \geq (q^k-4 \alpha ) N \geq ((1-p)^k- 4 \alpha )(p^{-k}+\varepsilon )n \geq cn, \end{equation*}

by the computation in (5). Therefore, in this case, there must exist some red $K_k$ with at least $cn$ red extensions, giving the desired red $B_{cn}^{(k)}$ . So we may assume instead that

\begin{equation*} \frac 1k \sum _{i=1}^k \sum _{v \in V} (1-y_i(v))^k \geq (1-q)^kN, \end{equation*}

which implies that for some $i \in [k]$ ,

\begin{equation*} \sum _{v \in V} (1-y_i(v))^k \geq p^k N. \end{equation*}

Thus, if $Q$ is a random blue $K_k$ inside this $C_i$ , we find that the expected number of blue extensions of $Q$ is at least

\begin{align*} \sum _{v \in V} \!\left [ (1-y_i(v))^k-4\alpha \right ] \geq (p^k-4 \alpha )N\geq (p^k-4 \alpha )(p^{-k}+\varepsilon )n \geq n, \end{align*}

by the same computation as in (4). This gives us our blue $B_n ^{(k)}$ , completing the proof.

Proof of Theorem 1.3. Given an integer $k \geq 2$ , let $c_1(k)$ be the infimum of $c \in (0,1]$ such that $k_2((c^{1/k}+1)^{-1}) \leq k$ , where $k_2$ is the constant from Lemma 4.3. Note that we declare this infimum to equal $1$ if no $c \in (0,1]$ satisfies this condition (as happens for $k=2$ ). In this case, there is nothing to prove, since Theorem 1.3 for $c=1$ is already known [Reference Conlon9]. We now fix $c\in [c_1,1]$ and $p = 1/(c^{1/k}+1) \in (\frac 12,1]$ , noting that we have $k \geq k_2(p)$ .

Fix $0\lt \varepsilon \lt \frac 12$ and suppose we are given a red/blue colouring of $E(K_N)$ where $N=(p^{-k}+\varepsilon )n$ . Our goal is to prove that if $n$ is sufficiently large in terms of $\varepsilon$ , then this colouring contains a red $B_{cn} ^{(k)}$ or a blue $B_n ^{(k)}$ . To do this, we fix parameters $\delta =(1-p)^{2k}\varepsilon/(4k)$ and $\eta =\min \{\delta ^{4k^2},(1-p)/(4k)\}$ depending on $c$ , $k$ , and $\varepsilon$ .

We apply Lemma 2.1 to the red graph from our colouring with parameters $\eta$ and $M_0=1/\eta$ to obtain an equitable partition $V(K_N)=V_1 \sqcup \dotsb \sqcup V_m$ , where each $V_i$ is $\eta$ -regular and, for each $i$ , there are at most $\eta m$ values $1 \leq j \leq m$ such that the pair $(V_i,V_j)$ is not $\eta$ -regular. Moreover, $M_0 \leq m \leq M=M(\eta,M_0)$ . Note that since the colours are complementary, the same properties also hold for the blue graph. Call a part $V_i$ blue if $d_B(V_i) \geq \frac 12$ and red otherwise.

Suppose first that at least $m' \geq pm$ of the parts are blue and rename the parts so that $V_1,\ldots,V_{m'}$ are these blue parts. We build a reduced graph $G$ whose vertex set is $v_1,\ldots,v_{m'}$ by making $\{v_i,v_j\}$ an edge if and only if $(V_i,V_j)$ is $\eta$ -regular and $d_R(V_i,V_j) \geq \delta$ . Suppose that some vertex in $G$ , say $v_1$ , has degree at most $(1-p^{k-1}-\eta/p)m'-1$ . Since $v_1$ has at most $\eta m \leq \eta m'/p$ non-neighbours coming from irregular pairs $(V_1,V_j)$ , this means that there are at least $p^{k-1} m'$ parts $V_j$ such that $(V_1,V_j)$ is $\eta$ -regular and $d_B(V_1,V_j) \geq 1- \delta$ . Let $J$ be the set of all these indices $j$ and $U=\bigcup _{j \in J}V_j$ be the union of all of these $V_j$ . We then have

(6) \begin{equation} e_B(V_1,U)=\sum _{j \in J} e_B(V_1,V_j) \geq \sum _{j \in J} (1- \delta ) |V_1||V_j|=(1- \delta ) |V_1||U|. \end{equation}

Let $V'_{\!\!1} \subseteq V_1$ denote the set of vertices $v \in V_1$ with $e_B(v,U) \geq (1-2 \delta )|U|$ . Then we may write

(7) \begin{equation} e_B(V_1,U)=\sum _{v \in V'_{\!\!1}} e_B(v,U)+\sum _{v \in V_1 \setminus V'_{\!\!1}} e_B(v,U) \leq |V'_{\!\!1}||U|+(1-2 \delta )|V_1 \setminus V'_{\!\!1}| |U|. \end{equation}

Combining inequalities (6) and (7), we find that $|V'_{\!\!1}| \geq \frac 12 |V_1|$ , where every vertex in $V'_{\!\!1}$ has blue density at least $1-2 \delta$ into $U$ . Moreover, since $\eta \lt \frac 16$ , we may apply the $\eta$ -regularity of $V_1$ to conclude that the internal blue density of $V'_{\!\!1}$ is at least $\frac 12- \eta \geq \frac 13$ , while the hereditary property of regularity implies that $V'_{\!\!1}$ is $2\eta$ -regular. Then the counting lemma, Lemma 2.2, implies that $V'_{\!\!1}$ contains at least

\begin{equation*} \frac {1}{k!}\!\left ( d_B(V'_{\!\!1})^{\left(\substack{ k\\[3pt] 2 }\right )}-2\eta \!\left (\begin {array}{c} k\\[-3pt] 2 \end {array}\right ) \right ) |V'_{\!\!1}|^k \geq \frac {1}{k!}\left ( 3^{-\left (\substack{ k\\[3pt] 2 }\right )}-2\eta \!\left (\begin {array}{c} k\\[-3pt] 2 \end {array}\right ) \right ) |V'_{\!\!1}|^k \gt 0 \end{equation*}

blue $K_k$ , so that $V'_{\!\!1}$ contains at least one blue $K_k$ . Every vertex of this blue $K_k$ has at least $(1-2 \delta )|U|$ blue neighbours in $U$ , so the blue $K_k$ has at least $(1-2 k \delta )|U|$ blue extensions into $U$ . Moreover, since we assumed that $|J| \geq p^{k-1} m' \geq p^{k} m$ and the partition is equitable, we find that $|U| \geq p^{k}N$ . Therefore,

\begin{align*} (1-2 k \delta )|U| &\geq (1-2 k \delta )p^k (p^{-k}+\varepsilon )n\\[5pt] &= (1-2 k \delta )(1+p^k \varepsilon )n\\[5pt] &\geq (1+p^k \varepsilon -4 k \delta )n\\[5pt] &\geq n, \end{align*}

since our choice of $\delta$ yields $4k \delta = (1-p)^{2k} \varepsilon \leq p^k \varepsilon$ . Thus, we find that any blue $K_k$ inside $V'_{\!\!1}$ must have at least $n$ blue extensions, giving us our blue $B_n ^{(k)}$ .

So we may assume that every vertex in $G$ has degree at least $(1-p^{k-1}-\eta/p)m'$ . Recall from (2) that $f(1-p,k) = p^{1-k}/(e^2k) \geq 1$ for $k \geq k_1(1-p)$ . Since we assume that $k \geq k_2(p) = k_1(1-p)$ , this implies that

(8) \begin{equation} p^{k-1} \leq \frac{1}{e^2k}\leq \frac 1{3(k-1)}. \end{equation}

Additionally, by our choice of $\eta \leq (1-p)/(4k) \leq p/(4k)$ , we know that

\begin{equation*} \frac \eta p \leq \frac {1}{3(k-1)}. \end{equation*}

The previous two inequalities imply that

\begin{equation*} 1-p^{k-1}-\frac \eta p \gt 1- \frac {1}{k-1}, \end{equation*}

so that $G$ contains a $K_k$ by Turán’s theorem. Let $v_{i_1},\ldots,v_{i_k}$ be the vertices of this $K_k$ and let $C_j=V_{i_j}$ for $1 \leq j \leq k$ . Then we claim that $C_1,\ldots,C_k$ is a $(k,\eta,\delta )$ -blue-blocked configuration. The fact that each $C_i$ is $\eta$ -regular follows immediately from our application of Lemma 2.1 and the fact that $d_B(C_i) \geq \delta$ follows from the fact that we assumed $d_B(C_i) \geq \frac 12$ . Finally, the definition of edges in $G$ implies that $(C_i,C_j)$ is $\eta$ -regular with $d_R(C_i,C_j) \geq \delta$ for all $i \neq j$ . Thus, our colouring contains a $(k,\eta,\delta )$ -blue-blocked configuration with $\delta \leq (1-p) \varepsilon$ and $\eta \leq \delta ^{4k^2}$ , so Lemma 4.3 implies that the colouring contains either a red $B_{cn} ^{(k)}$ or a blue $B_n ^{(k)}$ .

We have now finished the proof if at least $pm$ of the parts $V_i$ are blue. Therefore, we may assume instead that at least $m'' \geq (1-p) m$ of the parts are red and again rename the parts so that these red parts are $V_1,\ldots,V_{m''}$ . We construct a reduced graph $G$ on vertices $v_1,\ldots,v_{m''}$ by connecting $v_i$ to $v_j$ if $(V_i,V_j)$ is $\eta$ -regular with $d_B(V_i,V_j) \geq \delta$ . Suppose that some vertex in $G$ , say $v_1$ , has degree at most $(1-(1-p)^{k-1}-\eta/(1-p))m''-1$ . As before, $v_1$ has at most $\eta m \leq \eta m''/(1-p)$ non-neighbours coming from irregular pairs. Thus, if we let $J$ denote the set of indices $j$ for which $(V_1,V_j)$ is $\eta$ -regular with $d_R(V_1,V_j) \geq 1- \delta$ , then we find that $|J| \geq (1-p)^{k-1}m'' \geq (1-p)^k m$ . Thus, if $U=\bigcup _{j \in J}V_j$ , then we see that $|U| \geq (1-p)^k N$ , since the partition is equitable. Next, as above, we let $V'_{\!\!1} \subseteq V_1$ denote the set of vertices $v \in V_1$ with $e_R(v,U) \geq (1-2 \delta )|U|$ and find that $|V'_{\!\!1}| \geq \frac 12 |V_1|$ . Therefore, as above, we know that $V'_{\!\!1}$ contains at least one red $K_k$ and this red $K_k$ has at least $(1-2 k \delta )|U|$ red extensions in $U$ . Moreover,

(9) \begin{align} (1-2k \delta )|U|&\geq (1-2k \delta ) (1-p)^k N\notag \\[5pt] &=(1-2 k \delta )(1-p)^k (p^{-k}+\varepsilon )n \notag \\[5pt] &=(1-2 k \delta )(c+(1-p)^k \varepsilon )n \end{align}

(10) \begin{align} &\qquad\;\;\; \geq (c+(1-p)^k \varepsilon -4 k \delta )n\notag \\[5pt] & \qquad\;\;\;\geq cn, \end{align}

where in (9) we used the definition of $p$ , which implies that $((1-p)/p)^k=c$ , and in (10) we used our choice of $\delta$ to see that $\delta \leq (1-p)^k \varepsilon/(4k)$ . Thus, in this case, we can find a red $B_{cn}^{(k)}$ .

We may therefore assume that every vertex in $G$ has degree at least $(1- (1-p)^{k-1}-\eta/(1-p))m''$ . As before, we know that, since $k \geq k_2(p)$ ,

\begin{equation*} (1-p)^{k-1} \leq p^{k-1} \leq \frac {1}{3(k-1)} \end{equation*}

and our choice of $\eta \leq (1-p)/(4k)$ implies that

\begin{equation*} \frac {\eta }{1-p} \leq \frac {1}{3(k-1)}. \end{equation*}

Thus, by Turán’s theorem, $G$ must contain a $K_k$ , with vertices $v_{i_1},\ldots,v_{i_k}$ . If we let $C_j=V_{i_j}$ , then $C_1,\ldots,C_k$ will be a $(k,\eta,\delta )$ -red-blocked configuration, by the definition of edges in $G$ and the assumption that $N$ is sufficiently large in terms of $\varepsilon$ . Thus, by Lemma 4.3, we can again conclude that the colouring contains either a red $B_{cn}^{(k)}$ or a blue $B_n ^{(k)}$ .

To finish, we note that, as claimed, we may take $c_1(k) \leq \!\left((1+o(1))\frac{\log k}{k}\right)^k$ . Indeed, for any $c$ and $k$ , let $p(c,k) = (c^{1/k}+1)^{-1}$ and $y = y(c,k) = 1/\log [1/p(c,k)]$ . Then, from Lemma 4.3 and 4.1, we see that $k_2(p(c,k)) = 1+y(5+\log y+\log \log y)$ . Thus, if $y\leq (1+o(1))k/\log k$ , then $k \geq k_2(p(c,k))$ . Since $y = 1/\log \!(1+c^{1/k})$ , this condition is equivalent to $c^{1/k} \geq \exp\!\left((1+o(1))\frac{\log k}{k}\right)-1= (1+o(1))\frac{\log k}{k}$ , which yields the desired bound.

Proof. Let

\begin{equation*} F(x_1,\ldots,x_k)=p^{1-k} \prod _{i=1}^k x_i+\frac {(1-p)^{1-k}}{k}\sum _{i=1}^k (1-x_i)^k \end{equation*}

and $\varphi (y)=(1-e^y)^k$ . The goal is to apply Hölder’s defect formula, Theorem 2.6, using the strict convexity of the function $\varphi$ . However, $\varphi$ is only strictly convex on the interval $(\!\log \frac 1k,0)$ and, in order to apply Theorem 2.6, we in fact need a positive lower bound on $\varphi ''$ , but no such bound exists for the whole interval $(\!\log \frac 1k,0)$ . Because of this, we need to separately analyze the cases where all the variables are inside a large subinterval of $\big(\frac 1k,1\big)$ and when one of them is outside such a subinterval.

First, suppose that one of the variables, say $x_1$ , is in the interval $\big[0,\frac{1+\varepsilon _1}k\big]$ , for some small constant $\varepsilon _1\gt 0$ . Then we have that

\begin{align*} F(x_1,\ldots,x_k) &\geq \frac{(1-p)^{1-k}}{k} \!\left ( 1-x_1 \right ) ^k\geq \frac{(1-p)^{1-k}}{k} \!\left ( 1-\frac{1+\varepsilon _1}k \right ) ^k. \end{align*}

From the proof of Lemma 4.1, we see that this quantity is strictly larger than $1$ for all $k \geq k_1(p)$ , so, by choosing $\delta _0$ appropriately, we see that $F(x_1,\dots,x_k) \geq 1+\delta _0$ in this case. We may therefore assume from now on that all the variables are at least $\frac{1+\varepsilon _1}k$ .

Next, suppose that there exist values $x_1,\ldots,x_{k-1} \in \big[\frac{1+\varepsilon _1}k,1\big]$ such that $F(x_1,\ldots,x_{k-1},1)=1$ . We observe that

\begin{align*} \left . \mathchoice{\frac{\partial F}{\partial x_k}}{\partial F/\partial x_k}{\partial F/\partial x_k}{\partial F/\partial x_k} \right |_{x_k=1}=\left [ p^{1-k} \prod _{i=1}^{k-1}x_i-(1-p)^{1-k}(1-x_k)^{k-1} \right ] _{x_k=1}=p^{1-k} \prod _{i=1}^{k-1} x_i \gt 0. \end{align*}

This implies that if we move from $x_k=1$ to $x_k=1- \varepsilon _2$ for some sufficiently small $\varepsilon _2$ , the value of $F$ will decrease. Therefore, there will exist a vector $(x_1,\ldots,x_k)$ for which $F(x_1,\ldots,x_k)\lt 1$ , contradicting Lemma 4.1 as long as $k \geq k_1(p)$ . Thus, for every choice of $x_1,\ldots,x_{k-1} \in \big[\frac{1+\varepsilon _1}k,1\big]$ , we have that $F(x_1,\ldots,x_{k-1},1)\gt 1$ . Since the space $\big[\frac{1+\varepsilon _1}k,1\big]^{k-1} \times \{1\}$ is compact, we in fact find that $F(x_1,\ldots,x_{k-1},1) \geq 1+\delta' _{\!\!1}$ for all $x_1,\ldots,x_{k-1} \in \big[\frac{1+\varepsilon _1}k,1\big]$ , for some sufficiently small $\delta'_{\!\!1}$ depending on $p$ and $k$ . Finally, by continuity of $F$ , we have that $F(x_1,\ldots,x_k) \geq 1+\delta _1$ whenever $x_k \geq 1- \varepsilon _2$ for some other $\delta _1,\varepsilon _2\gt 0$ . Since $F$ is a symmetric function of its variables, the same conclusion holds if $x_i \geq 1- \varepsilon _2$ for any $i$ . Thus, as long as we take the $\delta _0$ in the lemma statement to be smaller than $\delta _1$ , we can assume from now on that $x_i \in [\frac{1+\varepsilon _1}k, 1- \varepsilon _2]$ for all $i$ .

By Lemma 2.5, subject to the constraint $\prod _{i=1}^k x_i = z$ , the term $\frac 1k \sum _{i=1}^k \!(1-x_i)^k$ is minimized when $x_i= z^{1/k}$ for all $i$ . As in the proof of Lemma 4.1, this shows that $F(x_1,\dots,x_k) \geq \psi (z^{1/k})$ , where $\psi (w) = p^{1-k}w^k + (1-p)^{1-k}(1-w)^k$ . The function $\psi$ has a global minimum at $w=p$ , where its value is $1$ . This shows that $F(x_1,\dots,x_k)\geq 1+\delta _0$ if $\lvert z^{1/k}-p\rvert \geq \varepsilon _3$ , for some $\varepsilon _3\gt 0$ depending on $p,k$ , and $\delta _0$ . Moreover, by picking $\delta _0$ sufficiently small, we can make $\varepsilon _3$ as small as we wish. Therefore, we may now assume that $z^{1/k} = p \pm \varepsilon _3$ , which implies that $\log (z^{1/k}) = (\!\log p) \pm \varepsilon _4$ for some $\varepsilon _4\gt 0$ , which can also be made arbitrarily small by picking $\delta _0$ appropriately.

We are now ready to apply Hölder’s defect formula. First, we observe that for $y \in [\!\log \frac{1+\varepsilon _1}k,\log (1-\varepsilon _2)]$ , we have

\begin{equation*} \varphi ''(y) = k e^y (1 - e^y)^{k - 2} (k e^y - 1)\geq k \cdot \frac {1+\varepsilon _1}k \cdot \varepsilon _2^{k-2} \cdot \varepsilon _1 \;=\!:\; m, \end{equation*}

where $m$ is a fixed, strictly positive constant. Let $y_i = \log x_i$ for $1 \leq i \leq k$ , so that $\frac 1k \sum _{i=1}^k y_i = \log (z^{1/k})$ . We assumed that $\lvert x_j-p\rvert \geq \varepsilon _0$ for some $j$ , which implies that $\lvert y_j-\log p\rvert \geq \varepsilon _0$ as well, since the derivative of $\log x$ is bounded below by $1$ on the interval $(0,1)$ . Therefore, choosing $\delta _0$ small enough that $\varepsilon _4\lt \varepsilon _0$ , we see that

\begin{equation*} \frac 1k \sum _{i=1}^k \big(y_i-\log \big(z^{1/k}\big)\big)^2 \geq \frac 1k \big(y_j-\log \big(z^{1/k}\big)\big)^2 \geq \frac 1k (\varepsilon _0-\varepsilon _4)^2, \end{equation*}

since $\log (z^{1/k}) = (\!\log p)\pm \varepsilon _4$ and $\lvert y_j-\log p\rvert \geq \varepsilon _0$ . Hence, by Theorem 2.6, we have that

\begin{align*} F(x_1,\dots,x_k) &= p^{1-k} z +\frac{(1-p)^{1-k}}k \sum _{i=1}^k (1-x_i)^k\\[5pt] &= p^{1-k}z + (1-p)^{1-k}\cdot \frac 1k \sum _{i=1}^k \varphi (y_i) \\[5pt] &\geq p^{1-k}z + \varphi \big(\!\log \big(z^{1/k}\big)\big)+ \frac m{2k}(\varepsilon _0-\varepsilon _4)^2\\[5pt] &=\psi (z^{1/k}) + \frac m{2k} (\varepsilon _0-\varepsilon _4)^2\\[5pt] &\geq 1+\delta _0, \end{align*}

where we use the fact that $\psi (w)\geq 1$ for all $w \in [0,1]$ and take $\delta _0$ sufficiently small.

Proof. We may assume without loss of generality that $|B_1| \geq \varepsilon _0 N$ . As in the proof of Lemma 4.3, we need to split into two cases, depending on whether $C_1,\ldots,C_k$ is blue-blocked or red-blocked. We begin by assuming that it is $(k,\eta,\delta )$ -red-blocked.

First, as in the proof of Lemma 4.3, observe that each $C_i$ contains at least one red $K_k$ and there is at least one blue $K_k$ spanning $C_1,\ldots,C_k$ . Moreover, if we assume that $|C_i| \geq \tau N$ for all $i$ , then Lemma 2.2 shows that the number of blue $K_k$ spanning $C_1,\ldots,C_k$ is at least

\begin{equation*} \left ( \prod _{1 \leq i \lt j \leq k}d_B(C_i,C_j)- \eta \!\left (\begin {array}{c} k\\[-3pt] 2 \end {array}\right ) \right ) \prod _{i=1}^k |C_i| \geq \!\left ( \delta ^{\left (\substack{k \\[3pt] 2 }\right )}-\eta \!\left (\begin {array}{c} k\\[-3pt] 2 \end {array}\right ) \right ) (\tau N)^k \geq \!\left ( \frac {\delta ^{\left (\substack{k\\[3pt] 2 }\right )}\tau ^k}{2} \right ) N^k \end{equation*}

and similarly, with an additional factor of $1/k!$ , for the number of red $K_k$ inside each $C_i$ .

For a vertex $v$ and $i \in [k]$ , let $x_i(v)=d_B(v,C_i)$ . Lemma 4.1 implies that, for any $v \in V$ ,

\begin{equation*} p \!\left ( p^{-k}\prod _{i=1}^k x_i(v) \right ) +(1-p) \left ( \frac {(1-p)^{-k}}{k}\sum _{i=1}^k (1-x_i(v))^k \right ) \geq 1. \end{equation*}

Additionally, if $v \in B_1$ , then $|x_1(v)-p|\geq \varepsilon _0$ , so Lemma 5.2 implies that, for $v \in B_1$ ,

\begin{equation*} p \!\left ( p^{-k}\prod _{i=1}^k x_i(v) \right ) +(1-p) \left ( \frac {(1-p)^{-k}}{k}\sum _{i=1}^k (1-x_i(v))^k \right ) \geq 1+\delta _0. \end{equation*}

Adding these two equations up over all $v \in V$ shows that

\begin{equation*} p \!\left ( p^{-k} \sum _{v \in V}\prod _{i=1}^k x_i(v) \right ) +(1-p) \left ( \frac {(1-p)^{-k}}{k}\sum _{i=1}^k \sum _{v \in V}(1-x_i(v))^k \right ) \geq N+\delta _0|B_1| \geq (1+\delta _0 \varepsilon _0)N. \end{equation*}

That is, a $(p,1-p)$ -weighted average of two quantities is at least $(1+\delta _0 \varepsilon _0)N$ , which implies that one of the quantities must itself be at least $(1+\delta _0 \varepsilon _0)N$ . Suppose first that

\begin{equation*} p^{-k} \sum _{v \in V} \prod _{i=1}^k x_i(v) \geq (1+\delta _0 \varepsilon _0)N. \end{equation*}

Let $Q$ be a uniformly random blue $K_k$ with one vertex in each of $C_1,\ldots,C_k$ . Let $\alpha =\delta ^{k^2} \leq \prod _{i\lt j} d_B(C_i,C_j)$ , so that $\eta \leq \delta ^{4k^2} = \alpha ^4 \leq \alpha ^3/k^2$ . Therefore, applying Lemma 2.3 to each $v$ and summing up the result, we find that the expected number of blue extensions of $Q$ is at least

\begin{align*} \sum _{v \in V} \!\left ( \prod _{i=1}^k x_i(v)-4 \alpha \right ) &\geq (p^k+ p^k\delta _0 \varepsilon _0-4 \alpha )N. \end{align*}

Next, observe that

(11) \begin{equation} 4 \alpha =4 \delta ^{k^2} \leq \frac{\delta ^k }2 \leq \frac{((1-p)\delta _0 \varepsilon _0)^k}{2} \leq \frac{(1-p)^k \delta _0 \varepsilon _0}{2} \leq \frac{p^k \delta _0 \varepsilon _0}{2}, \end{equation}

which implies that the expected number of blue extensions of $Q$ is at least $(p^k+\beta )N$ , where $\beta =(1-p)^k \delta _0 \varepsilon _0/2$ . Thus, there exists a blue $B_{(p^k+\beta )N}^{(k)}$ , proving (a) in this case. Moreover, if we assume that $|C_i| \geq \tau N$ for all $i$ , then our earlier computation shows that $Q$ is chosen uniformly at random from a set of at least $\kappa N^k$ monochromatic cliques, where $\kappa =\delta ^{ \left(\substack{ k\\[3pt] 2} \right)} \tau ^k/2$ . We may therefore apply Lemma 2.4 with $\xi =p^k+\beta$ and $\nu =p^k+\gamma$ , for some appropriately chosen $0\lt \gamma \lt \beta$ , to conclude that in this case our colouring contains at least $\gamma N^k$ blue cliques, each with at least $(p^k+\gamma )N$ blue extensions, proving (b).

Therefore, we may assume that the second term in the weighted average is the large one, that is, that

\begin{equation*} \frac {(1-p)^{-k}}{k}\sum _{i=1}^k \sum _{v \in V}(1-x_i(v))^k\geq (1+\delta _0 \varepsilon _0)N, \end{equation*}

which implies that, for some $i$ ,

\begin{equation*} \sum _{v \in V} (1-x_i(v))^k \geq (1-p)^k(1+ \delta _0 \varepsilon _0)N. \end{equation*}

Therefore, if $Q$ is now a random red $K_k$ inside this $C_i$ , Lemma 2.3 implies that the expected number of red extensions of $Q$ is at least

\begin{equation*} \sum _{v \in V} \!\left [ (1-x_i(v))^k-4 \alpha \right ] \geq \left [ (1-p)^k+(1-p)^k \delta _0 \varepsilon _0-4 \alpha \right ] N. \end{equation*}

But, by (11), $4 \alpha \leq (1-p)^k \delta _0 \varepsilon _0/2$ , which implies that the expected number of red extensions of $Q$ is at least $((1-p)^k+\beta )N$ , proving (a). As before, if we also assume that $|C_i| \geq \tau N$ for all $i$ , then we may apply Lemma 2.4 with $\kappa =\delta ^{ \left(\substack{ k \\[3pt] 2} \right)}\tau ^k/2k!$ , $\xi =(1-p)^k+\beta$ , and $\nu = (1-p)^k+\gamma$ to find that our colouring contains at least $\gamma N^k$ red $K_k$ , each with at least $((1-p)^k+\gamma )N$ red extensions for some appropriately chosen $\gamma \in (0,\beta )$ , yielding (b). This concludes the proof of the lemma in the case where $C_1,\ldots,C_k$ is a $(k,\eta,\delta )$ -red-blocked configuration.

As in the proof of Lemma 4.3, the other case, where $C_1,\ldots,C_k$ is a $(k,\eta,\delta )$ -blue-blocked configuration, follows in an almost identical fashion. We define $y_i(v)=d_R(v,C_i)$ for all $v \in V$ and $i \in [k]$ and let $q=1-p$ . We then apply Lemma 4.1 and 5.2 with these $y$ variables and with $q$ instead of $p$ . The remaining details are exactly the same.