Proof. Let $k,q\in \mathbb N$ and $r\in \mathbb Z$ such that $n=(k+3)q+r$ , where $0\leq r\leq k+2$ . Define integers $r_1,\ldots , r_{k+1}$ as equally as possible so that $2 \geq r_1 \geq r_2 \geq \ldots \geq r_{k+1} \geq 0$ and $r=\sum _{i=1}^{k+1}r_i$ .

Let $G$ be the $n$ -vertex digraph consisting of $k-1$ independent sets $V_1,\ldots , V_{k-1}$ and two other classes $V_k$ and $V_{k+1}$ so that there are all possible double edges going out of the $k-1$ independent sets; all possible double edges inside of $V_{k}$ and inside of $V_{k+1}$ ; all possible directed edges from $V_k$ to $V_{k+1}$ (but none from $V_{k+1}$ to $V_k$ ). Moreover, we choose the classes so that $|V_i|=q+r_i$ for all $i \in [k-1]$ , and $|V_i|=2q+r_i$ for all $i \in \{k,k+1\}$ .

Let $v\in V_i$ . If $i\in [k-1]$ , then $d_G(v)=2((k+2)q+r)-2r_i=2\lceil (1-\frac {1}{k+3})n\rceil -2r_i$ . If $i\in \{k,k+1\}$ , then $d_G(v)=2((k+2)q+r)-r_j-2=2\lceil (1-\frac {1}{k+3})n\rceil -r_j-2$ where $j\! \in\! \{k,k+1\} {\setminus} \{i\}$ . Therefore, $\delta (G)$ is as in the statement of the proposition.

Suppose for a contradiction that there is a $k$ th power of a Hamilton cycle $C$ in $G$ . Notice that every transitive tournament on $k+1$ vertices in $G$ contains at least two vertices from $V_k \cup V_{k+1}$ . By following the same ordering of vertices in $V_k \cup V_{k+1}$ appearing in $C$ , we deduce that $C[V_k \cup V_{k+1}]$ , and therefore $ G[V_k \cup V_{k+1}]$ , contains a Hamilton cycle. However, $G[V_k \cup V_{k+1}]$ is not strongly connected, a contradiction.

Proof. Set $m \,:\!= \vec {r}(k+1)-1$ and let $n=(3m+2)t$ for some $t \in \mathbb N$ . Consider the $n$ -vertex oriented graph $G_k$ defined as follows (see also Figure 1 for $k=2$ ). The vertex set of $G_{k}$ consists of sets $V_1, V_2, V_3$ where either (i) $|V_1|= m t$ and $|V_2|=|V_3|=(m+1)t$ or (ii) $|V_1|= m t -1$ , $|V_2|=(m+1)t+1$ and $|V_3|=(m+1)t$ . We choose the sizes of the $V_i$ such that $|V_1|$ is not divisible by $k$ . Add all edges directed from $V_1$ to $V_{2}$ , from $V_2$ to $V_{3}$ , and from $V_3$ to $V_{1}$ . Both $G_k[V_2]$ and $G_k[V_3]$ are semi-regular tournaments. Finally, let $G_k[V_1]$ be the $t$ -blow-up of the regular tournament on $m$ vertices that contains no copy of $\vec {T}_{k+1}$ (where one vertex is deleted if $|V_1|=mt-1$ ). It is easy to check that the desired minimum semi-degree condition holds.

Consider any $k$ th power of a cycle $C$ in $G_k$ . Notice that any copy of $\vec {T}_{k+1}$ in $G_k$ must contain vertices from at most two of the classes $V_1$ , $V_2$ , and $V_3$ . This implies that every time $C$ enters $V_1$ (from $V_3$ ), it must traverse at least $k$ vertices before leaving $V_1$ (and entering $V_2$ ). In fact, since $G_k[V_1]$ does not contain a copy of $\vec {T}_{k+1}$ , precisely $k$ vertices in $V_1$ are covered in each such step. Thus, the number of vertices in $V_1$ covered by $C$ is a multiple of $k$ . Therefore, $C$ cannot contain all of $V_1$ since $|V_1|$ is not divisible by $k$ .

Proof. Let $k\geq 15$ and set $t \,:\!= \lceil 2^{(k-5)/2}\rceil$ . We first shall prove that there is a (randomly generated) $t$ -vertex tournament $T(k)$ with $\delta ^0(T(k)) \geq \big (\frac {1}{2}-\frac {3/2}{t^{2/5}}-\frac {1}{2t}\big )t$ that is $\vec {T}_{k+1}$ -free. Then we will get the desired oriented graph by taking a blow-up of such a tournament.

Consider a random tournament $T$ on $t$ vertices; that is, the orientation of each edge is selected uniformly at random, independently of all other edges. By the union bound,

(3.1) \begin{align} \mathbb P(T\text{ contains a copy of }\vec {T}_{k+1}) \leq \frac {\binom {t}{k+1}(k+1)!}{2^{\binom {k+1}{2}}} \lt \frac {t^{k+1}}{2^{k(k+1)/2}}. \end{align}

By Chernoff’s bound,Footnote ³ for each $v \in V(T)$ , we have

\begin{equation*}\mathbb{P}\left (d^{\pm }_T(v)+\frac {1}{2}\leq \frac {t}{2}-(3/2)t^{3/5} = \left (\frac {1}{2} - \frac {3/2}{t^{2/5}}\right )t\right )\leq \exp \left (\frac {-(3/2)^2t^{6/5}}{t}\right ) = \exp \big (-(3/2)^2t^{1/5}\big ).\end{equation*}

Then, by the union bound, we have

(3.2) \begin{align} \mathbb P \bigg (\delta ^0(T)+\frac {1}{2} \leq \Big (\frac {1}{2} - \frac {3/2}{t^{2/5}}\Big )t\bigg ) \leq 2t\exp \big (-(3/2)^2t^{1/5}\big ). \end{align}

Thus, the probability that one of the events (3.1) or (3.2) holds is at most

\begin{equation*}\frac {t^{k+1}}{2^{k(k+1)/2}} + 2t\exp \big (-(3/2)^2t^{1/5}\big ) \lt 1,\end{equation*}

where the inequality holds for all $k\geq 15$ (in fact, one can see that the inequality holds for all $k\geq 2$ , but we are assuming $k\geq 15$ in this context). Hence, there is a tournament $T(k)$ on $t$ vertices with $\delta ^0(T(k))\gt \big (\frac {1}{2} - \frac {3/2}{t^{2/5}}-\frac {1}{2t}\big )t$ not containing a copy of $\vec T_{k+1}$ .

Finally, $R_k$ is obtained by blowing up each vertex of $T(k)$ into a set of size $\lfloor \frac {n}{t}\rfloor$ or $\lceil \frac {n}{t}\rceil$ so that it contains $n$ vertices in total. Thus, $R_k$ does not contain a copy of $\vec {T}_{k+1}$ and

\begin{align*}\delta ^0(R_k)\gt \bigg (\frac {1}{2}-\frac {3/2}{t^{2/5}}-\frac {1}{2t}\bigg )t\cdot \left \lfloor \frac {n}{t}\right \rfloor \geq \bigg (\frac {1}{2}-\frac {3/2}{t^{2/5}}-\frac {1}{2t}\bigg )(n-(t-1)) & \geq \bigg (\frac {1}{2}-\frac {2}{t^{2/5}}\bigg )n \\& \geq \bigg (\frac {1}{2}-\frac {4}{2^{k/5}}\bigg )n ,\end{align*}

as desired (where we used the fact that $n$ is sufficiently large in the second to last inequality and the fact that $t=\lceil 2^{(k-5)/2}\rceil \geq 2^{(k-5)/2}$ in the last inequality).

Proof. Define constants $\varepsilon , \xi , d\gt 0$ and $n_0,t_0,T \in \mathbb N$ such that

(5.1) \begin{align} \frac {1}{n_0} \ll \frac {1}{T} \leq \frac {1}{t_0} \ll \varepsilon \ll \xi \ll d \ll \frac {1}{k}, \eta , \end{align}

and where $T$ is the output of Lemma 4.1 on input $\varepsilon$ and $t_0$ .

Let $G$ be a digraph on $n \geq n_0$ vertices as in the statement of the lemma. Apply Lemma 4.1 with parameters $\varepsilon$ , $d$ , and $t_0$ to obtain the reduced digraph $R$ of $G$ with $t_0\lt |R|\lt T$ . Let $m$ denote the size of the clusters of $G$ . By Proposition 4.2 and (5.1),

\begin{equation*}\delta (R) \geq 2\left (1-\frac {1}{k+2}+\frac {\eta }{2}\right )|R|.\end{equation*}

Thus, Theorem5.1 yields a $\overline {K}_{k+2}$ -tiling $\mathcal T$ in $R$ covering all but at most $\eta |R| /2$ of the $V_i \in V(R)$ .

Recall that $C_{k+2}^{k}\subseteq \overline {K}_{k+2}$ . Therefore, for each tile $\overline {K}_{k+2}$ in $\mathcal T$ formed by clusters $V_{i_1},\ldots , V_{i_{k+2}}$ , we may apply Lemma 4.8 with $\xi$ playing the role of $\eta$ , to obtain a $k$ -path in $G[\bigcup _{j\in [k+2]} V_{i_j}]$ covering all but at most $\xi (k+2)m$ vertices from $\bigcup _{j\in [k+2]} V_{i_j}$ .

Together, these $k$ -paths form a collection of size at most $|R|\lt T$ . Moreover, all but at most

\begin{equation*}|V_0|+ \xi m |R|+\frac {\eta |R|}{2}m\leq (\varepsilon +\xi +\eta /2)n \stackrel {({\hbox{5.1}})}{\leq } \eta n\end{equation*}

vertices of $G$ are covered by these $k$ -paths, as desired.

Proof. Let $\xi _1 \gt 0$ be the output of Theorem5.3 on input $k+2$ , $\eta$ , and $t=2$ . Let $n_0 \in \mathbb N$ be sufficiently large and let $G$ be a digraph on $n \geq n_0$ vertices as in the statement of the lemma.

Given any $v\in V(G)$ note that

\begin{align*} |N^{+}_G(v)\cap N^-_G(v)| \geq 2 \left (1-\frac {1}{k+3}+\eta \right )n-n = \left (\frac {k+1}{k+3}+2 \eta \right )n . \end{align*}

Let $U$ be a subset of $N^{+}_G(v)\cap N^-_G(v)$ of size $\tfrac {k+1}{k+3}n$ . Observe that

\begin{align*} \delta (G[U]) &\geq 2 |U| - 2 \left ( \frac {1}{k+3}- \eta \right ) n = 2 \left ( \frac {k}{k+3} + \eta \right )n \geq 2\left (\frac {k}{k+1} + \eta \right )|U| , \end{align*}

and in particular $e(G[U])\geq \big (\tfrac {k}{k+1}+\eta )|U|^2$ . Thus, Theorem5.3 yields at least $\xi _1 |U|^{2(k+2)} \geq \xi _1 (n/2)^{2(k+2)}$ copies of $\overline {K}_{k+2}(2)$ in $G[U]$ . Set $\xi \,:\!= \xi _1 /2^{2(k+2)}$ .

As $C^{k}_{k+2} \subseteq \overline {K}_{k+2}$ , it is easy to see that each such copy of $\overline {K}_{k+2}(2)$ in $G[U]$ is spanned by a $k$ -path. Moreover, as $U \subseteq N^{+}_G(v)\cap N^-_G(v)$ , each of these $k$ -paths is an absorber for $v$ , as desired.

Proof. For a contradiction, assume that $G$ is an $n$ -vertex digraph and $ \{ A, B \}$ is a partition of $V(G)$ that together form a counterexample to the lemma. By considering the digraph with all orientations reversed, we may assume without loss of generality that $|A| \ge |B|$ . Define

\begin{align*} E_A & \,:\!= \{ ab \in E(A, B)\; :\; \text{there exists $a' \in A$ such that $a'ab$ is a $2$-path} \}, \\ E_B & \,:\!= \{ ab \in E(A, B)\; :\; \text{there exists $b' \in B$ such that $abb'$ is a $2$-path} \}. \end{align*}

By our assumption, $E_A \cap E_B = \emptyset$ . Note that, for every $ab \in E(A, B)$ , we have $d_{=4}(ab,V(G)) \ge 2\delta (G) - 3n \ge n/5 + 2\gamma n$ and so $ab \in E_A \cup E_B$ . Therefore, $\{E_A, E_B\}$ is a partition of $E(A, B)$ .

Proof. Suppose not. Let $C$ be a maximal tournament in $G[B]$ . Since $B$ is not an independent set, we have $|C| \ge 2$ . Note that $E_B = \emptyset$ implies that $d_G(a, C) \le |C| + 1$ for every $a \in A$ . Furthermore, because $C$ is a maximal tournament in $G[B]$ , we have $d_G(b, C) \le 2|C| - 2$ for every $b \in B$ . Recall that $|A| \ge |B|$ , so $|B| - 3n/5 \lt 0$ . Hence,

\begin{align*} |C|\delta (G) &\le \sum _{v \in C} d _G(v) \le |B|(2|C| - 2) + |A|(|C| + 1) = |C|(n + |B|) - 2|B| + |A| \\ &= 8n|C|/5 + |C|(|B| - 3n/5) - 2|B| + |A| \\ &\le 8n|C|/5 + 2(|B| - 3n/5) - 2|B| + |A| \lt 8n|C|/5, \end{align*}

a contradiction to the minimum total degree condition.

By Claim 5.7, there exists $ab \in E_B$ . Set

\begin{align*} B_4 & \,:\!= N_{=4}(ab, B), & B_3 & \,:\!= N_{=3}(ab, B), &A_3 & \,:\!= N_{=3}(ab, A), &c & \,:\!= |B_4| - n/5. \end{align*}

Since $ab \in E_B$ , we have $N_{=4}(ab, A) = \emptyset$ and so

\begin{equation*} 16n/5 + 2\gamma n \le d_G(a)+d_G(b) \le 2n + |B_3| + |A_3| + 2|B_4|. \end{equation*}

This implies that $c \gt 0$ and $|B_3| + |A_3| \ge 4n/5 - 2c$ . Since $|B_4| + |B_3| \le |B| \le n/2$ and $|B_4| = n/5 + c$ , we have $|B_3| \le 3n/10 - c$ and $c \le 3n/10$ , so $|A_3| \ge n/2 - c \gt 0$ . Hence,

(5.2) \begin{align} |A_3 \cup B_4| \ge n/2 - c + n/5 + c = 7n/10. \end{align}

Note that $\delta (G^{\pm }) \ge \delta (G) - n \ge 3n/5 + \gamma n$ . Therefore, by (5.2) and Claim 5.8, there exist $a_1a_2 \in E(G^{\pm }[A_3])$ and $b_1b_2 \in E(G^{\pm }[B_4])$ . For $i \in \{7,8\}$ set $V_i \,:\!= N_{=i}(\{a_1,a_2, b_1, b_2\}, V(G))$ . Note that $32n/5 + 4 \gamma n \le 2|V_8| + |V_7| + 6n$ , so

\begin{align*} 2|V_8| + |V_7| \ge 2n/5 + 4 \gamma n. \end{align*}

By Claim 5.8, $V_7 \cup V_8$ is disjoint from $A_3 \cup B_4$ , so

\begin{equation*} |V_7| + | V_8| \le n - |A_3 \cup B_4| \overset {{({\hbox{5.2}})}}\le 3n/10. \end{equation*}

Therefore,

\begin{equation*} |V_8| = (2|V_8| + |V_7|) - (|V_7| + |V_8|) \ge n/10 + 4 \gamma n. \end{equation*}

Let $x \in V_8$ . Since we have assumed that $G$ is a counterexample to the lemma, in $G$ , $x$ has no neighbours in $V_8$ , and in $G^{\pm }$ , $x$ has no neighbours in $V_7$ . Thus,

\begin{equation*} \delta (G) \le d_G(x) \le 2n - 2|V_8| - |V_7| \lt 8n/5, \end{equation*}

a contradiction.

Proof. Let $\sigma \gt 0$ be such that $1/n\ll \sigma \ll \gamma$ . Take $G$ , $ab$ , and $yz$ as in the statement of the lemma. For $\ell \ge 3$ , let

\begin{equation*} \overrightarrow {{}^{ab}{}{V}}_\ell \,:\!= \{v \in V(G)\; :\; \text{$\exists $ at least $(\sigma n)^{\ell - 3}$ $2$-walks of length $\ell $ that start with $ab$ and end with $v$}\} \end{equation*}

and

\begin{equation*} \overrightarrow {V^{yz}}_\ell \,:\!= \{v \in V(G)\; :\; \text{$\exists $ at least $(\sigma n)^{\ell - 3}$ $2$-walks of length $\ell $ that start with $v$ and end with $yz$}\}. \end{equation*}

Proof. We will only show that $|\overrightarrow {{}^{ab}{}{V}}_\ell | \ge n/5$ since the proof that $|\overrightarrow {V^{yz}}_\ell | \ge n/5$ follows analogously.

Let $\Phi _{2}\, :\!= \{ab\}$ . For $t \ge 3$ , let $\Phi _{t}$ be the collection of $2$ -walks of length $t$ in $G$ that start with the edge $ab$ . Since $\delta ^+(G) \ge (3n/5 + \gamma ) n$ , for all $t \geq 3$ we have that

(5.3) \begin{align} |\Phi _{t}| \ge (1/5 + 2\gamma )n |\Phi _{t - 1}| \ge (1/5 + 2\gamma )^{t-2}n^{t-2}. \end{align}

For a contradiction, suppose that $|\overrightarrow {{}^{ab}{}{V}}_\ell | \lt n/5$ . Hence the number of $2$ -walks in $\Phi _{\ell }$ that end with a vertex in $\overrightarrow {{}^{ab}{}{V}}_\ell$ is less than $|\Phi _{\ell -1}|n/5$ . So the number of $2$ -walks in $\Phi _\ell$ that end at a vertex not in $\overrightarrow {{}^{ab}{}{V}}_\ell$ is at least

\begin{align*} |\Phi _{\ell }| - |\Phi _{\ell -1}|n/5 & \overset {{({\hbox{5.3}})}}{\ge } 2\gamma n |\Phi _{\ell - 1}| \overset {{({\hbox{5.3}})}}{\ge } 2\gamma n (1/5 + 2\gamma )^{\ell -3}n^{\ell -3} \ge \sigma ^{\ell -3} n^{\ell -2}. \end{align*}

By an averaging argument, there exists a vertex $v \in V(G) {\setminus} \overrightarrow {{}^{ab}{}{V}}_\ell$ that is the end vertex of at least $\sigma ^{\ell -3} n^{\ell -3}$ $2$ -walks in $\Phi _{\ell }$ , a contradiction.

Let $\{V^+, V^-\}$ be the partition of $V(G)$ such that for every $v \in V^-$ we have $d^-_G(v) \ge d^+_G(v)$ and for every $v \in V^+$ we have $d^+_G(v) \gt d^-_G(v)$ .

Proof. We will only show that $V^- \subseteq \overrightarrow {{}^{ab}{}{V}}_\ell$ as the proof that $V^+ \subseteq \overrightarrow {V^{yz}}_\ell$ is analogous. Let $v \in V^-$ . By Claim 5.11 and the fact that $v \in V^-$ , we have

\begin{equation*} d^-_G\Big(v, \overrightarrow {{}^{ab}{}{V}}_{\ell - 2}\Big) \ge \delta (G)/2 + n/5 - n \ge \gamma n/2. \end{equation*}

Therefore, there are at least $(\sigma n)^{\ell - 5}\times \gamma n/2 \geq (\sigma n)^{\ell - 4}$ $2$ -walks $abv_3 \dotsc v_{\ell -3} v_{\ell -2}$ in $G$ with $v_{\ell -2} \in N^-_G(v)$ . Furthermore, for every such $2$ -walk, we have

\begin{equation*} |N^+_G(v_{\ell -3}) \cap N^+_G(v_{\ell -2}) \cap N^-_G(v)| \ge 2(\delta (G) - n) + \delta (G)/2 - 2n \ge 2\gamma n, \end{equation*}

and for every $v_{\ell - 1} \in N^+_G(v_{\ell -3}) \cap N^+_G(v_{\ell -2}) \cap N^-_G(v)$ , the sequence

\begin{equation*} abv_3 \dotsc v_{\ell -3} v_{\ell -2} v_{\ell -1} v \end{equation*}

is a $2$ -walk in $G$ . Thus, there are at least $(\sigma n)^{\ell - 4} \times 2 \gamma n \geq (\sigma n)^{\ell - 3}$ $2$ -walks in $G$ that start with $ab$ and end at $v$ . Hence, $v \in \overrightarrow {{}^{ab}{}{V}}_\ell$ and so $V^- \subseteq \overrightarrow {{}^{ab}{}{V}}_\ell$ , as desired.

For $\ell \ge 4$ , define

\begin{equation*} \overrightarrow {{}^{ab}{}{E}}_\ell \,:\!= \{e \in E(G)\; :\; \text{$\exists $ at least $(\sigma n)^{\ell - 4}$ $2$-walks of length $\ell $ that start with $ab$ and end with $e$}\}. \end{equation*}

and

\begin{equation*} \overrightarrow {E^{yz}}_\ell \,:\!= \{e \in E(G)\; :\; \text{$\exists $ at least $(\sigma n)^{\ell - 4}$ $2$-walks of length $\ell $ that start with $e$ and end with $yz$}\}. \end{equation*}

Let $X^- \subseteq V^-$ be such that $|X^-| = \min \{\sigma n, |V^-|\}$ and $d^+_G(x) \ge d^+_G(y)$ for every $x \in X^-$ and $y \in V^- {\setminus} X^-$ . Similarly, let $X^+ \subseteq V^+$ be such that $|X^+| = \min \{\sigma n, |V^+|\}$ and $d^-_G(x) \ge d^- _G(y)$ for every $x \in X^+$ and $y \in V^+ {\setminus} X^+$ . Set $Y^- \,:\!= V^- {\setminus} X^-$ and $Y^+ \,:\!= V^+ {\setminus} X^+$ .

Proof. We will only show that $E(G[Y^-]) \subseteq \overrightarrow {{}^{ab}{}{E}}_\ell$ since the proof that $E(G[Y^+]) \subseteq \overrightarrow {E^{yz}}_\ell$ is analogous. We may assume that $E(G[Y^-]) \neq \emptyset$ and so $|X^-| = \sigma n$ . Let $cd \in E(G[Y^-])$ . Pick $v_{\ell -4} \in X^-$ ; by Claim 5.12, $v_{\ell -4} \in \overrightarrow {{}^{ab}{}{V}}_{\ell -4}$ . Therefore, there are at least $(\sigma n)^{\ell -7}$ $2$ -walks $W$ of length $\ell -4$ in $G$ that start with $ab$ and end with $v_{\ell -4}$ . Let $v_{\ell -5}$ be the penultimate vertex on any such $2$ -walk $W$ . There are more than $2(\delta (G) - n) + \delta (G)/2 - 2n \ge 2\gamma n$ vertices in $N^+_G(v_{\ell -5}) \cap N^+_G(v_{\ell -4}) \cap N^-_G(c)$ . Let $v_{\ell -3}$ be such a vertex.

There are at least $\gamma n$ vertices $v_{\ell -2}$ in $N^+_G(v_{\ell -4}) \cap N^+_G(v_{\ell -3}) \cap N^-_G(c) \cap N^-_G(d)$ . To see this, recall that $v_{\ell - 4} \in X^-$ and $c,d \in Y^- \subseteq V^-$ , so $d^+_G(v_{\ell -4}) \ge d^+_G(c)$ and $d^-_G(d) \ge \delta (G)/2$ . Therefore,

\begin{align*} & | N^+_G(v_{\ell -4}) \cap N^+_G(v_{\ell -3}) \cap N^-_G(c) \cap N^-_G(d)| \ge d^+_G(v_{\ell -4}) + d^+_G(v_{\ell -3}) + d^-_G(c) + d^-_G(d) - 3n \\ & \ge d^+_G(v_{\ell -4}) + (\delta (G) - n) + (\delta (G) - d^+_G(c)) + \delta (G)/2 -3n \ge 5\delta (G)/2 - 4n \ge \gamma n. \end{align*}

In summary, there are $\sigma n$ choices for $v_{\ell -4}$ ; at least $(\sigma n)^{\ell -7}$ choices for the $2$ -walk $W$ ; at least $2 \gamma n$ choices for $v_{\ell -3}$ ; at least $\gamma n$ choices for $v_{\ell -2}$ . By the choice of $v_{\ell -3}$ and $v_{\ell -2}$ , adding $v_{\ell -3}v_{\ell -2}cd$ to the end of $W$ yields a $2$ -walk of length $\ell$ in $G$ that starts with $ab$ and ends with $cd$ . In total, this process gives rise to at least

\begin{equation*}\sigma n \times (\sigma n)^{\ell -7} \times 2 \gamma n \times \gamma n \geq (\sigma n)^{\ell - 4} \end{equation*}

such $2$ -walks. Thus, $cd \in \overrightarrow {{}^{ab}{}{E}}_\ell$ and so $E(G[Y^-]) \subseteq \overrightarrow {{}^{ab}{}{E}}_\ell$ , as desired.

Let $Y^-_1 \,:\!= Y^- {\setminus} \{a, b, y, z\}$ and $Y^+_1 \,:\!= Y^+ {\setminus} \{a, b, y, z\}$ . Suppose that $Y^-_1$ is an independent (or empty) set in $G$ . Note that $|Y^-_1 \cup Y^+_1| \ge (1- 2 \sigma )n-4$ and $\delta ^0(G) \ge \delta (G) - n \ge (3/5+\gamma ) n$ . Thus, there exist $v_1,v_2,v_3,v_4 \in Y^-_1 \cup Y^+_1$ such that $abv_1v_2v_3v_4$ is a $2$ -path in $G$ . Since $Y^-_1$ is an independent (or empty) set, there exists some $i \in [3]$ such that $v_i,v_{i+1} \in Y^+_1$ . By Claim 5.13, $v_i v_{i+1} \in E(G[Y^+]) \subseteq \overrightarrow {E^{yz}}_{12-i}$ . Therefore, there are at least $(\sigma n)^{8-i}$ $2$ -walks of length $13$ in $G$ that start with $abv_1 \ldots v_iv_{i+1}$ and end with $yz$ . By a simple counting argument, one such $2$ -walk is in fact a $2$ -path. Thus, the conclusion of the lemma holds in this case.

An analogous argument holds in the case when $Y^+_1$ is an independent (or empty) set in $G$ . Thus, we may assume that neither $Y^-_1$ nor $Y^+_1$ is an independent or empty set.

Let $G' \,:\!= G {\setminus} ( X^- \cup X^+ \cup \{a, b, y, z\})=G[Y^-_1 \cup Y^+ _1]$ , so $\delta (G') \ge \delta (G) - 2 \sigma n - 4 \ge (8/5+\gamma /2) n$ . Then by applying Lemma 5.6 to $G'$ and the partition $\{Y^-_1, Y^+_1\}$ of $V(G')$ , there exists a $2$ -path on at most $6$ vertices in $G'$ that starts with an edge in $E(G[Y^-_1])$ and ends with an edge in $E(G[Y^+_1])$ . By Claim 5.13, $E(G[Y^-_1]) \subseteq \overrightarrow {{}^{ab}{}{E}}_9$ and $E(G[Y^+_1]) \subseteq \overrightarrow {E^{yz}}_9$ . Combining these facts, and using an averaging argument, one obtains a $2$ -path on at most $20$ vertices in $G$ that starts with $ab$ and ends with $yz$ , as desired.

Proof. As $\delta (G {\setminus} U)\geq (8/5 + \gamma /2)n$ , this follows immediately by applying Lemma 5.10 to $G{\setminus} U$ .

Proof. Let $X \,:\!= V(T)$ , $Y \,:\!= \{v\in V(G)\; :\; d^-_G(v,X)\leq k-1\}$ , and $Z \,:\!= \{v\in V(G)\; :\; d^-_G(v,X)\geq k\}$ . We have

\begin{align*} (\delta n-2k+1) (2k-1)& = (\delta n-|X|) |X| \leq e_G (X, V(G){\setminus} X) \\ & \leq (k-1)|Y|+(2k-1)|Z| = (k-1)n+k|Z|, \end{align*}

so

\begin{align*} |Z| & \geq \frac {( (2k-1)\delta -k+1 )n-(2k-1)^2}{k}. \end{align*}

Thus, there exists a $k$ -set $X'\subseteq X$ such that

\begin{align*} \left |\bigcap _{x\in X'}N^+_G(x) \right | & \geq \frac {|Z|}{\binom {2k-1}{k}} \geq \frac { ( (2k-1)\delta -k+1)n-(2k-1)^2}{\binom {2k-1}{k}k} \geq \frac {(2k-1)\delta -k+1}{k2^{2k-1}}n, \end{align*}

where the last inequality follows as $n$ is sufficiently large. Therefore, $X'$ induces a $\delta$ -out-good tournament on $k$ vertices. The proof for $\delta$ -in-good tournaments is analogous.

Proof. We have

\begin{align*} e_G(A,B)&\geq \sum _{v\in A} \left ( d^+_G(v) - d^+_G(v, A) - (n-|A|-|B|) \right ) \geq |A|\delta n-e(G[A])-|A|(n-|A|-|B|)\\ &\geq |A|\delta n-\frac {|A|^2}{2}-|A|(n-|A|-|B|) = |A|\left (\frac {|A|}{2}+|B|-(1-\delta )n\right ). \end{align*}

When $|A|\geq |B| \geq \delta n$ , the desired inequality follows easily since

\begin{align*} |A|\left (\frac {|A|}{2}+|B|-(1-\delta )n\right ) = |A||B| \left (\frac {|A|}{2|B|}+1-\frac {(1-\delta )n}{|B|}\right ) \geq |A||B| \left (\frac {5}{2} - \frac {1}{\delta }\right ). \end{align*}