Proof. We prove the lemma by induction on $k$ . The base case $k=1$ is trivial. Let $k \geq 2$ . By the induction hypothesis, the process continues for at least $k-1$ steps. It remains to show that $N(x_1,\dots,x_{k-1})$ contains at least $a$ edges, because this would also imply that $N(x_1,\dots,x_{k-1}) \neq \emptyset$ and hence the process continues for at least $k$ steps. For $1 \leq i \leq k-1$ , let $S_i$ be the set of vertices which are adjacent to $x_1,\dots,x_{i-1}$ but not adjacent to $x_i$ . In particular, $S_1$ is just the set of vertices not adjacent to $x_1$ and $x_i \in S_i$ for all $i$ . Then $V(G) = S_1 \cup \dots \cup S_{k-1} \cup N(x_1,\dots,x_{k-1})$ . Put $S \;:\!=\; S_1 \cup \dots \cup S_{k-1}$ , $N \;:\!=\; N(x_1,\dots,x_{k-1})$ , $s_i \;:\!=\; |S_i|$ , $s = |S|$ and $d_i \;:\!=\; d(x_i)$ . Note that $s_i \leq n - d_i$ . Also, all vertices in $S_i$ have degree at most $d_i$ . We have

(1) \begin{equation} e(N,S) + 2e(S) = \sum _{v \in S}d(v) \leq \sum _{i = 1}^{k-1}{s_i \cdot d_i} \leq \sum _{i=1}^{k-1}{s_i(n-s_i)}. \end{equation}

Since $e(G) = e(N) + e(S) + e(N,S)$ , we have $2e(S) = 2e(G) - 2e(N) - 2e(N,S)$ . Plugging this into (1) and rearranging, we get

(2) \begin{equation} e(N) \geq e(G) - \frac{1}{2}e(N,S) - \frac{1}{2} \cdot \sum _{i = 1}^{k-1}{s_i(n-s_i)}. \end{equation}

We have $e(N,S) \leq |N| \cdot |S| = (n - s)s$ . Also, by Cauchy–Schwarz,

\begin{equation*}\sum _{i = 1}^{k-1}{s_i(n-s_i)} = ns - \sum _{i=1}^{k-1}{s_i^2} \leq ns - \frac {s^2}{k-1}. \end{equation*}

Plugging this into (2) gives

\begin{equation*} e(N) \geq e(G) - \frac {1}{2}(n-s)s - \frac {1}{2}\left ( ns - \frac {s^2}{k-1} \right ) = e(G) - ns + \frac {ks^2}{2(k-1)}. \end{equation*}

The maximum of $ns - \frac{ks^2}{2(k-1)}$ is obtained at $s = \frac{(k-1)n}{k}$ and equals $\frac{(k-1)n^2}{2k}$ . Hence, $e(N) \geq e(G) - \frac{(k-1)n^2}{2k} \geq a$ , as required.

Proof. Let $C_1,\dots,C_m$ be the connected components of $G$ with $|C_1| \geq \dots \geq |C_m|$ . Let $\ell \geq 1$ be the minimal integer satisfying $|C_1| + \dots + |C_{\ell }| \geq s$ . If $\ell \leq 1 + \frac{sn}{a}$ then take $F$ to be a forest contained in $C_1 \cup \dots \cup C_{\ell }$ having $s$ vertices and $\ell$ connected components. Suppose now by contradiction that $\ell \gt 1 + \frac{sn}{a}$ . Set $r = |C_1| + \dots + |C_{\ell -1}|$ . Then $r \lt s$ and $|C_{\ell -1}| \leq \frac{r}{\ell -1}$ . We have $e(G) \leq \binom{r}{2} + \sum _{i = \ell }^{m}{\binom{|C_i|}{2}}$ . By convexity, the sum $\sum _{i = \ell }^{m}{\binom{|C_i|}{2}}$ is maximised when all except maybe one of the $|C_i|$ ’s are equal to their maximal value, which is $\frac{r}{\ell -1}$ . So

\begin{align*} e(G) \leq \binom{r}{2} + \left \lceil \frac{n-r}{r/(\ell -1)} \right \rceil \cdot \binom{r/(\ell -1)}{2} &\leq \binom{r}{2} + \left ( 1 + \frac{n-r}{r/(\ell -1)} \right ) \cdot \binom{r/(\ell -1)}{2} \\[5pt] &\leq \binom{r}{2} + \binom{r/(\ell -1)}{2} + \frac{nr}{2(\ell -1)} \\[5pt] &\lt 2\binom{s}{2} + \frac{sn}{2sn/a} \leq a, \end{align*}

where the last inequality uses $a \geq 2s^2$ . We got a contradiction to $e(G) = a$ .

Proof of Theorem 2.1 . The proof is by induction on $n$ . Fix constants $k \ll c \ll c_1 \ll C$ , to be chosen implicitly later. Suppose first that $a \leq (ck + 1)^2n$ . In this case we proceed as follows. If $\delta (G) \geq \lfloor \frac{(k-1)n}{k} \rfloor - c_1\sqrt{n}$ , then take a $(k+1)$ -clique $x_1,\dots,x_{k+1} \in V(G)$ and take $H$ to consist of all edges that touch $x_1,\dots,x_{k+1}$ . Then $H$ is chordal and

\begin{align*} e(H) &= \sum _{i=1}^{k+1}{d(x_i)} - \binom{k+1}{2} \geq (k+1) \cdot \left ( \frac{(k-1)n}{k} - 2c_1\sqrt{n} \right ) - \binom{k+1}{2} \\[5pt] &= (k-1/k)n - 2(k+1)c_1\sqrt{n} - \binom{k+1}{2} \geq (k-1/k)n + \sqrt{2(k+1)a/k} - C\sqrt{n} - \binom{k+1}{2}, \end{align*}

where the last inequality holds as $C \gg c_1,c$ and $a \leq (ck + 1)^2n$ . Suppose now that there is $v \in V(G)$ with $d(v) \leq \lfloor \frac{(k-1)n}{k}\rfloor - c_1\sqrt{n}$ . Let $G' = G - v$ . Then

\begin{equation*}e(G') \geq t_k(n) + a - \left \lfloor \frac {(k-1)n}{k}\right \rfloor + c_1\sqrt {n} = t_k(n-1) + a + c_1\sqrt {n}. \end{equation*}

By the induction hypothesis with parameter $a' = a + c_1\sqrt{n}$ , $G'$ contains a chordal subgraph $H'$ with

\begin{equation*} e(H') \geq (k - 1/k)(n-1) + \sqrt {2(k+1)a'/k} - C\sqrt {n} - \binom {k+1}{2}.\end{equation*}

As $(k-1/k)(n-1) \geq (k-1/k)n - k$ , it suffices to show that $\sqrt{2(k+1)a'/k} \geq \sqrt{2(k+1)a/k} + k$ . Squaring and plugging in the value of $a'$ , we get

\begin{equation*}2(k+1)/k \cdot (a+c_1\sqrt {n}) = 2(k+1)a'/k \geq 2(k+1)a/k + 2k\sqrt {2(k+1)a/k} + k^2.\end{equation*}

Cancelling the term $2(k+1)a/k$ from both sides and rearranging, we see that it is enough to have

\begin{equation*}c_1\sqrt {n} \geq \frac {k^2}{k+1}\sqrt {2(k+1)a/k} + \frac {k^3}{2(k+1)},\end{equation*}

which holds because $a \leq (ck + 1)^2n$ and $c_1 \gg c$ .

For the rest of the proof we assume that $a \geq (ck + 1)^2n$ . Note that $e(G) \geq t_k(n) + a \geq \frac{(k-1)n^2}{2k} + \frac{a}{2}$ because $t_k(n) \geq \frac{(k-1)n^2}{2k} - O_k(1)$ and $a \geq c \gg k$ . Let $x_1,\dots,x_{k-1}$ be as in Lemma 2.2 and put $N = N(x_1,\dots,x_{k-1})$ . By Lemma 2.2 we have $e(N) \geq \frac{a}{2}$ . Also, the choice of $x_1,\dots,x_{k-1}$ in Lemma 2.2 implies that $d(y) \leq d(x_{k-1}) \leq \dots \leq d(x_1)$ for every $y \in N$ . For convenience, we set

\begin{equation*} d_0 \;:\!=\; \frac {(k-1)n}{k} + \sqrt {\frac {2a}{k(k+1)}} - c\sqrt {n}. \end{equation*}

Note that $d_0 \geq \frac{(k-1)n}{k}$ by our assumption that $a \geq (ck + 1)^2n$ .

Claim 2.4. If the statement of the theorem does not hold, then $G[N]$ contains a forest $F$ with $v(F) = \lfloor \sqrt{n} \rfloor$ , $e(F)\geq v(F) - 1 - \frac{2n^{3/2}}{a}$ , and

(3) \begin{equation} \sum _{y \in V(F)}{d(y)} \leq v(F) \cdot d_0 + n/k. \end{equation}

Proof. We consider two cases. Suppose first that there is $x_k \in N$ such that $d(x_k) \geq d_0$ . Then $d(x_i) \geq d_0$ for every $1 \leq i \leq k-1$ . Hence, $d(x_1) + \dots + d(x_k) \geq k \cdot d_0$ . This means that $x_1,\dots,x_k$ have at least

\begin{equation*} kd_0 - (k-1)n = \sqrt {\frac {2ka}{(k+1)}} - ck\sqrt {n} \geq \sqrt {a} - ck\sqrt {n} \geq \sqrt {n} \end{equation*}

common neighbours, where the last inequality holds by the assumption $a \geq (ck + 1)^2n$ . Take $F$ to be the star whose centre is $x_k$ and whose leaves are $\lfloor \sqrt{n} \rfloor - 1$ common neighbours of $x_1,\dots,x_k$ . Let $y \in N(x_1,\dots,x_k)$ . If $d(y) \geq d_0$ then

\begin{equation*} d(x_1) + \dots + d(x_{k}) + d(y) \geq (k+1)d_0 = (k-1/k)n + \sqrt {2(k+1)a/k} - (k+1)c\sqrt {n}, \end{equation*}

and then the subgraph consisting of all edges touching $\{x_1,\dots,x_k,y\}$ is a chordal graph with at least $(k- 1/k)n + \sqrt{2(k+1)a/k} - (k+1)c\sqrt{n} - \binom{k+1}{2}$ edges, so the assertion of the theorem holds. Hence, we may assume that $d(y) \leq d_0$ for every $y \in N(x_1,\dots,x_k)$ . This means that

\begin{equation*} \sum _{v \in V(F)}{d(v)} \leq d(x_k) + (v(F) - 1) \cdot d_0 \leq n/k + v(F) \cdot d_0, \end{equation*}

as required by the claim. Also, $F$ has the right number of edges, as $e(F)=v(F)-1$ .

The second case is that $d(y) \leq d_0$ for every $y \in N$ . Since $e(N) \geq \frac{a}{2} \geq 2n$ , we can apply Lemma 2.3 to $G[N]$ with parameters $\frac{a}{2}$ and $s = \lfloor \sqrt{n} \rfloor$ to obtain a forest $F$ with $\lfloor \sqrt{n} \rfloor$ vertices and at least $v(F) - 1 - \frac{2n^{3/2}}{a}$ edges. All vertices in $F$ have degree at most $d_0$ , so (3) holds.

We continue with the proof of the theorem. Let $F$ be the forest given by Claim 2.4. Let $G'$ be the graph obtained from $G$ by deleting the $t \;:\!=\; k-2 + v(F)$ vertices $T \;:\!=\; \{ x_2,\dots,x_{k-1} \} \cup V(F)$ . By (3), we have

\begin{equation*} \sum _{v \in T}{d(v)} \leq d(x_2) + \dots + d(x_{k-1}) + v(F) \cdot d_0 + n/k \leq (v(F) + k - 2) \cdot d_0 + n = t \cdot d_0 + n, \end{equation*}

where the second inequality uses that $d_0 \geq \frac{(k-1)n}{k}$ . As $e(G) \geq t_k(n) + a \geq \frac{(k-1)n^2}{2k} - O_k(1) + a$ , we have that

\begin{align*} e(G') \geq e(G) - t \cdot d_0 - n &\geq \frac{(k-1)n^2}{2k}- O_k(1) + a - t \cdot d_0 - n \\[5pt] &= \frac{(k-1)(n-t)^2}{2k} + \frac{(k-1)nt}{k} - \frac{(k-1)t^2}{2k} - O_k(1) + a - t \cdot d_0 - n \\[5pt] &= \frac{(k-1)(n-t)^2}{2k} - O_k(1) + a - \frac{(k-1)t^2}{2k} - t \cdot \sqrt{\frac{2a}{k(k+1)}} + ct\sqrt{n} - n \\[5pt] &\geq t_k(n-t) + a - \frac{(k-1)t^2}{2k} - t \cdot \sqrt{\frac{2a}{k(k+1)}} + \frac{c}{2}t\sqrt{n}, \end{align*}

where the last inequality uses that $t \geq \lfloor \sqrt{n} \rfloor$ and $c \gg k$ , so that $\frac{c}{2}t\sqrt{n} \geq O_k(1) + n$ . Set

\begin{equation*}\label {eq:r'} a' \;:\!=\; a - \frac {(k-1)t^2}{2k} - t \cdot \sqrt {\frac {2a}{k(k+1)}} + \frac {c}{2}t\sqrt {n}, \end{equation*}

so that $e(G') \geq t_k(n-t)+a'$ . We have $a' \geq 1$ because $t \leq \sqrt{n}+k-2$ , $a \geq c^2n$ (say) and $c \gg k$ . By the induction hypothesis, $G'$ contains a chordal subgraph $H'$ of size at least

\begin{equation*} e(H') \geq (k-1/k) \cdot (n-t) + \sqrt {2(k+1)a'/k} - C\sqrt {n-t} - \binom {k+1}{2}. \end{equation*}

Let $H$ be the subgraph of $G$ obtained by adding to $H'$ the edges of the clique $x_1,\dots,x_{k-1}$ , the edges between $x_1,\dots,x_{k-1}$ and $V(F)$ , and the edges of $F$ . Then $H$ is chordal. To complete the proof, it suffices to verify that

\begin{equation*}e(H) \geq (k-1/k)n + \sqrt {2(k+1)a/k} - C\sqrt {n} - \binom {k+1}{2}.\end{equation*}

By the definition of $H$ , we have

\begin{equation*}e(H) = e(H') + \binom {k-1}{2} + (k-1) \cdot v(F) + e(F).\end{equation*}

Note that

\begin{equation*}(k-1) \cdot v(F) + e(F) \geq k \cdot v(F) - 1 - \frac {2n^{3/2}}{a} = k \cdot (t - k + 2) - 1 - \frac {2n^{3/2}}{a},\end{equation*}

and therefore,

\begin{equation*}\binom {k-1}{2} + (k-1) \cdot v(F) + e(F) \geq tk - \frac {(k+1)(k-2)}{2} - 1 - \frac {2n^{3/2}}{a}.\end{equation*}

For convenience, set $h \;:\!=\; \frac{(k+1)(k-2)}{2} + 1 + \frac{2n^{3/2}}{a}$ . Then,

(4) \begin{equation} e(H) \geq (k-1/k) \cdot (n-t) + \sqrt{2(k+1)a'/k} - C\sqrt{n-t} - \binom{k+1}{2} + tk - h. \end{equation}

So it remains to verify that the right-hand side of (4) is at least as large as

\begin{equation*}(k-1/k)n + \sqrt {2(k+1)a/k} - C\sqrt {n} - \binom {k+1}{2}.\end{equation*}

Cancel the terms $(k-1/k)n$ , $\binom{k+1}{2}$ which appear in both expressions. Also, we may drop the terms $C\sqrt{n-t}$ , $C\sqrt{n}$ . After rearranging, we get the inequality

\begin{equation*} \sqrt {2(k+1)a'/k} \geq \sqrt {2(k+1)a/k} - \frac {t}{k} + h. \end{equation*}

By squaring and plugging in the value of $a'$ , we get:

(5) \begin{equation} \begin{split} & \frac{2(k+1)}{k} \cdot \left ( a - \frac{(k-1)t^2}{2k} - t \cdot \sqrt{\frac{2a}{k(k+1)}} + \frac{c}{2}t\sqrt{n} \right ) \geq \\[5pt] &\frac{2(k+1)a}{k} + \frac{t^2}{k^2} + h^2 - \frac{2t}{k} \cdot \sqrt{\frac{2(k+1)a}{k}} + 2h \cdot \sqrt{\frac{2(k+1)a}{k}} - \frac{2th}{k}. \end{split} \end{equation}

Both sides of the inequality (5) have the terms $\frac{2(k+1)}{k}a$ and $-\frac{2t}{k} \cdot \sqrt{\frac{2(k+1)a}{k}}$ . We can also drop the negative term $\frac{2th}{k}$ on the right-hand side. After rearranging, we get the inequality

(6) \begin{equation} \frac{2(k+1)}{k} \cdot \frac{c}{2}t\sqrt{n} \geq t^2 + h^2 + 2h \cdot \sqrt{\frac{2(k+1)a}{k}}. \end{equation}

We have $t \leq \sqrt{n} + k$ and $h \leq O_k(1) + \frac{2n^{3/2}}{a} \leq O_k(1) + \sqrt{n}$ , so $t^2,h^2 \leq O_k(n)$ . Also,

\begin{equation*}h \sqrt {a} \leq \left (O_k(1) + \frac {2n^{3/2}}{a}\right ) \cdot \sqrt {a} \leq O_k(n) + \frac {2n^{3/2}}{\sqrt {a}} = O_k(n),\end{equation*}

as $n \leq a \leq n^2$ . So the right-hand side of (6) is $O_k(n)$ . On the other hand, the left-hand side is larger than $\frac{cn}{2}$ because $t \geq \lfloor \sqrt{n}\rfloor \geq \sqrt{n}/2$ . So (6) holds because $c \gg k$ , as required.

Proof of Theorem 1.4 . For the upper bound, let $r \geq 1$ be the minimal integer satisfying $t_2(r) \geq m$ and take $G$ to be $T_2(r)$ with $n-r$ isolated vertices. Then $e(G) \geq m$ , but every chordal subgraph of $G$ has at most $r-1$ edges. For the lower bound, we prove by induction on the number of vertices that every graph $G$ with $m$ edges has a chordal subgraph $H$ with at least $g_1(m)-1$ edges, where $g_1(m) \;:\!=\; \min _{r: t_2(r) \geq m} r$ . For $m = 0$ , the assertion is trivial. Suppose $m \geq 1$ and let $xy \in E(G)$ . Fix $r$ such that $g_1(m) = r$ . If $d(x) + d(y) \geq r$ then take $H$ to be the subgraph consisting of all edges touching $x,y$ . This graph is chordal and has $d(x)+d(y)-1$ edges. Suppose now that $d(x) + d(y) \leq r-1$ ; without loss of generality, $d(x) \leq \lfloor \frac{r-1}{2} \rfloor$ . Let $G' = G-x$ . Then $e(G') = m - d(x) \geq m - \lfloor \frac{r-1}{2} \rfloor$ . We claim that $e(G') \gt t_2(r-2)$ . Indeed, if $e(G') \leq t_2(r-2)$ then $m \leq t_2(r-2) + \lfloor \frac{r-1}{2} \rfloor = t_2(r-1)$ , in contradiction to the choice of $r$ . So $g_1(e(G')) \geq r-1$ . By the induction hypothesis, $G'$ contains a chordal subgraph $H'$ with at least $r-2$ edges. Now, $H' + \{xy\}$ is a chordal subgraph of $G$ with at least $r-1$ edges, as required.

Proof. Fix $t,r$ which achieve the minimum in the definition of $g_2(n,m)$ ; so $t(n-t)+t_2(r) \geq m$ and $g_2(n,m) = 2n-t+r$ . Put $h(t,r) \;:\!=\; 2t - n - \frac{r}{2}$ . If $h(t,r) \in \{-\frac{1}{2},0,\frac{1}{2}\}$ then we are done. If not, we try replacing $(t,r)$ with $(t-1,r-1)$ or $(t+1,r+1)$ . Suppose first that $h(t,r) \geq 1$ . Replace $(t,r)$ with $(t-1,r-1)$ . We have $2n - (t-1) + (r-1) = 2n-t+r$ and $(t-1)(n-t+1) + t_2(r-1) = t(n-t) + t_2(r) - n + 2t - \lfloor \frac{r}{2} \rfloor - 1 \geq t(n-t) + t_2(r) \geq m$ , where the penultimate inequality uses $h(t,r) \geq 1$ . So $(t-1,r-1)$ also achieves the minimum in the definition of $g_2(n,m)$ . Moreover, $h(t-1,r-1) = 2(t-1) - n - \frac{r-1}{2} = h(t,r) - \frac{3}{2}$ . So as long as $h(t,r) \geq 1$ , we can replace $(t,r)$ with $(t-1,r-1)$ and decrease $h$ by $\frac{3}{2}$ . At the last step, we decrease $h$ to be in $\left\{-\frac{1}{2}, 0, \frac{1}{2}\right\}$ .

Similarly, suppose that $h(t,r) \leq -1$ . Replace $(t,r)$ with $(t+1,r+1)$ . We have $2n-(t+1) + (r+1) = 2n-t+r$ and $(t+1)(n-t-1) + t_2(r+1) = t(n-t) + t_2(r) + n - 2t + \lceil \frac{r}{2} \rceil - 1 \geq t(n-t) + t_2(r) \geq m$ , where the penultimate inequality uses $h(t,r) \leq -1$ . So $(t+1,r+1)$ also achieves the minimum in the definition of $g_2(n,m)$ . Also, $h(t+1,r+1) = 2(t+1) - n - \frac{r+1}{2} = h(t,r) + \frac{3}{2}$ . So as long as $h(t,r) \leq -1$ , we can replace $(t,r)$ with $(t+1,r+1)$ and increase $h$ by $\frac{3}{2}$ . At the last step, we increase $h$ to be in $\left\{-\frac{1}{2}, 0, \frac{1}{2}\right\}$ .

Proof. Let $t,r$ be such that $t(n-t) + t_2(r) \geq m-1$ and $2n-t+r = g_2(n,m-1)$ . Since $m-1 \geq t_2(n)+1$ we have $r \geq 2$ . Thus $t(n-t) + t_2(r+1) \geq m$ , so $g_2(n,m) \leq 2n-t+(r+1) = g_2(n,m-1) + 1$ , as required.

Proof. The first part in both items follows from $t_2(n-2) = t_2(n) - n + 1$ . Let $t,r$ such that $t(n-2-t) + t_2(r) \geq m-n+1$ and $g_2(n-2,m-n+1) = 2(n-2) - t + r$ . We have $(t+1) (n-1-t) + t_2(r) = t(n-2-t) + n-1 + t_2(r) \geq m$ . Hence, $g_2(n,m) \leq 2n - (t+1) + r = g_2(n-2,m-n+1) + 3$ . This proves the first item. For the second item, $g_2(n-2,m-n) \geq g_2(n-2,m-n+1) - 1 \geq g_2(n,m) - 4$ , where the first equality uses Lemma 3.3 since $m-n+1\geq t_2(n-2)+2$ .

Proof of Lemma 3.5 . First we show that $m-d \gt t_2(n-1)$ . Set $a \;:\!=\; m - t_2(n)$ . Since $3d \leq g_2(n,m) - 1$ , it is enough to show that

(7) \begin{equation} g_2(n,m) \lt 3(m - t_2(n-1)) + 1 = 3 \cdot \left \lfloor \frac{n}{2} \right \rfloor + 3a + 1. \end{equation}

For $a=1$ , $g_2(n,m) = n + \lfloor \frac{n}{2} \rfloor + 2 \lt 3 \cdot \lfloor \frac{n}{2} \rfloor + 4$ , and the last expression equals the right-hand side of (7). Suppose now that $a \geq 2$ . Set $k = \lceil \sqrt{\frac{a}{3}} \rceil$ , so that $3k^2 \geq a$ . Set $t \;:\!=\; \lfloor \frac{n}{2} \rfloor + k$ and $r \;:\!=\; 4k$ . Then $t_2(r) = 4k^2$ and

\begin{equation*}t(n-t) = \left ( \left \lfloor \frac {n}{2} \right \rfloor + k \right )\left ( \left \lceil \frac {n}{2} \right \rceil - k \right ) \geq t_2(n) - k^2,\end{equation*}

so $t(n-t) + t_2(r) \geq t_2(n) + 3k^2 \geq t_2(n) + a = m$ . Hence, $g_2(n,m) \leq 2n - t + r = 2n - \lfloor \frac{n}{2} \rfloor + 3k \leq 3 \cdot \lfloor \frac{n}{2} \rfloor + 2 + 3k$ . So to prove (7), it suffices to show that $3k+1 \lt 3a$ . As $k \leq \sqrt{\frac{a}{3}}+1$ , it suffices to show that $3\left ( \sqrt{\frac{a}{3}}+1 \right ) \lt 3a-1$ . Rearranging and squaring, we get the inequality $9a^2-27a+16 \gt 0$ , which holds for all $a \geq 3$ . For $a=2$ we simply note that $k = 1$ and so $3k+1 = 4 \lt 6 = 3a$ .

Now we show that $g_2(n-1,m-d) \geq g(n,m) - 2$ . Fix $t,r$ that achieve the minimum in the definition of $g_2(n,m)$ ; so $t(n-t)+t_2(r) \geq m$ and

(8) \begin{equation} g_2(n,m) = 2n-t+r = 3t - 2 \cdot \left (2t - n - \frac{r}{2}\right ). \end{equation}

By Lemma 3.2, we may assume that $-\frac{1}{2} \leq 2t-n-\frac{r}{2} \leq \frac{1}{2}.$ Fix also $t',r'$ such that $t'(n-1-t') + t_2(r') \geq m-d$ and $g_2(n-1,m-d) = 2(n-1) - t' + r'$ . Suppose by contradiction that $g_2(n-1, m-d) \leq g_2(n,m)-3$ . Then $2(n-1) - t' + r' \leq 2n - t + r - 3$ , so $t' - t \geq r' - r + 1$ . Put $c \;:\!=\; t'-t$ , so that $r' \leq r+c-1$ . Then $t_2(r') \leq t_2(r+c-1)$ . So we can write

(9) \begin{equation} \begin{split} m-d &\leq t'(n-1-t') + t_2(r') \leq (t+c)(n-t-c-1) + t_2(r+c-1) \\[5pt] &= t(n-t) + cn - (2c+1)t - (c+1)c + t_2(r+c-1). \end{split} \end{equation}

Since $(t,r)$ achieves the minimum in the definition of $g_2(n,m)$ , we must have $m-1 \geq t(n-t) + t_2(r-1)$ , so $t(n-t) \leq m-1-t_2(r-1)$ . Plugging this into (9) and rearranging, we get

\begin{equation*}d \geq t + c(2t - n) + (c+1)c + 1 - t_2(r+c-1) + t_2(r-1).\end{equation*}

Note that

\begin{equation*}t_2(r+c-1) - t_2(r-1) \leq \frac {(r+c-1)^2}{4} - \frac {(r-1)^2-1}{4} = \frac {cr}{2} + \frac {(c-1)^2}{4}.\end{equation*}

So we get

(10) \begin{equation} d \geq t + c\left (2t-n-\frac{r}{2}\right ) + (c+1)c + 1 - \frac{(c-1)^2}{4}. \end{equation}

We now complete the proof by considering the three possible values of $2t-n-\frac{r}{2}$ . Suppose first that $2t-n-\frac{r}{2} = 0$ . Then $g_2(n,m) = 3t$ by (8). By (10), we have $d \geq t + (c+1)c + 1 - \frac{(c-1)^2}{4} = t + \frac{3}{4}(c+1)^2 \geq t = g_2(n,m)/3$ , in contradiction to our assumption on $d$ .

Suppose now that $2t-n - \frac{r}{2} = \frac{1}{2}$ . Then $g_2(n,m) = 3t-1$ by (8). By (10), we have $d \geq t + \frac{c}{2} + (c+ 1)c + 1 - \frac{c^2-2c+1}{4} = t+\frac{3c^2}{4} + 2c+\frac{3}{4} \gt t-1$ , where the last inequality holds for every $c$ . So $d \geq t$ and hence $3d \geq 3t \geq g_2(n,m)$ , a contradiction.

Finally, suppose that $2t-n - \frac{r}{2} = -\frac{1}{2}$ . Then $g_2(n,m) = 3t+1$ by (8). By (10), we have $d \geq t - \frac{c}{2} + (c+ 1)c + 1 - \frac{c^2-2c+1}{4} = t+\frac{3c^2}{4} + c + \frac{3}{4} \gt t$ , where the last inequality holds for every $c$ . So $d \geq t+1$ and hence $3d \geq 3t+3 \geq g_2(n,m)$ , a contradiction.

Proof of Theorem 3.1 . The proof is by induction on $n$ . The base cases $n=1,2$ are trivial because for these $n$ there is no graph on $n$ vertices with $t_2(n)+1$ edges. The case $n = 3$ is also easy to verify. So from now on let $n \geq 4$ . Let $x,y,z$ be a triangle in $G$ . We consider several cases. After dealing with each case, we will assume in all subsequent cases that this case does not hold.

Case 1: $d(x) + d(y) + d(z) \geq g_2(n,m)$ . In this case, take $H$ to be the graph consisting of all edges touching $x,y,z$ . This graph is chordal and clearly contains the edges of the triangle $x,y,z$ . Also, $e(H) = d(x) + d(y) + d(z) - 3 \geq g_2(n,m) - 3$ , as required.

Case 2: There are distinct $u,v \in \{x,y,z\}$ such that $d(u) + d(v) \leq n$ . Without loss of generality, suppose that $u = x, v = y$ . Let $G' = G - \{x,y\}$ . Then $ e(G') = e(G) - d(x) - d(y) + 1 \geq m - n + 1$ . By the induction hypothesis (applied to any arbitrary triangle in $G'$ ), $G'$ contains a chordal subgraph $H'$ with $e(H') \geq g_2(n-2, m-n+1) - 3 \geq g_2(n,m) - 6$ , by Lemma 3.4. Let $H \;:\!=\; H' + \{xy,xz,yz\}$ . Then $H$ is chordal, contains the edges of the triangle $x,y,z$ , and satisfies $e(H) = e(H') + 3 \geq g_2(n,m) - 3$ .

We claim that if cases - do not hold then $m \geq t_2(n)+2$ . Indeed, suppose by contradiction that $m = t_2(n)+1$ . We have $g_2(n,t_2(n)+1) = 2n - \lceil \frac{n}{2} \rceil + 2$ . Since case does not hold, $d(x) + d(y) + d(z) \leq g_2(n,t_2(n)+1) - 1 \leq \frac{3n}{2} + 1$ . But then there are distinct $u,v \in \{x,y,z\}$ with $d(u) + d(v) \leq \frac{2}{3} \cdot \left ( \frac{3n}{2} + 1 \right ) \leq n + \frac{2}{3}$ . Hence $d(u) + d(v) \leq n$ , contradicting that case does not hold. So $m \geq t_2(n)+2$ .

Case 3: There are distinct $u,v \in \{x,y,z\}$ such that the edge $uv$ is on exactly one triangle (namely the triangle $x,y,z$ ). Without loss of generality, suppose that $u = x, v = y$ . Since case does not hold, we have $d(x) + d(y) \geq n+1$ . On the other hand, since $xy$ is on exactly one triangle, it must be that $d(x) + d(y) = n+1$ , $N(x) \cap N(y) = \{z\}$ and every vertex in $V(G) \setminus \{x,y,z\}$ is adjacent to exactly one of $x,y$ . Let $G' = G - \{x,y\}$ . Then $e(G') = e(G) - d(x) - d(y) + 1 = m - n$ . By Lemma 3.4, $e(G') \geq t_2(v(G')) + 1$ , which means that $G'$ contains a triangle. By the induction hypothesis (applied to an arbitrary triangle in $G'$ ), $G'$ contains a chordal subgraph $H'$ with $ e(H') \geq g_2(n-2, m-n) - 3 \geq g_2(n,m) - 7$ , by Lemma 3.4. So it is enough to show that $G$ contains a chordal graph $H$ which contains the edges of the triangle $x,y,z$ and satisfies $e(H) = e(H') + 4$ . Suppose first that $z$ is isolated in $H'$ . We claim that there exists $w \in N_G(z) \setminus \{x,y\}$ . Indeed, if not, then $d_G(z) = 2$ . Also, as $d(x) + d(y) = n+1$ , we have $d(x) \leq \lfloor \frac{n+1}{2} \rfloor$ or $d(y) \leq \lfloor \frac{n+1}{2} \rfloor$ . Suppose this holds for $y$ . Then $d(y) + d(z) \leq \lfloor \frac{n+1}{2} \rfloor + 2 \leq n$ for $n \geq 4$ , so case holds, contradiction. This proves our claim that there exists $w \in N_G(z) \setminus \{x,y\}$ . Now take $H = H' + \{zw,xy,xz,yz\}$ . By adding the vertices in the order $z,x,y$ , we always add a simplicial vertex. Hence, $H$ is chordal.

Suppose now that $z$ is not isolated in $H'$ , and let $w \in V(G) \setminus \{x,y,z\}$ such that $zw \in E(H')$ . As mentioned above, $w$ is adjacent in $G$ to either $x$ or $y$ ; without loss of generality it is adjacent to $x$ . Take $H = H' + \{xw,xy,xz,yz\}$ . By adding to $H'$ first $x$ and then $y$ , we always add a simplicial vertex. Hence, $H$ is chordal.

Case 4: Cases - do not hold. Since case does not hold, we have $d(x) + d(y) + d(z) \leq g_2(n,m) - 1$ . Assume that $d \;:\!=\; d(x) \leq d(y) \leq d(z)$ ; then $3d \leq g_2(n,m) - 1$ . Since case does not hold, there exists $w \in V(G) \setminus \{x\}$ which is a common neighbour of $y,z$ . Set $G' = G - \{x\}$ . We have $e(G') = m - d$ . By the induction hypothesis, $G'$ has a chordal subgraph $H'$ which contains the edges of the triangle $y,z,w$ and satisfies $ e(H') \geq g_2(n-1,m-d) - 3 \geq g_2(n,m) - 5,$ by Lemma 3.5. Let $H = H' + \{xy,xz\}$ . Then $H$ is chordal, contains the edges of the triangle $x,y,z$ , and satisfies $e(H) = e(H') + 2 \geq g_2(n,m) - 3$ , as required.

Proof. For $i = 1,2,3$ , $T_3(n-i)$ is obtained from $T_3(n)$ by deleting one vertex from each of the $i$ largest classes of $T_3(n)$ . For $i = 1$ , the deleted vertex has degree $\lfloor \frac{2n}{3} \rfloor$ . For $i = 2$ , the sum of degrees of the deleted vertices is $\lfloor \frac{4n}{3} \rfloor$ , and these vertices are adjacent. For $i = 3$ , the sum of degrees of the deleted vertices is $2n$ , and they form a triangle. Finally, $t_3(n) - t_3(n-4) = t_3(n) - t_3(n-1) + t_3(n-1) - t_3(n-4) = \lfloor \frac{2n}{3} \rfloor + 2(n-1)-3 = 3n-\lceil \frac{n}{3} \rceil - 5 = g_3(n) - 7$ .

Proof. $g_3(n) - g_3(n-1) = 3 - \lceil \frac{n}{3} \rceil + \lceil \frac{n-1}{3} \rceil$ , and it is easy to see that $\lceil \frac{n}{3} \rceil - \lceil \frac{n-1}{3} \rceil$ equals $0$ if $n \equiv 0,2 \pmod{3}$ and equals $1$ if $n \equiv 1 \pmod{3}$ .

Proof. The proof is by induction on $n$ . For the base case $n = 5$ , take a $4$ -clique and notice that the remaining vertex must send at least 3 edges to this $4$ -clique as $t_3(5)+1 = 9$ . Let $n \geq 6$ . Take a $4$ -clique $x_1,\dots,x_4$ . If there is a vertex outside of $x_1,\dots,x_4$ which has three neighbours in $x_1,\dots,x_4$ then we are done. Else, $d(x_1) + \dots + d(x_4) \leq 2n+4$ , so there is $i \in [4]$ such that $d(x_i) \leq \lfloor \frac{n}{2}\rfloor + 1 \leq \lfloor \frac{2n}{3}\rfloor$ , where the last inequality holds for all $n \geq 6$ . So $G' \;:\!=\; G-x_i$ has at least $t_3(n-1)+1$ edges, and we can apply induction.

Proof of Theorem 4.1 . The proof is by induction on $n$ . The claim is trivial if $n \leq 4$ so suppose that $n \geq 5$ . We proceed by a sequence of claims.

Proof of Claim 4.5 . Suppose that there is no such $y$ . We will show that the assertion of Theorem 4.1 holds. For $0 \leq j \leq 2$ , let $A_j$ be the set of vertices in $V(G) \setminus \{x_1,\dots,x_4\}$ which are adjacent to $j$ of the vertices $x_1,\dots,x_4$ , and let $a_j = |A_j|$ . Then $a_0 + a_1 + a_2 = n-4$ and $a_1 + 2a_2 = d(x_1) + \dots + d(x_4) - 12$ . So

(11) \begin{equation} d(x_1) + \dots + d(x_4) = n + 8 + a_2 - a_0. \end{equation}

We consider three cases:

Case 1: $0 \lt a_2 \leq n-5$ . Then $d(x_1) + \dots + d(x_4) \leq 2n+3$ by (11). Let $G'$ be the graph obtained from $G$ by deleting $x_2,x_3,x_4$ and all edges touching $x_1$ . Then $v(G') = n-3$ and $e(G') = e(G) - (d(x_1) + \dots + d(x_4)) + 6 \geq e(G) - 2n + 3 = t_3(n) + 1 - 2n + 3 = t_3(n-3) + 1.$ In particular, $n-3 \geq 4$ (else $G'$ cannot have more than $t_3(n-3)$ edges). By the induction hypothesis (applied to an arbitrary $4$ -clique in $G'$ ), $G'$ contains a chordal subgraph $H'$ with $e(H') \geq g_3(n-3)-6$ . Take $z \in A_2$ and assume without loss of generality that $z$ is adjacent to $x_1,x_2$ . Let $H = H' + \{x_1z,x_2z\} + \{x_ix_j \;:\; 1 \leq i \lt j \leq 4\}$ . Then $e(H) = e(H') + 8 \geq g_3(n-3) + 8 - 6 = g_3(n) - 6$ . Also, $H$ is chordal: adding $x_1z$ to $H'$ keeps it chordal because $x_1$ is isolated in $H'$ , and then by adding $x_2,x_3,x_4$ in this order we always add a simplicial vertex.

Case 2: $a_2 = n-4$ . Then every vertex outside of $\{x_1,\dots,x_4\}$ is adjacent to two of the vertices $x_1,\dots,x_4$ . By (11), $d(x_1) + \dots + d(x_4) = 2n+4$ . Pick an arbitrary $v \in V(G) \setminus \{x_1,\dots,x_4\}$ and suppose without loss of generality that $v$ is adjacent to $x_1$ . Let $G'$ be the graph obtained from $G$ by deleting $x_2,x_3,x_4$ and all edges touching $x_1$ except for $x_1v$ . Then $v(G') = n-3$ and $e(G') = e(G) - (d(x_1) + \dots + d(x_4)) + 7 = e(G) - 2n + 3 = t_3(n-3) + 1$ . Pick an arbitrary $4$ -clique $y_1,\dots,y_4$ in $G'$ . By the induction hypothesis, $G'$ contains a chordal subgraph $H'$ with $e(H') \geq g_3(n-3) - 6$ such that $H'$ contains the edges of the clique $y_1,\dots,y_4$ . Each $y_i$ has two neighbours in $x_1,\dots,x_4$ . By pigeonhole, two of the $y_i$ ’s have a common neighbour. Without loss of generality, we may assume that $y_1,y_2$ are both adjacent to $x_k$ for some $k \in [4]$ . Suppose that $y_1$ is also adjacent to $x_{\ell }$ . Let $H \;:\!=\; H' - x_1v + \{x_ky_1,x_ky_2,x_{\ell }y_1\} + \{x_ix_j \;:\; 1 \leq i \lt j \leq 4\}$ , see Figure 2(a). Then $e(H) \geq e(H') + 8$ (since we add $9$ edges and delete at most $1$ ), so $e(H) \geq g_3(n-3) + 8 - 6 = g_3(n) - 6$ . Also, $H' - x_1v$ is chordal because $x_1$ is a leaf or an isolated vertex in $H'$ . Now observe that $H$ is chordal: adding first $x_k$ , then $x_{\ell }$ and then the other two vertices among $x_1,\dots,x_4$ , we add a simplicial vertex at each step.

Figure 2. Illustrations to proofs of claims. The red edges are added after applying induction. The dashed edges are deleted.

Case 3: $a_2 = 0$ . Then there are at most $n-4$ edges between $x_1,\dots,x_4$ and $V(G) \setminus \{x_1,\dots,x_4\}$ . Also, $d(x_1) + \dots + d(x_4) \leq n+8$ by (11). Let $G'$ be the graph obtained from $G$ by deleting $x_2,x_3,x_4$ and all edges touching $x_1$ . Then $v(G') = n-3$ and $e(G') \geq e(G) - (n + 2) = t_3(n) + 1 - (n + 2) = t_3(n-3) + 1 + n - 5$ . Note that $n \geq 8$ because else $e(G) \leq \binom{n-4}{2} + n - 4 + 6 \lt t_3(n)$ (where the last inequality holds for $n \leq 7$ ), a contradiction. So $v(G') \geq 5$ . By Lemma 4.4, there are distinct $v_1,v_2,w_1,w_2,w_3 \in V(G')$ such that $v_i,w_1,w_2,w_3$ is a $4$ -clique for $i = 1,2$ . Let $G''$ be the graph obtained from $G'$ by deleting all edges touching $v_1$ . Then $v(G'') = n-3$ and $e(G'') \geq e(G') - (n-5)$ because $v_1$ is not adjacent in $G'$ to $x_1$ or to itself, leaving $n-5$ potential neighbours. So $e(G'') \geq t_3(n-3) + 1$ . By the induction hypothesis, $G''$ contains a chordal subgraph $H'$ with $e(H') \geq g_3(n-3) - 6$ , such that $H'$ contains the edges of the clique $v_2,w_1,w_2,w_3$ . Let $H = H' + \{v_1w_i \;:\; 1 \leq i \leq 3\} + \{x_ix_j \;:\; 1 \leq i \lt j \leq 4\}$ . Then $e(H) = e(H') + 9 \geq g_3(n) - 6$ , and it is easy to check that $H$ is chordal.

Proof of Claim 4.6 . Without loss of generality, $i = 1$ . Suppose that $d(x_1) \geq \lfloor \frac{2n}{3} \rfloor + 1$ but there is no $y$ as above. We will show that the assertion of Theorem 4.1 holds. For $0 \leq j \leq 3$ , let $A_j$ be the set of vertices $z \in V(G) \setminus \{x_2,x_3,x_4\}$ which are adjacent to exactly $j$ of the vertices $x_2,x_3,x_4$ , and let $a_j = |A_j|$ . Note that $x_1 \in A_3$ . By Claim 4.5, we may assume that there exists $z \in V(G) \setminus \{x_1,\dots,x_4\}$ which is adjacent to three of the vertices $x_1,\dots,x_4$ . By assumption, these three vertices cannot include $x_1$ , so we must have $z \in A_3$ . Hence, $A_3 \setminus \{x_1\} \neq \emptyset$ .

We have $a_0 + a_1 + a_2 + a_3 = n-3$ and $a_1 + 2a_2 + 3a_3 = d(x_2) + d(x_3) + d(x_4) - 6$ . So $a_2 + 2a_3 = d(x_2) + d(x_3) + d(x_4) - n - 3 + a_0$ . By assumption, there is no $y \in V(G) \setminus \{x_1,\dots,x_4\}$ which is adjacent to $x_1$ and belongs to $A_2 \cup A_3$ . Hence, $d(x_1) \leq n - a_2 - a_3 = 2n - d(x_2) - d(x_3) - d(x_4) + 3 + a_3 - a_0$ . So we get

(12) \begin{equation} d(x_1) + \dots + d(x_4) \leq 2n+3 + a_3 - a_0. \end{equation}

Pick a set of edges $F$ as follows:

• • Suppose first that $a_0 = 0$ . We claim that $A_1 \cup A_2 \neq \emptyset$ . Indeed, if $A_1 \cup A_2 = \emptyset$ then $A_3 = V(G) \setminus \{x_2,x_3,x_4\}$ . But then $N(x_1) = \{x_2,x_3,x_4\}$ so $d(x_1) = 3$ . On the other hand, $d(x_1) \geq \lfloor \frac{2n}{3} \rfloor + 1$ by assumption. So $\lfloor \frac{2n}{3} \rfloor \leq 2$ , which is false for every $n \geq 5$ . Now pick an arbitrary $v \in A_1 \cup A_2$ , and suppose without loss of generality that $v$ is adjacent to $x_4$ . Take $F$ to be the set of all edges between $x_4$ and $N(x_4) \setminus (A_3 \cup \{x_2,x_3,v\})$ . Note that $|F| = d(x_4) - a_3 - 3$ .

• • If $a_0 \geq 1$ then $F$ is the set of all edges between $x_4$ and $N(x_4) \setminus (A_3 \cup \{x_2,x_3\})$ . Then $|F| = d(x_4) - a_3 - 2$ .

In both cases we have $|F| \leq d(x_4) - a_3 - 3 + a_0$ . Let $G'$ be the graph obtained from $G$ by deleting $x_1,x_2,x_3$ and all edges in $F$ . So $v(G') = n-3$ . To count the deleted edges, note that there are $d(x_1) + d(x_2) + d(x_3) - 3$ edges touching $x_1,x_2,x_3$ , and that the edges in $F$ do not touch $x_1,x_2,x_3$ . So the number of deleted edges is

\begin{equation*} d(x_1) + d(x_2) + d(x_3) - 3 + |F| \leq d(x_1) + \dots + d(x_4) - a_3 + a_0 - 6 \leq 2n-3, \end{equation*}

where the last inequality uses (12). So $e(G') \geq e(G) - 2n + 3 = t_3(n) + 1 - 2n + 3 = t_3(n-3) + 1$ . By the induction hypothesis, $G'$ contains a chordal subgraph $H'$ with $e(H') \geq g_3(n-3) - 6$ . Define a subgraph $H''$ of $G'$ as follows:

• • If there is $z \in A_3 \setminus \{x_1\}$ such that $x_4z \in E(H')$ , then $H'' = H'$ ;

• • Else, pick any $z \in A_3 \setminus \{x_1\}$ and set $H'' = H' - x_4v + x_4z$ .

We claim that $e(H'') \geq e(H')$ . In the first item this is obvious. Suppose we are in the second item. By the definition of $F$ , every neighbour of $x_4$ in $G'$ belongs to $\{v\} \cup A_3 \setminus \{x_1\}$ . Since we are in the second item, the only possible neighbour of $x_4$ in $H'$ is $v$ . So indeed $e(H'') \geq e(H')$ . Also, $H''$ is chordal; in the second case, this is because $x_4$ is a leaf in both $H'$ and $H''$ .

By the definition of $H''$ , there is $z \in A_3 \setminus \{x_1\}$ such that $x_4z \in E(H'')$ . Now put $H = H'' + \{x_2z,x_3z\} + \{x_ix_j \;:\; 1 \leq i \lt j \leq 4\}$ , see Figure 2(b). Then $e(H) = e(H'') + 8 \geq g_3(n-3) + 8 - 6 = g_3(n) - 6$ . Also, $H$ is chordal: adding the vertices $x_2,x_3,x_1$ in this order, we always add a simplicial vertex. This proves Claim 4.6.

We now continue with the proof of the theorem. For $i \in [4]$ , we say that $x_i$ is deletable if $d(x_i) \leq \lfloor \frac{2n}{3} \rfloor$ , and there is $y \in V(G) \setminus \{x_1,\dots,x_4\}$ which is adjacent to $x_j$ for all $j \in [4] \setminus \{i\}$ .

By Claim 4.5, we may assume that there is $y_0 \in V(G) \setminus \{x_1,\dots,x_4\}$ which is adjacent to at least 3 of the vertices $x_1,\dots,x_4$ ; say to $x_2,x_3,x_4$ . By Claim 4.7, we may assume that $x_i$ is not deletable for any $i \in [4]$ . Since $x_1$ is not deletable, $d(x_1) \geq \lfloor \frac{2n}{3} \rfloor + 1$ . By Claim 4.6, there is $z_0 \in V(G) \setminus \{x_1,\dots,x_4\}$ which is adjacent to $x_1$ and to at least two of the vertices $x_2,x_3,x_4$ ; without loss of generality, $z_0$ is adjacent to $x_3,x_4$ . Since $x_2$ is not deletable, we have $d(x_2) \geq \lfloor \frac{2n}{3} \rfloor + 1$ . Also, we may assume that $y_0 \neq z_0$ because else we would have a $5$ -clique containing $x_1,\dots,x_4$ .

If $d(y_0) \leq \lfloor \frac{2n}{3} \rfloor$ then we can delete $y_0$ , apply induction to the remaining graph to find a chordal subgraph $H'$ with at least $g_3(n-1)-6$ edges which contains the edges of the clique $x_1,\dots,x_4$ and then add the edges $x_2y_0,x_3y_0,x_4y_0$ to $H'$ to conclude the proof. So suppose that $d(y_0) \geq \lfloor \frac{2n}{3} \rfloor +1$ . Let $v \notin \{x_3,x_4\}$ be a common neighbour of $x_2$ and $y_0$ . Such a vertex $v$ exists because $d(x_2) + d(y_0) \geq 2 \cdot \left ( \lfloor \frac{2n}{3} \rfloor + 1 \right ) \geq n+3$ , where the last inequality holds for every $n \geq 5$ . Note that $v \neq x_1$ because $x_1$ is not adjacent to $y_0$ , as otherwise $x_1,\dots,x_4,y_0$ would be a $5$ -clique. Similarly, $v \neq z_0$ because $v$ is adjacent to $x_2$ and $z_0$ is adjacent to $x_1,x_3,x_4$ (so otherwise $x_1,\dots,x_4,z_0$ would be a $5$ -clique).

Suppose first that $x_2,y_0,v$ have no common neighbour. Then $d(x_2) + d(y_0) + d(v) \leq 2n$ . Let $G' = G - \{x_2,y_0,v\}$ . Then $e(G') = e(G) - d(x_2) - d(y_0) - d(v) + 3 \geq t_3(n) + 1 - 2n + 3 = t_3 (n-3) + 1$ . By the induction hypothesis, $G'$ contains a chordal subgraph $H'$ with $e(H') \geq g_3(n-3) - 6$ such that $H'$ contains the edges of the clique $x_1,x_3,x_4,z_0$ . Take $H = H' + \{x_2x_1,x_2x_3,x_2x_4,y_0x_2,y_0x_3,y_0x_4,vx_2,vy_0\}$ , see Figure 3(a). Then $e(H) = e(H') + 8 \geq g_3(n-3) + 8 - 6 = g_3(n) - 6$ . Also, $H$ is chordal: adding the vertices $x_2,y_0,v$ in this order, we always add a simplicial vertex.

Figure 3. Applying induction after deleting certain vertices. The added edges are red.

Suppose now that $x_2,y_0,v$ have a common neighbour $w$ . First, suppose that $w \in \{x_3,x_4\}$ , say $w = x_3$ . Let $G' = G - \{x_1,x_4\}$ . We have $d(x_1) + d(x_4) \leq 2n - d(x_2) \leq 2n - \lfloor \frac{2n}{3} \rfloor - 1 \leq \lfloor \frac{4n}{3} \rfloor$ , where the first inequality is by Claim 4.10. So $e(G') = e(G) - d(x_1) - d(x_4) + 1 \geq t_3(n) + 1 - \lfloor \frac{4n}{3} \rfloor + 1 = t_3(n-2) + 1$ . By the induction hypothesis, $G'$ has a chordal subgraph $H'$ with $e(H') \geq g_3(n-2) - 6$ such that $H'$ contains the edges of the clique $x_2,y_0,v,x_3$ . Take $H = H' + \{x_4x_2,x_4x_3,x_4y_0,x_1x_2,x_1x_3,x_1x_4\}$ , see Figure 3(b). Then $e(H) = e(H') + 6 \geq g_3(n-2) + 6 - 6 \geq g_3(n) - 6$ . Also, $H$ is chordal: adding $x_4$ and then $x_1$ , we always add a simplicial vertex. Now suppose that $w \notin \{x_3,x_4\}$ . Let $G' = G - \{x_1,x_3,x_4\}$ . By Claims 4.10 and 4.11, we have $d(x_1) + d(x_3) \leq 2n - d(x_2) \leq \frac{4n}{3}$ and $d(x_4) \leq \frac{2n}{3}$ , so $d(x_1) + d(x_3) + d(x_4) \leq 2n$ . Hence, $e(G') = e(G) - d(x_1) - d(x_3) - d(x_4) + 3 \geq t_3(n) + 1 - 2n + 3 = t_3(n-3) + 1$ . By the induction hypothesis, $G'$ has a chordal subgraph $H'$ with $e(H') \geq g_3(n-3) - 6$ such that $H'$ contains the edges of the clique $x_2,y_0,v,w$ . Now take $H = H' + \{x_3y_0,x_4y_0\} + \{x_ix_j \;:\; 1 \leq i \lt j \leq 4\}$ , see Figure 3(c). Then $e(H) = e(H') + 8 \geq g_3(n-3) + 8 - 6 = g_3(n) - 6$ . Also, $H$ is chordal: adding the vertices $x_3,x_4,x_1$ in this order, we always add a simplicial vertex. This completes the proof of the theorem.