gpt pro 5.2 attempt on Math-978

FORMAL RESTATEMENT.
All graphs are finite, simple, undirected. For graphs $G,H$ write $H\to (G,G)$ if for every red/blue coloring of $E(H)$ there is a monochromatic copy of $G$ in $H$. The size Ramsey number of $G$ is \[ \hat R(G):=\min\{|E(H)|: H\to (G,G)\}. \] The question asks whether there exists a function $f:(0,\infty)\to (0,\infty)$ with $\lim_{x\to\infty} f(x)/x=\infty$ such that for all sufficiently large real $C$, every graph $G$ with $n:=|V(G)|$ and $e:=|E(G)|\ge Cn$ satisfies \[ \hat R(G) > f(C)\, e. \]

QUICK LITERATURE/CONTEXT CHECK.
No literature is quoted in the problem text. I will not import external results about size Ramsey numbers.

ATTACK PLAN.
Proof track. Derive general lower bounds on $\hat R(G)$ in terms of $n,e$ and (possibly) other parameters controlled by the hypothesis $e\ge Cn$.

Disproof track. Try to find, for arbitrarily large $C$, graphs $G$ with $e\ge Cn$ but with size Ramsey number $\hat R(G)$ only on the order of $C\,e$ (or smaller). Any universal upper bound of the form $\hat R(G)\le K C e$ for such a family would contradict the desired superlinear $f(C)$.

WORK.
Phase 1: fast reality check (small instance). For $G=K_3$ one can brute-force check that the smallest triangle-Ramsey graph has $6$ vertices and $15$ edges (namely $K_6$), i.e. $\hat R(K_3)=15$. (I verified this by enumerating all graphs on at most $6$ vertices and checking all $2^{|E(H)|}$ two-colorings for a monochromatic triangle.) This sanity-check is consistent with the trivial lower bound $\hat R(G)\ge e(G)$ below.

Lemma 911.1 (trivial edge-count lower bound). For every graph $G$, $\hat R(G)\ge |E(G)|$.

Proof. Let $e=|E(G)|$. If a graph $H$ has $|E(H)|<e$, then $H$ cannot contain a copy of $G$ as a (not-necessarily-induced) subgraph because any copy of $G$ uses $e$ distinct edges. Color the edges of $H$ arbitrarily red/blue; then there is no monochromatic copy of $G$ (indeed no copy at all). Hence no such $H$ is Ramsey for $G$, so the minimum defining $\hat R(G)$ is at least $e$. $\square$

Lemma 911.2 (vertex Ramsey lower bound). Let $R(G)$ denote the usual (two-color) Ramsey number of $G$, i.e. the least integer $N$ such that $K_N\to (G,G)$. If $H\to (G,G)$ then $|V(H)|\ge R(G)$.

Proof. Let $N:=|V(H)|$. Suppose for contradiction that $N<R(G)$. By the definition of $R(G)$, there exists a red/blue coloring of the complete graph $K_N$ with no monochromatic copy of $G$. Restrict this coloring to the edge set $E(H)\subseteq E(K_N)$; since any monochromatic copy of $G$ in $H$ would also be a monochromatic copy of $G$ in $K_N$, the restricted coloring has no monochromatic copy of $G$ in $H$. This contradicts $H\to (G,G)$. $\square$

Lemma 911.3 (first-moment obstruction via random coloring). Fix graphs $G$ and $H$. Let $e:=|E(G)|$ and let $N_G(H)$ be the number of (labeled) subgraphs of $H$ isomorphic to $G$ (equivalently, the number of injections $\varphi:V(G)\hookrightarrow V(H)$ such that $\varphi$ maps edges of $G$ to edges of $H$).

In a uniformly random red/blue coloring of $E(H)$, the expected number of monochromatic copies of $G$ equals $N_G(H)\,2^{1-e}$. In particular, if $N_G(H)<2^{e-1}$ then $H\not\to (G,G)$.

Proof. Color each edge of $H$ independently red/blue with probability $1/2$. Fix a labeled copy $\varphi$ of $G$ in $H$. The event that this copy is monochromatic is the event that all $e$ edges of $\varphi(G)$ are red or all are blue. Each has probability $(1/2)^e$, and the two events are disjoint, so \[\mathbb{P}(\varphi(G)\text{ is monochromatic}) = 2\cdot (1/2)^e = 2^{1-e}. \] Let $X$ be the random variable counting monochromatic labeled copies of $G$ in $H$. Then \[\mathbb{E}X = \sum_{\varphi} \mathbb{P}(\varphi(G)\text{ is monochromatic}) = N_G(H)\,2^{1-e} \] by linearity of expectation. If $N_G(H)<2^{e-1}$ then $\mathbb{E}X<1$, hence there exists at least one coloring with $X=0$, i.e. with no monochromatic copy of $G$; thus $H$ is not Ramsey for $G$. $\square$

VERIFICATION.

• Lemma 911.1 uses only the subgraph containment requirement implicit in $H\to (G,G)$.
• Lemma 911.2 correctly handles the "sparser host" issue by extending a coloring to the complete graph and then restricting.
• Lemma 911.3 uses a labeled copy count; unlabeled versions differ by $|\mathrm{Aut}(G)|$ but the conclusion "$\mathbb{E}X<1$ implies existence of a coloring with $X=0$" is unaffected.

FINAL.
UNRESOLVED.

(i) Strongest proved partial result: general lower bounds $\hat R(G)\ge |E(G)|$ (Lemma 911.1) and $|V(H)|\ge R(G)$ for any Ramsey host $H$ (Lemma 911.2). Also, Lemma 911.3 gives a clean probabilistic criterion: any candidate host $H$ with too few copies of $G$ (specifically $N_G(H)<2^{e-1}$) cannot be Ramsey for $G$.

(ii) First gap (crisp): Prove or disprove the existence of a function $f(C)$ with $f(C)/C\to\infty$ such that every graph $G$ with $e(G)\ge C|V(G)|$ satisfies $\hat R(G) > f(C)\,|E(G)|$ for all sufficiently large $C$.

(iii) Top 3 next moves:

1. Disproof attempt: Construct, for arbitrarily large $C$, explicit graphs $G_C$ with $|E(G_C)|\ge C|V(G_C)|$ and prove an upper bound $\hat R(G_C)\le K\,C\,|E(G_C)|$ with an absolute constant $K$. Any such family would rule out superlinear $f$.
2. Proof attempt: Fix $C$ and attempt to show that every $H$ with $|E(H)|\le g(C)\,e$ has $N_G(H)<2^{e-1}$ for some $g(C)$ growing faster than $C$. Combining with Lemma 911.3 would imply the desired lower bound.
3. Computation: For small $C$ and moderate $n$, brute-force search for graphs $G$ with $e\approx Cn$ and compute/estimate $\hat R(G)$ to guess scaling in $C$ (even partial data could suggest whether $\hat R(G)/e$ can stay $O(C)$).

(iv) Minimal counterexample structure:

• To the existence of such $f$: a family $\{G_C\}$ with average degree $\Theta(C)$ but with $\hat R(G_C)=O(C\,e(G_C))$.
• To the nonexistence: a fixed large $C$ and a graph $G$ with $e\ge Cn$ such that every Ramsey host has $\hat R(G)\ge f(C)e$ for some $f(C)\gg C$.