Solution

Theorem J is a direct consequence of Theorem L.

Theorem L states that if numbers $\theta_i$ can be chosen so that

$$ p_i\leq \theta_i\prod_{i\mathbin{-}j\text{ in }G}(1-\theta_j) $$

for every vertex $i$ of $G$, then the probability vector $(p_1,\ldots,p_n)$ belongs to $R(G)$.

In Theorem J, every vertex of $G$ has degree at most $d-1$, and the probability vector is symmetric:

$$ p_1=p_2=\cdots=p_n=p. $$

Choose the parameters in Theorem L symmetrically by setting

$$ \theta_1=\theta_2=\cdots=\theta_n=\frac1d . $$

For a vertex $i$, the product in Theorem L contains at most $d-1$ factors because the degree of $i$ is at most $d-1$. Hence

$$ \theta_i\prod_{i\mathbin{-}j\text{ in }G}(1-\theta_j) \geq \frac1d\left(1-\frac1d\right)^{d-1}. $$

The right-hand side simplifies to

$$ \frac1d\left(\frac{d-1}{d}\right)^{d-1}

\frac{(d-1)^{d-1}}{d^d}. $$

Therefore, whenever

$$ p\leq \frac{(d-1)^{d-1}}{d^d}, $$

the hypotheses of Theorem L are satisfied, and the conclusion of Theorem L gives

$$ (p,\ldots,p)\in R(G). $$

This is precisely the assertion of Theorem J. Hence Theorem J is obtained by applying Theorem L with the uniform choice $\theta_i=1/d$.

This completes the proof.

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