Obtaining an estimate for the Lagrangian $L=H^*$ from the Hamiltonian $H$ This is from C. Evans' PDE book, page 130. The convex function $H:\mathbb{R}^n\to\mathbb{R}$ is $C^2$ and satisfies $$ H\big(\frac{p_1+p_2}{2}\big) \leq \frac{1}{2}H(p_1) + \frac{1}{2}H(p_2) - \frac{\theta}{8}|p_1-p_2|^2. $$ The lagrangian $L$ is defined as the Legendre transform of $H$: $$ L(v)=(H^*)(v)=\max_{p\in\mathbb{R}^n}p\cdot v - H (p). $$  It's a fact that the statements $p\cdot v = L(v) + H(p), $ $ p=DL(v), $ $v=DH(p)$ are equivalent. 
One must show that from the above estimate it follows that $$ \frac{1}{2}L(v_1)+\frac{1}{2}L(v_2)\leq L\big(\frac{v_1+v_2}{2}\big) + \frac{1}{8\theta}|v_1-v_2|^2.$$ I haven't been able to show this. Below is a screenshot of the book's page.
Thank you.


As your hint says, for every $q_1, q_2 \in \mathbb R^n$ there exist $p_1, p_2 \in \mathbb R^n$ such that $L(q_j) = p_j \cdot q_j - H(p_j)$ for $j \in \{1,2\}$. Combining with the inequality for $H$,
\begin{align}
\frac{1}{2} L(q_1) + \frac{1}{2} L(q_2)
&= \frac{p_1 \cdot q_1 + p_2 \cdot q_2}{2} - \frac{1}{2} H(p_1) - \frac{1}{2} H(p_2) \\
&\leq \frac{p_1 \cdot q_1 + p_2 \cdot q_2}{2} - H\left(\frac{p_1+p_2}{2}\right) - \frac{\theta}{8} \lvert p_1 - p_2\rvert^2.
\end{align}
Note that
\begin{align}
\frac{p_1 \cdot q_1 + p_2 \cdot q_2}{2} - H\left(\frac{p_1+p_2}{2}\right)
&= \frac{p_1 - p_2}{2} \cdot \frac{q_1 - q_2}{2} + \frac{p_1 + p_2}{2} \cdot \frac{q_1 + q_2}{2} - H\left(\frac{p_1+p_2}{2}\right) \\
&\leq \frac{p_1 - p_2}{2} \cdot \frac{q_1 - q_2}{2} + \max_{p \in \mathbb R^n} \left(p \cdot \frac{q_1 + q_2}{2} - H(p)\right) \\
&= \frac{p_1 - p_2}{2} \cdot \frac{q_1 - q_2}{2} + L\left(\frac{q_1 + q_2}{2}\right),
\end{align}
which means
$$
\frac{1}{2} L(q_1) + \frac{1}{2} L(q_2)
\leq L\left(\frac{q_1 + q_2}{2}\right) - \frac{\theta \lvert p_1 - p_2\rvert^2 - 2(p_1 - p_2) \cdot (q_1 - q_2)}{8}.
$$
The desired inequality for $L$ is now immediate from the fact that
$$
\left\lvert \sqrt\theta (p_1 - p_2) - \frac{q_1 - q_2}{\sqrt\theta}\right\rvert^2
\geq 0.
$$
