# 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. • I would have expected $L(v_i)$ in equation $(37)$ of the book. – mvw Jan 1 '16 at 23:11
• @mvw newer editions have $v_1, v_2$ instead of $q_1, q_2.$ – Weltschmerz Jan 1 '16 at 23:21
• If one applies the Legendre transformation to $(36)$, one arrives closer to $(37)$. Alas I have no idea yet how to handle $\lVert p_1 - p_2\rVert^2$. – mvw Jan 1 '16 at 23:38
• Isn't it straightforward to show that the hessian $L_{v_i,v_j} = H_{p_i,p_j}^{-1}$? So the lower bound on the norm of $H_{p_i,p_j}$ becomes an upper bound on the norm of $L_{v_i,v_j}$ from which the convexity follows. – user226970 Jan 4 '16 at 1:08
• @epimorphic using $v_1$ and $v_2$ such that $L(v_j)=p_j\cdot v_j - H(p_j), j=1,2.$ – Weltschmerz Jan 11 '16 at 0:32

## 1 Answer

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

$$\left\lvert \sqrt\theta (p_1 - p_2) - \frac{q_1 - q_2}{\sqrt\theta}\right\rvert^2 \geq 0.$$

• Wonderful, thanks. – Weltschmerz Jan 11 '16 at 19:36
• @Weltschmerz Apart from the sign error, that is :P – epimorphic Jan 11 '16 at 19:37