I need some help with 2 things in Evan's Partial Differential Equations. Both being part of chapter 3.3.3, proof of Lemma 4. Our prerequisite is that H is uniformly convex, i.e.: $$\sum_{i.j=1}^{n} H_{p_ip_j}(p)\xi_i \xi_j \geq \theta |\xi|^2 \;\;\forall p,\xi \in \mathbb{R}^n. \;\;\;\;\;(1)$$ Now Evans claims that this implies (with Taylor's formula) $$H\left(\frac{p_1+p_2}{2}\right) \leq \frac{1}{2}H(p_1)+\frac{1}{2}H(p_2)-\frac{\theta}{8}|p_1-p_2|^2. \;\;\;\;\;(2)$$ But I don't know how to prove it. Here's what I have: $$ H(x + \xi) \overset{Taylor}{=} H(x) + \sum_{i=1}^{n} H_{p_i} (x) \xi_i + 2 \sum_{|\alpha|=2} \int_{0}^{1} (1-t) \frac{D^\alpha H(x+t\xi)}{\alpha!} \xi^\alpha dt $$ $$ = H(x) + \sum_{i=1}^{n} H_{p_i} (x) \xi_i + \int_{0}^{1} (1-t) \sum_{i,j=1}^{n} H_{p_i p_j} (x+t\xi) \xi_i \xi_j dt $$ $$ \overset{(1)}{\geq} H(x) + \sum_{i=1}^{n} H_{p_i} (x) \xi_i + \frac{\theta}{2}|\xi|^2. $$ So now we choose $\xi = \frac{p_1-p_2}{2}$, $x = \frac{p_1+p_2}{2}$: $$ \Rightarrow H(p_1) \geq H\left(\frac{p_1+p_2}{2}\right) + \nabla H\left(\frac{p_1+p_2}{2}\right) \cdot \frac{1}{2}\left(p_1-p_2\right) + \frac{\theta}{8}\left| p_1-p_2 \right|^2. $$ My problem is how to eliminate the gradient term in the middle. I thought about using the mean value theorem but $\frac{p_1+p_2}{2}$ doesn't have to be the right value for it.
After that Evans goes on and claims that $(2)$ implies
$$
\frac{1}{2}L(q_1)+\frac{1}{2}L(q_2) \leq L\left(\frac{q_1+q_2}{2}\right) + \frac{1}{8\theta}|q_1-q_2|^2. \;\;\;\;\;(3)
$$
The only idea I have is that the direct coherence between $H$ and $L$ can be expressed as
$$
\forall q_1, q_2 \text{ there exist }p_1, p_2\text{ such that }H(p_i) + L(q_i) = p_i \cdot q_i.
$$
I really hope you can help me proving formula $(2)$ and $(3)$.
Thanks in advance,
Zhorrid.
My answer is nearly 4 years too late, but I thought that I would share it, just encase anybody else has trouble with this.
$H$ is $C^2$ and so $\forall p_1,p_2 \in \mathbb{R}^n$ we have by Taylor's Theorem (with Lagrange remainder) that $\exists t \in (0,1)$ s.t. \begin{align*} H(p_1) = & H \left (\frac{p_1 + p_2}{2} + \frac{p_1 - p_2}{2} \right ) \\ = & H \left (\frac{p_1 + p_2}{2} \right ) + DH \left (\frac{p_1 + p_2}{2} \right ) \cdot \left (\frac{p_1 - p_2}{2} \right ) \\ & + \frac{1}{2} \left (\frac{p_1 - p_2}{2} \right )^TD^2H \left (\frac{p_1 + p_2}{2} + t\frac{p_1 - p_2}{2} \right ) \left (\frac{p_1 - p_2}{2} \right ) \\ \geq & H \left (\frac{p_1 + p_2}{2} \right ) + DH \left (\frac{p_1 + p_2}{2} \right ) \cdot \left (\frac{p_1 - p_2}{2} \right ) + \frac{\theta}{8}|p_1 - p_2|^2. \end{align*} Where the last line follows from the fact that $H$ is uniformly convex. Similarly we get $$ H(p_2) \geq H \left (\frac{p_1 + p_2}{2} \right ) + DH \left (\frac{p_1 + p_2}{2} \right ) \cdot \left (\frac{p_2 - p_1}{2} \right ) + \frac{\theta}{8}|p_1 - p_2|^2. $$ Adding these two inequalities and dividing by 2, we get $$ \frac{1}{2} H(p_1) + \frac{1}{2} H(p_2) \geq H \left (\frac{p_1 + p_2}{2} \right ) + \frac{\theta}{8}|p_1 - p_2|^2. $$ Thus by taking away $\frac{\theta}{8}|p_1 - p_2|^2$ from both sides, we get the desired inequality. For the next part, we use the fact that $L = H^*$ and that the supremum is attained because of the conditions on $H$ ($H$ is continuous and $\lim_{|p| \to \infty} \frac{H(p)}{|p|} = \infty$), that is $$ L(v) = \max_{p \in \mathbb{R}^n} \left \{v \cdot p - H(p) \right \}. $$ We also use Young's inequality to get that $\forall p_1,p_2,v_1,v_2 \in \mathbb{R}^n$, we have \begin{align*} \left( \frac{p_1 - p_2}{2} \right ) \cdot \left( \frac{v_1 - v_2}{2} \right ) &= \left( \sqrt{\theta} \frac{p_1 - p_2}{2} \right ) \cdot \left( \frac{v_1 - v_2}{2\sqrt{\theta}} \right )\\ &\leq \frac{\theta}{8} |p_1 - p_2|^2 + \frac{1}{8\theta} |v_1 - v_2|^2. \end{align*} So for each $v_1,v_2 \in \mathbb{R}^n, \exists p_1,p_2 \in \mathbb{R}^n$ s.t. $$ L(v_i) = v_i \cdot p_i - H(p_i) $$ for $i = 1,2$. Now \begin{align*} \frac{1}{2} L(v_1) + \frac{1}{2} L(v_2) = & \frac{1}{2} v_1 \cdot p_1 + \frac{1}{2} v_2 \cdot p_2 - \frac{1}{2}H(p_1) - \frac{1}{2}H(p_2) \\ \leq & \frac{1}{2} v_1 \cdot p_1 + \frac{1}{2} v_2 \cdot p_2 - H \left (\frac{p_1 + p_2}{2} \right ) - \frac{\theta}{8}|p_1 - p_2|^2. \end{align*} Where we have applied the first inequality for $H$. Next \begin{align*} &= \frac{1}{2} v_1 \cdot p_1 + \frac{1}{2} v_2 \cdot p_2 - H \left (\frac{p_1 + p_2}{2} \right ) + \frac{1}{8\theta} |v_1 - v_2|^2 - \left ( \frac{\theta}{8}|p_1 - p_2|^2 + \frac{1}{8\theta} |v_1 - v_2|^2 \right ) \\ &\leq \frac{1}{2} v_1 \cdot p_1 + \frac{1}{2} v_2 \cdot p_2 - H \left (\frac{p_1 + p_2}{2} \right ) + \frac{1}{8\theta} |v_1 - v_2|^2 - \left( \frac{p_1 - p_2}{2} \right ) \cdot \left( \frac{v_1 - v_2}{2} \right ) \end{align*} Where we have applied Young's inequality, next we multiply out the brackets to get \begin{align*} & = \left( \frac{p_1 + p_2}{2} \right ) \cdot \left( \frac{v_1 + v_2}{2} \right ) - H \left (\frac{p_1 + p_2}{2} \right ) + \frac{1}{8\theta} |v_1 - v_2|^2 \\ & \leq L\left( \frac{v_1 + v_2}{2} \right ) + \frac{1}{8\theta} |v_1 - v_2|^2. \end{align*} Where the last line follows from the fact that $L = H^*$. So we are done.