Evans PDE Ch. 3.5 exercise with subdifferential of convex function I am very stuck on this question in Evans PDE book. The question is as follows:

Here  $L(q) = \sup_{s \in \mathbb{R}^n}\left\{  s \cdot q - H(q) \right\}. $
Let $(1)$ be the statement that
$$ v \in \partial H(p), $$
$(2)$ be the statement that
$$  p \in \partial L(v), $$
and $(3)$ be the statement that
$$ L(v) + H(p) = p \cdot v.  $$
I was able to show $$ (3) \Rightarrow (1) \Rightarrow (2), $$
but I can't seem to show $(2) \Rightarrow (3)$, or find any other way to complete the chain.
Can someone please help me?
Thanks!
 A: I think it is easier to prove $(3) \iff (1) \iff (2).$
I'll try to give you some hints because it all proofs of these kind are based on a similar way of reasoning and it is useful to assimilate it.
Let's see $(1) \Rightarrow (3)$.
A good way to prove $$L(v) + H(p) = p \cdot v$$
is to prove both inequalities (usually one of the $2$ is trivial). In this case $\leq$ is trivial as $L$ is a $\sup$.
The other inequality is "trickier" as you have the $\sup$ on one of the $2$ sides and this doesn't allow you to really replace $L(v)$ with something equal to it. But you don't need to replace $L$ with an equality, as an inequality is sufficient for your purposes. Using the definition of $\sup$, you know that $\forall\varepsilon$ there is at least one $r$ such that $$L(v) - \varepsilon \leq v \cdot r - H(r),$$ or equivalently $$L(v) \leq v \cdot r - H(r) + \varepsilon.$$ Now if you use $(1)$ you can play with the r.h.s. to obtain an inequality that holds for every $\varepsilon$ (and with no $r$).
Let's see $(2) \Rightarrow (1)$.
You want to prove $$H(r) \geq H(p) + v\cdot(r -p)\ \ \ \ \forall r.$$
Using the fact that $H$ is convex you know that $H^{*}=L$, thus
$$ H(r) \geq t\cdot r -L(t),$$
for every $t$, in particular for $t=v$, i.e. $H(r) \geq v\cdot r - L(v)$. Now if you use $p \in \partial L(v)$ to play with $- L(v)$ you should be able to get the result.
