PDEs: Subdifferentials

Problem taken out of Evans, Partial Differential Equations, Chapter 3 Problem 11. (replace all the v's with q's)
PROBLEM
Let $H: \mathbb{R}^n\rightarrow\mathbb{R}^n$ be convex. We say that $q$ belongs to the subdifferential of $H$ at $p$,written $q\in \partial H(p)$ if $H(r)\geq H(p)+q\cdot(r-p)$ for all $r\in\mathbb{R}^n$.

Prove $q\in\partial H(p) $ if and only if $p\in \partial L(q)$ if and only if $p\cdot q = H(p)+L(q)$ where $L=H^*$, i.e. $L$ is the Legendre (Fenchel) transform of $H$.

PROOF

Assume first that $p\cdot q=H(p)+L(q)$. Clearly $H(p)=p\cdot q - L(q)$. Now observe that since $H(p)+L(r)\geq p\cdot r$ for all $r\in \mathbb{R}^n$ we have that $H(p)\geq p\cdot r - L(r)$. This implies that $p\cdot q - L(q)\geq p\cdot r - L(r)$, which when rearranged gives $L(r)\geq L(q)+p\cdot (r-q)$ for all $r$. So $p\in \partial L(q)$.

Again, assuming that $p\cdot q=H(p)+L(q)$, using a similar argument as above, we claim that $q\in \partial H(p)$. This follows since $L(q)=p\cdot q - H(p)$, and $H(r)+L(q)\geq q \cdot r$ for all $r\in\mathbb{R}^n$, we have that $p\cdot q - H(p)\geq q\cdot r - H(r)$, which when rearranged gives us $H(r)\geq H(p)+q\cdot (r-p)$ for all $r$, which shows the claim is true.

Conversely, assume, that $q\in \partial H(p)$. Then it follows that $H(r)\geq H(p)+q\cdot(r-p)$. Then
\begin{eqnarray*}
q\cdot p&\geq& H(p)+q\cdot r - H(r)\;\; \forall r\in \mathbb{R}^n\\
&\geq& H(p)+\displaystyle\sup_{r\in\mathbb{R}^n}[q\cdot r - H(r)]\\
&\geq& H(p)+L(q)
\end{eqnarray*}
By definition of $L$ and $H$, we have the reverse inequality$H(p)+L(q)\geq p\cdot q$. Hence $H(p)+L(q)=p\cdot q$. The case $p\in \partial L(q)$ is similar. This completes the proof.