About Legendre transform in Evan's PDE: Show that (i) $v\in \partial H(p)$ iff (ii) $p\in \partial H^*(v)$ iff (iii) $v\cdot p=H(p)+H^*(v)$

Question

Let $H: \mathbb{R}^n\to \mathbb{R}$ be convex. We say $v\in \partial H(p)$ if $$H(r)\geq H(p)+v\cdot(r-p)$$ for all $r\in \mathbb{R}^n$

Prove that (i) $v\in \partial H(p)$ iff (ii) $p\in \partial H^*(v)$ iff (iii) $v\cdot p=H(p)+H^*(v)$.

Here $H^*(v):=\sup_{p}\{v\cdot p-H(p)\}$

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My idea:

I just try to prove that (i) $\to$ (ii) $\to $ (iii) $\to $ (i) .

But For (i) $\to$ (ii), we know $$H(r)\geq H(p)+v\cdot(r-p)$$. How to show that $$H^*(r)\geq H^*(p)+v\cdot(r-p)?$$

I am stuck in $$rp-H(p)\geq vr-H(r)+p(r-v)$$ So $\sup[rp-H(p)]\geq \sup[vr-H(r)+p(r-v)]$

Answer 1

We know $$H(r)\geq H(p)+v\cdot(r-p) \quad \forall r \in \mathbb{R}$$ so $$v\cdot p - H(p) \geq v\cdot r - H(r) \quad \forall r \in \mathbb{R}$$ so $$v\cdot p - H(p) \geq \sup_r \{ v\cdot r - H(r)\} = H^*(v)$$ Since $v\cdot p - H(p) \leq \sup_p \{ v\cdot p - H(p) \}=H^*(v)$, we have $v\cdot p - H(p) = H^*(v)$. Showing (ii) is now trivial: the right hand side simplifies to $pr-H(p)$, and taking the supremum over $p$ gives the desired result.