1

Assume that $H_1$, $H_2$ : $ R^{n} \rightarrow R $ are convex, coercive and smooth. Prove that $min_{p \in R ^{n}}[H_1(p)+H_2(p)]=\max_{\nu \in R^{n}}[-L_1(\nu)-L_2(-\nu)]$ where $L_1=H_{1}^{*}$, $L_2=H_{2}^{*}$.

We define $H(p)$ and $L(q)$ as follows

$H(p)=\sup_{q \in R ^{n}}[pq-L(q)]$; $L(q)=\sup_{p \in R ^{n}}[qp-H(p)]$

I proved one side of the equality. That is $min_{p \in R ^{n}}[H_1(p)+H_2(p)] \ge \max_{\nu \in R^{n}}[-L_1(\nu)-L_2(-\nu)]$. This part is easy. I am trying to prove $min_{p \in R ^{n}}[H_1(p)+H_2(p)] \le \max_{\nu \in R^{n}}[-L_1(\nu)-L_2(-\nu)]$. This is my question actually. Thanks for any hint.

Yang
  • 521
  • 1
  • 9
  • 18

1 Answers1

1

Let $p_0$ be a point where $H_1+H_2$ attains its maximum. (As an aside: since $L_1,L_2$ are smooth, it follows that $H_2,H_2$ are strictly convex, so $p_0$ is unique.) Let $\nu = \nabla H_1(p_0)$. Note that $\nabla H_2(p_0)=-\nu$.

Since the concave function $p\mapsto \nu p - H_1(p)$ has a stationary point at $p=p_0$, it attains its maximum there. Hence $L_1(\nu)=\nu p_0-H_1(p_0)$. Similarly, $L_2(-\nu)=-\nu p_0-H_2(p_0)$. Conclude with $$-L_1(\nu)-L_2(-\nu)= H_1(p_0)+H_2(p_0)$$