Legendre transform of max of two convex functions

272 Views Asked by At

Suppose $H_{1}, H_{2} : \mathbb{R} \to \mathbb{R}$ are convex functions with Legendre transforms $L_{1},L_{2}$. This means $L_{i}(q) = \sup\{qp - H_{i}(p) \, \mid \, p \in \mathbb{R}\}$. Define $H(p) = \max\{H_{1}(p),H_{2}(p)\}$.

Almost tautologically, $H$ can be written as \begin{equation*} H(p) = \sup\{(1 - \lambda)(q_{1}p - L_{1}(q_{1})) + \lambda (q_{2}p - L_{2}(q_{2})) \, \mid \, \lambda \in [0,1], \, q_{1},q_{2} \in \mathbb{R}\}. \end{equation*} In particular, this has the very nice (not tautological) consequence that \begin{equation*} \min \{H(p)\, \mid \, p \in \mathbb{R}\} = \sup\{-(1 - \lambda)L_{1}(q_{1}) - \lambda L_{2}(q_{2}) \, \mid \, \lambda \in [0,1], \, \, q_{1},q_{2} \in \mathbb{R}, \, \, (1 - \lambda)q_{1} + \lambda q_{2} = 0\}. \end{equation*}

My question is: is there a similarly nice expression for the Legendre transform of $H$? That is, if $L(q) = \sup \{pq - H(p) \, \mid \, p \in \mathbb{R}\}$, is there a tractable formula for $L$ in terms of $H_{1}$ and $H_{2}$?

1

There are 1 best solutions below

2
On

$$\begin{align} L(q) &= \sup_p \{pq - \max\{H_1(p), H_2(p)\} \} \\ & = \sup_p \inf_{\lambda \in [0,1]} \{pq - (\lambda H_1(p) + (1-\lambda) H_2(p)) \} \\ & = \inf_{\lambda \in [0,1]} \sup_p \{pq - (\lambda H_1(p) + (1-\lambda) H_2(p)) \} \\ &= \inf_{\lambda \in [0,1]} (\lambda H_1 + (1-\lambda) H_2)^*(q) \\ &= \inf_{\lambda \in [0,1]} \inf_{q_i} \{ (\lambda H_1)^*(q_1) + ((1-\lambda) H_2)^*(q_2) : q_1+q_2 = q \} \\ &= \inf_{\lambda \in [0,1]} \inf_{q_i} \{ \lambda L_1^*(q_1/\lambda) + (1-\lambda) L_2^*(q_2/(1-\lambda)) : q_1 + q_2 = q \} \end{align}$$