Let $$F(m, b) = H_1(m) + H_2(m, b)$$ where $H_2(m, b) = m G\left(\frac{b}{m}\right)$ and $m > 0$.
I want to compute the Legendre Transform of $F$. The Legendre Transform of $H_2$ is well known : $$H_2^\star(\phi, \psi) = \begin{cases}+\infty & \text{if } \phi + G^\star(\psi) > 0 \\\ 0 & \text{otherwise} \end{cases}.$$
A simple way to compute this Legendre transform is as this : $$\begin{align} F^\star(\phi, \psi) & = \sup_{m > 0, b} \phi m + \psi b - H_1(m) - H_2(m, b)\\\\ & = \sup_{m > 0, b} \phi m + \psi b - H_1(m) - m G\left(\frac{b}{m}\right) \\\\ & = \sup_{m > 0} \left(\phi m - H_1(m) + m\sup_{b} \psi \frac{b}{m} - G\left(\frac{b}{m}\right) \right) \\\\ & = \sup_{m > 0} \left(\phi m - H_1(m) + m G^\star(\psi) \right) \\\\ & = H_1^\star(\phi + G^\star(\psi)) \end{align}$$
Another way is using inf-conv which is where I fail : $$\begin{align} F^\star(\phi, \psi) & = \inf \left\{ H_1^\star(\phi - \phi^\prime) + \iota_0(\psi - \psi^\prime) + H_2^\star(\phi^\prime, \psi^\prime) | \phi^\prime, \psi^\prime\right\} \\\\ & = \inf \left\{ H_1^\star(\phi - \phi^\prime) | \phi^\prime + G^\star(\psi) \leq 0\right\} \end{align}$$
But then I don't understand how to show the constraint is saturated, unless $H_1$ is isotone.
What am I missing ?
$$\begin{align} F^\star(\phi, \psi) & = \inf \left\{ H_1^\star(\phi - \phi^\prime) + \iota_0(\psi - \psi^\prime) + H_2^\star(\phi^\prime, \psi^\prime) | \phi^\prime, \psi^\prime\right\} \\ & = \inf \left\{ H_1^\star(\phi - \phi^\prime) | \phi^\prime + G^\star(\psi) \leq 0\right\} \\ & = \inf_{\phi^\prime + G^\star(\psi) \leq 0} \sup_{m > 0} (\phi - \phi^\prime)m - H_1(m) \\ & \geq \sup_{m > 0} \inf_{\phi^\prime + G^\star(\psi) \leq 0} (\phi - \phi^\prime)m - H_1(m) \\ & \geq \sup_{m > 0} (\phi + G^\star(\psi))m - H_1(m) \\ & \geq H_1(\phi + G^\star(\psi)) \end{align}$$
Then the rest follows.