Show that if $\phi \oplus \psi \leq c \implies \inf_{\gamma \geq 0} \int c - \phi \oplus \psi d\gamma = 0$

In the book Optimal Transport for Applied Mathematicians, the original problem for Optimal Transport is written as: $$\min_\gamma \quad \int_{X \times Y} c d\gamma \quad + \quad \sup_{\phi,\psi} \quad \int_X \phi d\mu + \int_Y \psi d\nu - \int_{X \times Y}\phi(x) - \psi(y) d\gamma$$

Where $$\gamma \in \mathcal P(X\times Y)$$ and $$\mu,\nu \in \mathcal P(X),\mathcal P(Y)$$, respectively.

Then, the author writes the dual as:

$$\sup_{\phi,\psi} \quad \int_X \phi d\mu + \int_Y \psi d\nu \quad + \quad \inf_\gamma \quad \int_{X \times Y} c(x,y) - \phi(x) - \psi(y) d\gamma$$

Finally, the author then claims the following: $$\inf_\gamma \quad \int_{X \times Y} c(x,y) - \phi(x) - \psi(y) d\gamma = 0, \quad = \begin{cases} 0, & \text{if } \phi \oplus \psi \leq c \quad \text{on }X\times Y \\ -\infty, & \text{otherwise}. \end{cases}$$

My question is why if $$\phi \oplus \psi \leq c \implies \inf_{\gamma \geq 0} \int c - \phi \oplus \psi d\gamma = 0$$ ?

Since $$\gamma$$ is a probability measure, is not like I can make the measure as small as I'd like.

For the case where $$\gamma$$ is just a positive measure, then I can see how we could make $$\gamma=0$$. But I don’t see how this would be valid assuming that we are dealing with probability measures.

1 Answer

If you look at the book you can see that the infimum is taken over $$\mathscr{M}_+(X\times Y)$$ - meaning the space of positive finite measures. Therefore you can just choose $$\gamma$$ to be the zero measure.