Villiani writes (some notation changed) in Topics in Optimal Mass Transportation:

Theorem 1.9. Let $E$ be a normed VS, $E^*$ it topological dual. $\Theta$ and $\Psi$ are two convex functions on $E$ with values in $\mathbb{R}\cup{+\infty}$. Let $\Theta^*$ and $\Psi^*$ be Legendre-Fenchel transforms of $\Theta$ and $\Psi$ (Earlier definited as $R^*(z^*) = \sup_{z\in E}[<z^*,z>-R(z)]$). Assume that there exists $z_0\in E$ such that $\Theta(z_0)$ and $\Psi(z_0)$ are finite and $\Theta$ is continuous at $z_0$. Then: $$\inf_E[\Theta + \Psi]=\max_{z^*\in E^*}[-\Theta^*(-z^*)-\Psi^*(z^*)]$$

He begins his proof by saying: We want to prove: $$\sup_{z^*\in E^*} \inf_{x,y\in E} [\Theta(x) + \Psi(y) + <z^*,x-y>]=\inf_{x\in E}[\Theta + \Psi]$$ Then he says: The choice of $x=y$ shows that the LHS is not larger than the RHS.

  1. I'm confused about where his "sup inf" expression is coming from. I mean I get there's a substitution but that's it. Does someone understand where this expression is coming from?

  2. Doesn't a choice of $x=y$ establish equality?

  • 3
    $\begingroup$ 1. He changes the $\max$ to a $\sup$ but other than that, your 1. is just plugging in the definition and using $\sup -f = -\inf f$. 2. The infimum taken over all $x,y$ could potentially be lower than the infimum taken over only those $x,y$ satisfying $x=y$, hence we only get an inequality by this reasoning. $\endgroup$
    – Steve
    Oct 1, 2017 at 15:49
  • $\begingroup$ @Steve 1. I suspect I'm getting mixed up, but the negative signs aren't working out for me. 2. I see, thanks! $\endgroup$
    – yoshi
    Oct 1, 2017 at 16:06

1 Answer 1


More details based on Steve's comment: We have \begin{align} -\Theta^*(-z^*) &= - \sup_{x \in E} \big[ \langle-z^*,x \rangle - \Theta(x) \big] \\ &=\inf_{x \in E} \big[ \langle z^*,x \rangle + \Theta(x) \big] \end{align} and \begin{align} -\Psi^*(z^*) &= - \sup_{y \in E} \big[ \langle z^*,y \rangle - \Psi(y) \big] \\ &=\inf_{y \in E} \big[ \langle z^*,- y \rangle + \Psi(y) \big] \end{align} Then, \begin{align} -\Theta^*(-z^*) -\Psi^*(z^*) &= \inf_{x,y \in E} \big[ \langle z^*,x -y\rangle + \Theta(x) + \Psi(y)\big] \end{align}

  • $\begingroup$ indeed I was just multiplying through by negative sign without swapping sup for inf. thx! $\endgroup$
    – yoshi
    Oct 1, 2017 at 16:16
  • $\begingroup$ @yoshi, no problem. $\endgroup$
    – passerby51
    Oct 1, 2017 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.