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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?

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    $\begingroup$ 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 '17 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 '17 at 16:06
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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}

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

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