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 $x=y$ shoes 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?

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?

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-Frenchel 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 $x=y$ shoes 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?

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 $x=y$ shoes 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?