Optimal transport: how $\varphi^c$ can be written as $\varphi^c = \lim _{\ell \rightarrow \infty} \psi_{\ell}$?

Question

Let $X,Y$ be Polish spaces and $c:X \times Y \to [0, \infty]$ lower semi-continuous. There is a sequence $(c_\ell)_{\ell \in \mathbb N}$ with $c_\ell:X \times Y \to [0, \infty)$ of bounded Lipschitz continuous functions such that $c_\ell \uparrow c$ pointwise. Fix $\varphi:X \to \mathbb R$. The following is taken from page 33 of Villani's Topics in Optimal Transportation.

Remark 1.12 (c-concave functions). It follows from the proof that, when $c$ is bounded, one can restrict the supremum in the right-hand side of (1.4) to those pairs $\left(\varphi^{c c}, \varphi^c\right)$, where $\varphi$ is bounded and $$ (1.18) \quad \varphi^c(y)=\inf _{x \in X}[c(x, y)-\varphi(x)], \quad \varphi^{c c}(x)=\inf _{y \in Y}\left[c(x, y)-\varphi^c(y)\right] $$ An easy argument shows that $\left(\varphi^{c c}\right)^c=\varphi^c$ (see Exercise 2.35). The pair $\left(\varphi^{c c}, \varphi^c\right)$ is called a pair of conjugate $c$-concave functions. Note that $\varphi^c$ is measurable, since it can be written (exercise) as $\lim _{\ell \rightarrow \infty} \psi_{\ell}$, where $$ \psi_{\ell}(y)=\inf _{x \in X}\left[c_{\ell}(x, y)-\varphi(x)\right], $$ and $c_{\ell}$ is an increasing family of bounded uniformly continuous functions converging pointwise to $c$. Indeed, each $\psi_{\ell}$ is uniformly continuous, and therefore $\varphi^c$ is measurable. Similarly, $\varphi^{c c}$ is measurable.

I proved that $\psi_\ell$ is bounded Lipschitz continuous for each $\ell \in \mathbb N$.

Could you explain how $\varphi^c$ can be written as $\varphi^c = \lim _{\ell \rightarrow \infty} \psi_{\ell}$?

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I posted this question on MSE, but have not received any answer so far. So I post it here.

Answer 1

Are you sure there aren’t additional conditions on $\varphi$? Because otherwise taking $X = \mathbb R$ and $Y$ to be a one point space, the following gives a counterexample:

$c(x, y) = 0$ if $x = 0$; $c(x, y) = 1$ otherwise, and

$\varphi(x) = 0$ if $x = 0$, $\varphi(x)= 2$ otherwise.

Indeed, $\varphi^c = -1$, while $\psi_\ell \leq -2$ for all $\ell$.

My guess is that $\varphi$ should be restricted to be continuous.

Answer 2

Here is an example which shows that also the continuity of $\varphi$ does not help. Let $X = \mathbb R$ and $Y = \{0\}$. In the sequel, we will just drop the $Y$-argument. Let $c \equiv 1$, $\varphi \equiv 0$ and define $c_n \colon X \to \mathbb R$ via $$ c_n(x) = \exp(- x^2/n). $$ Note that $c_n$ is (uniformly) Lipschitz, (uniformly) bounded and converges pointwise monotonously to $c$. Nevertheless, $$ \psi_n = \inf\{ c_n(x) \mid x \in \mathbb R\} = 0 \ne 1 = \varphi^c = \inf\{c(x) \mid x \in \mathbb R \}. $$

Thus, the Remark 1.12 does not work as stated, also in case that $\varphi$ is continuous. It is not clear to me if it would work if we use the special construction (1.11) for $c_n$.