Generalised Lebesgue transform continuous wrt. strict topology?

Question

Let $X$ and $Y$ be $\sigma$-compact spaces, and $\mu$ [resp. $\nu$] be a regular Borel probability measure on $X$ [resp. $Y$].

For a bounded continuous $c:X\times Y\rightarrow\mathbb{R}$, consider the $c$-transform $$\tag{1} T_c \, : \, L^1(\mu) \,\ni\, \varphi \,\mapsto\, \varphi^c, \quad\text{where}\quad \varphi^c(y):= \inf_{x\in X} c(x,y) - \varphi(x) $$

(as known e.g. from the theory of optimal transport).

My question is if $T_c$ is continuous wrt. the (generalised) strict topology $\tau_X$ described here: $$\tag{2}\text{If $\varphi, (\varphi_\alpha)\in C_b(X)$ with $\varphi_\alpha\stackrel{\tau_X}{\rightarrow}\varphi$, does this imply that $T_c(\varphi_\alpha)\stackrel{L^1(\nu)}{\longrightarrow} T_c(\varphi)$ ?} $$

Any hints or references are welcome.

Answer 1

$\newcommand\vpi\varphi$The answer, which is a modification of the previous answer, is still no.

E.g., suppose that $X=Y=\mathbb R$, $c=0$, and $\vpi_a(x)=\min(1,\max(0,x-a))$ for $a\ge0$ and all real $x$.. Then, as $a\to\infty$, we have $\vpi_a\to0$ in the (generalised) strict topology, whereas $T_c(\vpi_a))=-1\not\to0=T_c(0)$ in $L^1(\nu)$ for any probability measure $\nu$ over $\mathbb R$.