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](https://mathoverflow.net/questions/449834/does-global-boundedness-ruin-stone-weierstrass-denseness):
$$\tag{2}\text{If $\varphi, (\varphi\_\alpha)\in C\_b(X)$ with $\varphi\_\alpha\stackrel{\tau\_X}{\rightarrow}\varphi$, does this imply that $T\_c(\varphi\_\alpha)\stackrel{L^1(\nu)}{\longrightarrow} T\_c(\varphi)$ ?}
$$


Any hints or references are welcome.



Generalised Lebesgue transform continuous wrt. strict topology?
$\newcommand\vpi\varphi$The answer, which is a modification of the [previous answer](https://mathoverflow.net/a/451596/36721), 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 $\mathbb R$.


