Let $(u_n)_{n \in \mathbb{N}}$ be a sequence of probability densities, which are uniformly bounded in $L^{\infty}(\mathbb{R}^d)$ and converge weakly (i.e. in duality with all $C_b(\mathbb{R}^d)$-fcts.) to a probability measure with density $u$. The boundedness in $L^{\infty} = {L^1}^*$ yields the existence of a subsequence (also denoted by $u_n$), which converges weak$^*$ to $u$, i.e. in duality with each $f \in L^1$.
Further, let $\beta: \mathbb{R} \to \mathbb{R}$ be $C^2$ with $$\beta(0)=0, \beta'(x) \geq a|x|^{\alpha-1}, \beta(x) \leq C|x|^{\alpha}$$ for some $\alpha \geq 1$ and $a, C >0$. Due to the growth-bound on $\beta$, $\beta$ leaves $L^{\infty}$ invariant and $(\beta(u_n))_n$ is bounded in $L^{\infty}$ as well. Thus there is a subsequence (again, I surpress change of notation) converging weak$^*$ in $L^{\infty}$ to some $g \in L^{\infty}$: $\beta(u_n) \underset{n}{\longrightarrow} g$. Can we show $g = \beta(u)?$
This would follow at once, if there was pointwise convergence $u_n \to u$ (due to the continuity of $\beta$), but I do not think we can obtain such strong kind of convergence from the assumptions. We may consider $\beta: u \mapsto \beta(u)$ as an operator in $L^{\infty}$ of which I'd like to show $w^*-w^*$-closedness. However, I do not know how to prove or disprove this. Any suggestions?