Let $\mathcal{B}(0,1)$ be the Borel sigma algebra on $(0,1)$ and $m$ be the Lebesgue measure. Consider $L^\infty((0,1),\mathcal{B}(0,1),m)$ equipped with weak-star topology and $B=\{f\in L^\infty(0,1):0\le f\le 1\}$.
Let $g\in L^1(0,1)$ and $\phi:[0,1]\rightarrow \mathbb{R}$ be a smooth function. By the definition of weak star topology, it is clear that the mapping $f\mapsto\int_0^1fgdx$ is continuous in $B$.
However, I want to know under what conditions can we conclude that $f\mapsto\int_0^1\phi(f)dx$ is also continuous in B?
Assume that $\phi$ is not affine. Then, there are $x,y \in [0,1]$ and $\lambda \in [0,1]$ with $$ \phi(\lambda x + (1-\lambda) y) \ne \lambda \phi(x) + (1-\lambda) \phi(y).$$
It is well known that there exists a sequence $(f_n) \subset B$ with the properties $$ f_n(t) \in \{ x , y \} \quad\text{f.a.a. } t \in (0,1),$$ $$m(\{t \in (0,1) \mid f_n(t) = x\}) = \lambda$$ for all $n \in \mathbb N$ and $f_n \rightharpoonup^* f$, where $f \equiv \lambda x + (1-\lambda) y$.
Thus, $$ \int_0^1 \phi(f(t)) \,\mathrm dt = \lambda \phi(x) + (1-\lambda) \phi(y), $$ but $$ \int_0^1 \phi(f(t)) \, \mathrm dt = \phi(\lambda x + (1-\lambda) y).$$
In the case that $\phi$ is convex and lower semicontinuous, it might be possible to show that $$ f \mapsto \int_0^1 \phi(f(t)) \mathrm dt$$ is sequentially weak* lower semicontinuous.