lower semicontinuous of the convex function Let $\Omega\subset\mathbb{R}$ be a convex set, $\phi :\Omega\to\mathbb{R}$ be a convex function and $\{f_n\}_{n=1}^\infty$ be a sequence in $L^1(\mathbb{R}^d)$ with value in $\Omega$.  Assume that
\begin{align*}
\lim_{n\to\infty}\|f_n-f\|_{L^1(\mathbb{R}^d)}=0\quad \text{and}\quad \sup_{n}\|\phi(f_n)\|_{L^1(\mathbb{R}^d)}<\infty.
\end{align*}
Can we show that
\begin{align*}
\int_{\mathbb{R}^d}\phi(f(x))dx\le\varliminf_{n\to\infty}\int_{\mathbb{R}^d}\phi(f_n(x))dx?
\end{align*}
Notice that the convex function $\phi$ may be negative, which makes Fatou's Lemma failed  . For example, if $f_n$ and $f$ are positive functions, and $\phi(x)=x\ln x$, we expect that
\begin{align*}
\int_{\mathbb{R}^d}f(x)\ln f(x)dx\le\varliminf_{n\to\infty}\int_{\mathbb{R}^d}f_n(x)\ln f_n(x)dx.
\end{align*}
 A: In general, the claim is not true.
Here is a counterexample for
$$
\phi(x) = \begin{cases} + \infty & \text{ if } x<0\\
-\sqrt x &  \text{ if } x\ge0.
\end{cases}
$$
Set $f_n = \frac1{n^2}\chi_{(0,n)}$. Then $f_n \to 0$ in $L^1$, $\phi(0)=0$, but $\int_{\mathbb R}\phi(f_n) dx = -1$.
Similarly, a counterexample can be constructed for $\phi(x)=x\log x$:
take $a_n$ to be a solution of $a \log a = -\frac1n$ such that $a_n\cdot n\to0$ for $n\to\infty$, which is possible using the Lambert $W$-function. Then set $f_n = a_n\chi_{(0,n)}$.

Assume $\phi:\mathbb R \to \mathbb R \cup\{+\infty\}$ to be convex and lower semicontinuous.
If $\phi(0)=0$ and $\partial \phi(0)\ne\emptyset$ then the claim can be proven as follows:
Take $s\in \partial \phi(0)$, then $\phi(x) \ge sx$. Now apply Fatou's Lemma to get
$$
\int \phi(f) - sf \ dx
\le \int \liminf (\phi(f_n)-sf_n)
\le \liminf \int \phi(f_n)-sf_n \\
= \liminf \int \phi(f_n)dx - \int sf\ dx,
$$
the integral $\int sf\ dx$ is finite, so can be cancelled.
(Here, first inequality is from lower semicontinuity of $f\mapsto \phi(f)-sf$, second is Fatou's lemma.)
