lower semicontinuous of the convex function

Question

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*}

Answer 1

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 $f_n = a_n\chi_{(0,n)}$.

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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.)