I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Let $(\Omega, \mathcal F, \mu)$ be a measure space with $\mu(\Omega) < \infty$. Let $p \in [1, \infty)$ and $j:\mathbb R \to \mathbb R$ be convex. We define $$ J:L^p(\Omega) \to (-\infty,+\infty], u \mapsto \int_\Omega j \circ u. $$ Prove that $J$ is convex and lower semi-continuous.
Could you confirm if my below attempt is correct?
Proof Notice that $\mu(\Omega) < \infty$ and Jensen's inequality imply $J(u)$ is well-defined for every $u \in L^p(\Omega)$. Notice that $L^\infty (\Omega) \subset L^{n+1} (\Omega) \subset L^n (\Omega)$ for all $n \in \mathbb N^*$.
- $J$ is convex.
Let $t \in [0, 1]$ and $u_1, u_2 \in L^p(\Omega)$. By convexity of $j$, $$ j (t u_1 (\omega)+(1-t)u_2(\omega)) \le tj\circ u_1 (\omega)+(1-t) j \circ u_2 (\omega) \quad \forall \omega \in \Omega. $$
Then $$ \int_\Omega j \circ (tu_1+(1-t)u_2) \le t\int_\Omega j \circ u_1 +(1-t) \int_\Omega j \circ u_2. $$
Then $J(tu_1+(1-t)u_2) \le tJ(u_1)+(1-t)J(u_2)$ and thus $J$ is convex.
- $J$ is l.s.c.
First, we assume $j$ is non-negative. Fix $\alpha \in \mathbb R$. To prove that $J$ is l.s.c., it suffices to prove $$ A:=\{u \in L^p(\Omega) : J(u) \le \alpha\} $$ is closed. Let $u_n \in A$ and $u \in L^p(\Omega)$ such that $u_n \to u$ in $L^p$. There is a sub-sequence $\varphi$ of $\mathbb N$ such that $u_{\varphi (n)} \to u$ $\mu$-a.e. Then $j \circ u_{\varphi (n)} \to j \circ u$ $\mu$-a.e. By Fatou's lemma, $$ 0 \le \int j \circ u \le \liminf_n \int j \circ u_{\varphi (n)} \le \alpha. $$
Hence $u \in A$. To solve the general case, we need
Lemma: Any proper convex lower semi-continuous function is bounded from below by a continuous affine function.
By above Lemma, there are $a,b \in \mathbb R$ such that $j' (x) :=j(x) - ax-b \ge 0$ for all $x \in \mathbb R$. Clearly, $j'$ is convex. Let $$ J':L^p(\Omega) \to (-\infty,+\infty], u \mapsto \int_\Omega j' \circ u. $$
As we have just proved above, $J'$ is l.s.c. On the other hand, $J'(u) = J(u) -a \int u \mathrm{~d} \mu-b \mu(\Omega)$. Clearly, $u \mapsto -a \int u \mathrm{~d} \mu-b \mu(\Omega)$ is continuous. Hence $J$ is l.s.c. This completes the proof.
