I have some problem with the weak l.s.c and coercivity of the following map
$J\colon\,H_{0}^{1}(0,1)\rightarrow\mathbb{R}$ given by
$J(x)=\dfrac{1}{2}\int_{0}^{1}(x'(t))^2dt-\int_{0}^{1}F(t, x(t))dt$,
where $F\in C^{1}(0,1)$ with respect to the second variable.
My considerations are as follows: assume that
$|F(t,x(t))|\leq A(t)\|x\|^2+B(t)\|x\|+C(t)|t|$ (norm is from $H_{0}^{1}(0,1)$). This hypotheses leads to the following inequality:
$\int_{0}^{1}|F(t,x(t))|\leq \int_{0}^{1}A(t)\|x\|^2+B(t)\|x\|+C(t)|t|dt\leq \max(\|x\|^2)\|A\|_{L^{1}}+\max(\|x\|)\|B\|_{L^{1}}+\\\|C\|_{L^{1}}$.
So I additionally want $A,B,C\in L^{1}$. My first question is: Can I write $\max(\|x\|)$? Does this statemant have any sens in $H_{0}^{1}(0,1)$?
The next part is the weak lsc. So I take a sequance $(x_{n})\subset H_{0}^{1}$ converging weakly to some $x_{0}$. Note that
$g(x)=\dfrac{1}{2}\int_{0}^{1}(x^{'}(t))^2dt=\dfrac{1}{2}\|x'\|_{L^{2}(0,1)}^{2}$. The norm function is continuous and since we consider $\|\cdot\|^{2}$, hence $g$ is convex as well. Does it imply that $g$ is weak lower semicontinuous? I think so; Fix $\alpha \in \mathbb{R}$. Then, by the continuity of $g$ the set $S_{\alpha}=\{x\in H_{0}^{1}\,|\, g(x)\leq \alpha\}$ is closed. Since $g$ is convex, then $S_{\alpha}$ is weakly closed, by the Mazur theorem and as a result $g$ is weak lsc. We have
$\liminf_{n}J(x_{n})\geq \liminf_{n}\dfrac{1}{2}\|x_{n}'\|_{L^{2}(0,1)}^{2}-\limsup_{n}\int_{0}^{1}F(t, x_{n}(t))dt$. limsup is a problem, so I need just a limit. I want to have
$\lim_{n} \int_{0}^{1}F(t,x_{n})dt=\int_{0}^{1}\lim_{n}F(t,x_{n})dt$. This however can only happen if $F$ is uniformly continuous. Can we deduce it form previous inequalities? Maybe the weak convergence implies the norm convergance in $C_{0}(0,1)$? If so, I can move the limit inside the integral I quess. This would solve the problem with weak. l.s.c. . How to deal with coercivity? I need inequalities:
$\dfrac{1}{2}\int_{0}^{1}(x'(t))^2dt\geq ...$
and
$\int_{0}^{1}F(t, x(t))dt\leq ...$