Using energy method to existence and uniquesness the following PDE . 
Approach : Find the energy functional : 
$\int_\Omega \nabla u \nabla v \int_\Omega |u| u.v -\int_\Omega fu=0 \forall u \in H_0^1(\Omega) \cap L^3(\Omega)$
$\implies E(u)=\int_\Omega\frac{1}{2} |\nabla u|^2+\frac{1}{3} |u|^3 -fu  dx$
If $u$ solves $\min_{u \in A} E(u)$ with $A={}u\in H_0^1( \Omega)\cap L^3(\Omega)$ then for any $v\in A$ we have 
$0=\frac{d}{d\epsilon}E(u+\epsilon v)=\int\frac {d}{d\epsilon}|\nabla(u+\epsilon v)|^2 +\frac{1}{3} |u+\epsilon v|^3 -f(u+\epsilon v)dx$
$$=\int(\nabla u. \nabla v +|u|v -fv)  dx$$
Remark: There seems to be a problem while differentiating with respect to $\epsilon$ the term $|u+\epsilon v|^3$ , how do I resolve it ? 
Next : Can I say that $u$ is a weak solution now ? If I could then would proceed further this way  If
$A$ is not a null set , then $\exists (u_k)_{k\in \mathbb N} \subset A$ 
$$\lim_{k\to\infty}u_k=\inf_{u\in A} E(u)$$
Assume $u_k$ is not bounded in $H_0^1 $
$$E(u_k)\ge \int_\Omega \frac {|\nabla u_k|^2}{2}-fu_k dx  \ge \frac{1}{2} ||\nabla u||_{L^2}-C ||f||_{L^2} ||\nabla u_k||_{L^2(\Omega)}$$
I am stuck now , How do I proceed further ? 
Thanks. 
 A: Concerning the differentiation of $\epsilon\mapsto |u+\epsilon v|^3$. The function $g(x)=|x|^3$ is in $C^2(\mathbb R)$ with $g\,'(x)=3x|x|$ and $g\,''(x)=6|x|$. Notably, $g$ is convex, which makes the entire functional $E$ convex; we need this for its lower semicontinuity with respect to the weak convergence in $A$.
But to begin with, we want to show that $E$ is bounded from below on $A$; otherwise the entire approach is doomed. The only question is what to do with $\int-fu$: suppressing the urge to use Holder's inequality, I would write $-fu\ge -\frac12f^2-\frac12 u^2$. Here the integral of $f^2$ is a finite constant and $u^2\le \frac12|u|^3+8$ pointwise. It follows that $E$ is bounded from below: $E(u)\ge \int (\frac12|\nabla u|^2+\frac{1}{12}|u|^3-\frac12f^2-8)$ or something of the kind. Note that I bounded $u^2$ by a small multiple of $|u|^3$ in order to have some of the cubic term left over. This will be useful in a moment. 
Pick a minimizing sequence $u_k\in A$ as you've done already. The lower bound for $E(u)$ in the previous paragraph tells us that $\int|\nabla u_k|^2$ and $\int|u_k|^3$ are uniformly bounded. Thus, the sequences is bounded in $H_0^1$ and in $L^3$. Pick a weakly convergent subsequence in $H_0^1$; extract from it a weakly convergent subsequence in $L^3$. The lower semicontinuity of $E$ implies $E(u)=\min_A E$ where $u$ is the weak limit. 
And yes, having $\int (\nabla u\cdot\nabla v +\frac13 u|u|v-fv)=0$ for all test functions $v$ (you only need to consider $C^\infty$-smooth $v$ with compact support) means that $u$ is a weak solution; this is what weak solution means. The boundary condition $u=0$ is automatically fulfilled since we work in $H_0^1$ all the time. 
