I am having trouble in seeing the following assertion. If someone could give me any hint I will be very grateful. May be some of the hypotheses could be ignored but I am not completely sure:
Consider $\dfrac{1}{q}+\dfrac{1}{q'}=1$. Let $\Omega \subset \mathbb{R}^n$ be a domain with finite volume. Let $f \in L_{q'}(\Omega)$ with $q'<2$. Assume there is a unique $u\in W^{1,2}(\Omega)$ such that: \begin{align} \int_\Omega \left(\displaystyle \sum_{i,j=1}^n a_{ij}\dfrac{\partial u}{\partial x_i}\dfrac{\partial v}{\partial x_j} + auvdx\right) = \int_\Omega fv dx \hspace{0.3cm} \textit{ for every } v \in W^{1,2}(\Omega) \end{align} where $a, a_{ij}=a_{ji}$ are bounded measurable functions on $\Omega$, $0<constant<a$ and $c^{-1}\vert \zeta \vert^2 \ \leq \ \displaystyle \sum_{i,j=1}^n a_{ij}(x)\zeta_i \zeta_j \ \leq \ c\vert \zeta \vert^2$ for some positive constant $c$ and any $\zeta = (\zeta_1, \cdots, \zeta_n) \in \mathbb{R}^n$. Then there is a constant $C$ such that:
\begin{align} \parallel v \parallel_{L_q(\Omega)} \ \leq \ C \parallel v \parallel_{W^{1,2}(\Omega)} \hspace{0.3cm} \text{for every } v \in W^{1,2}(\Omega) \end{align}
where $\parallel v \parallel_{W^{1,2}(\Omega)}=\parallel \nabla v \parallel_{L_2(\Omega)} + \parallel v \parallel_{L_2(\Omega)}$.