For $f \in W^{-1,p'}(\Omega)$, $f \geq 0$, we definr $v,$ $v_{\epsilon}$, $g_{\epsilon}$ by $$-\mathrm{div}(|Dv|^{p-2} Dv)=f \quad \mbox{in} \mathcal{D}'(\Omega), \quad v \in W^{1,p}_0(\Omega)$$ with the variational formulation associate $$\displaystyle\int |D v_{\epsilon}|^{p-2} Dv_{\epsilon} Dw dx + \dfrac{1}{\epsilon} \displaystyle\int (v_{\epsilon} - v) w dx = 0, \quad \forall w \in W^{1,p}_0(\Omega) \cap L^2(\Omega)$$ $$v_{\epsilon} \in W^{1,p}_0(\Omega), \quad v_{\epsilon} - v \in L^2(\Omega)$$ $g_{\epsilon}$ is defined by $$g_{\epsilon}=-\dfrac{1}{\epsilon} (v_{\epsilon} - v)= - \mathrm{div}(|Dv_{\epsilon}|^{p-2} Dv_{\epsilon}).$$
My problem is: how we prouve the existence of $v_{\epsilon}$ (with consider the two cases $p \geq 2$, $1 < p < 2$)?
Let $V=W^{1,p}_0(\Omega)$ ($\Omega$ is bounded domain in $R^n$). Clearly $V$ is convex and closed. Define the operator $A: V\to V^*$ by $$ \left<A(u),w\right>=\int |D u|^{p-2} Du Dw dx + \dfrac{1}{\epsilon}\int (u - v) w dx, u, w\in V. $$ Then $A$ is continuous, strictly monotone, bounded and coercive (you check!). Then for each $v$, there is $v_\varepsilon\in V$ in the associated variational problem.