I am trying to show that, given $f \in H^{-1}(U)$, there exists a unique $u \in H_0^1(U)$ such that: $$\int_U \nabla u\cdot\nabla v \, \mathrm{d}x= \langle f,v \rangle_{H^{-1}} \, , \quad \forall \, v \in H_0^1(U) \, .$$ To this end, I define the functional $J \colon H_0^1(U) \to \mathbb{R}$ by: $$J(v):=\int_U \left\| \nabla v \right\|^2 \, \mathrm{d}x- \langle f,v \rangle_{H^{-1}}.$$ Then I would like to do the following:
- Show that $J(v)$ is bounded from below;
- Show that there is a weakly convergent sequence whose limit in $H_0^1(U)$ minimises $J$ (hence the inf is achieved);
- Show that this limit satisfies the problem and it is unique.
Number 3 is quite easy, but I am stuck on 1 and 2. I am not sure how to proceed. So far I have oly been able to use the definition of $\|\cdot\|_{H^{-1}}$ to get: $$J(v) \geq \int_U \left\| \nabla v \right\|^2 \, \mathrm{d}x- \|f \|_{H^{-1}} \|v\|_{H_0^1} \, .$$ Any pointers would be very helpful.
P.S.: I am aware that this could be shown by applying Lax-Milgram but I need to take this direct approach.
Let $U$ be a bounded smooth domain and $\|u\|=\|\nabla u\|_2$ the usual norm in $H_0^1(U)$.
1) $J$ is coercive.
Indeed, we have that $$J(u)\ge \|u\|^2-\|f\|_{H^{-1}}\|u\|,$$
which implies that $J$ is coercive.
2) $J$ is weakly lower semicontinuous, or equivalently, if $u_n\to u$ weak then, $J(u)\le \liminf J(u_n)$.
In fact, if $u_n\to u$ weak then $\langle f,u_n\rangle\to \langle f,u\rangle$, because $f$ is a continuous linear functional in $H_0^1(U)$. On the other hand, the norm function $\|u\|$ is always weakly lower semicontinuous, therefore
\begin{equation} \label{eq1} \begin{split} J(u) & = \int |\nabla u|^2-\langle f,u\rangle \\ & \le \liminf \int |\nabla u_n|^2-\lim \langle f,u_n\rangle \\ & \le \liminf \left(\int |\nabla u_n|^2-\langle f,u_n\rangle\right) \\ & = \liminf J(u_n). \end{split} \end{equation}
Now you can combine 1) with 2) to guarantee the existence of a minimum for $J$. Also, the strictly convexity of $J$ will imply that the minimum is unique.