Minimizing a functional by using the Lax-Milgram theorem Let $(M,g)$ be a compact Riemannian manifold with nontrivial boundary and let $W_0^{1,2}(M)$ be the closure of $C_c^\infty(M)$ in the Sobolev space $W^{1,2}(M)$. Given $f\in C^\infty(M)$, I want to know how to minimize the functional
$$A(v):=\int_M\frac{1}{2}(|\nabla v|^2+fv)d\mu_g$$
over $v\in W_0^{1,2}(M)$. My book said that the Lax-Milgram theorem, together with the Poincaré inequality, can be used to show that a minimizer $u\in W_0^{1,2}(M)$ exists, but I don't know exactly how. To my knowledge, the cornerstone of the Lax-Milgram theorem is a bilinear form on a Hilbert space. Obviously, $W_0^{1,2 }$ is the Hilbert space to be considered, but where's the bilinear form that matters the most? Please give me a hint. Thank you very much.
 A: Too long for a comment. Hope this could help, or give some inspiration!
As far as I remember, there exists this.

Theorem
Let $\Omega$ be a bounded domain in $\mathbb{R}^n$, and $p \in [2, +\infty)$ with conjugate exponent satisfying $\frac{1}{p} + \frac{1}{q} = 1$. Let $f \in W^{-1, q}(\Omega)$, the dual of $W_0^{1, p}(\Omega)$, be given.
Then there exists a weak solution $u\in W_0^{1, p}(\Omega)$ of the boundary value problem
$$\begin{cases}
-\nabla\cdot (|\nabla u|^{p-2} \nabla u) = f \quad \text{in}\ \Omega \\\\
u = 0 \quad \text{on}\ \Omega
\end{cases}
$$
in the sense that $u$ satisfies the equation
$$\int_{\Omega} (\nabla u|\nabla u|^{p-2}\nabla \phi - f\phi)\ \text{d}x = 0 \qquad \forall \phi \in C_0^{+\infty}(\Omega)$$

Now I will try to give you a proof for what I still remember of.

Proof
First of all note that the left part of the last equation is the directional derivative of the $C^1$ functional:
$$E(u) = \frac{1}{p}\int_{\Omega}|\nabla u|^p\ \text{d}x - \int_{\Omega} f u\ \text{d}x$$
on the Banach space $V = H_0^{1, p}(\Omega)$ in direction $\phi$ which means that the above boundary value problem is of variational form.
Note also that $H_0^{1, p}(\Omega)$ is reflexive, and moreover $E$ is coercive. Infact:
$$E(u) \geq \frac{1}{p}||u||^p_{H_0^{1, p}} - ||f||_{H_0^{-1, q}} \geq \frac{1}{p}\left(||u||^p_{H_0^{1, p}} - c||u||_{H_0^{1, p}}\right) \geq \frac{1}{c}||u||^p_{H_0^{1, p}} - C$$
Finally, $E$ is sequentially weakly lower semi-continuous, so it suffices to show $u_m \to u$ in $H_0^{1, p}(\Omega)$, and we have
$$\int_{\Omega} f u_m\ \text{d}x = \int_{\Omega} f u\ \text{d}x$$
since $f\in H_0^{-1, q}(\Omega)$ (however this follows from the definition of weak convergence).
$\color{red}{\text{This means}}$ there exists a minimizer $u\in H_0^{1, p}(\Omega)$ of $E$ solving the equation above (that one for $E(u)$ in the Theorem).
Notice also that for $p > 2$ the Laplacian is strongly monotone in the sense that
$$\int_{\Omega}(|\nabla u|^{p-2}\nabla u - |\nabla v|^{p-2}\nabla v)\cdot (\nabla u - \nabla v)\ \text{d}x \geq c||u - v||^p_{H_0^{1, p}}$$
and also the solution $u$ is unique.
If $f$ is more regular, you can expect the solution to be more regular too.

Some details about $\color{red}{\text{This means}}$
In order to conclude that, we make use of another statement (another Theorem), which reads:
Suppose $V$ is a reflexive Banach space with norm $||\cdot ||$ and let $M\subset V$ be a weakly closed subset of $V$. Suppose $E : M\to \mathbb{R} \cup \{+\infty\}$ is coercive and sequentially weakly lower semi-continuous on $M$ with respect to $V$, that is: suppose the followings are fullfilled:

*

*$E(u) \to \infty\quad \text{as}\quad ||u||\to \infty,\quad u\in M$

*For any $u\in M$, any sequence $(u_m)$ in $M$ such that $u_m \to u$ weakly in $V$ there holds
$$E(u) \leq \lim_{m\to +\infty} \text{inf} E(u_m)$$
Then $E$ is bounded from below on $M$ and attains its infimum in $M$
A: This is what the area of calculus of variations is all about. I would start by reading the book by Dacorogna, which covers the question you are asking in much more detail than you probably ever wanted to know!
