# Intuition for the Lax-Milgram Theorem

I have seen the proof of the Lax-Milgram theorem (and can replicate it with most of the details), I've also applied it for a bi-linear form defined as the weak formulation of a PDE to show the existence of a solution. I still don't get the idea behind it. Is there an equivalent in finite dimensional linear algebra to give me some intuition?

Lax milgram says: $$a:H \times H \rightarrow \mathbb{R}$$ continuous, coercive, symmetric, then:

1. $$\forall \phi \in H^{\ast}, \exists ! u \in H$$ such that $$a(u,v) = \phi(v)$$ for all $$v \in H$$.
2. This $$u$$ is the only minimiser of $$H$$ of the problem $$\inf_{v \in H}\left\{\frac{1}{2}a(v,v)-\phi(v)\right\}$$

More concretely, my questions are:

• What is (1), some kind of dimension reduction?
• How did (2) come about? It's so useful in the context of PDEs but its mysterious to me.
• What's the finite dimensional intuition I can draw from?

EDIT:

Take $$H:=\mathbb{R}^n$$. In this simple finite dimensional context, $$a(u,v) := \langle u,v\rangle$$ is the natural choice for a continuous, coercive, and symmetric bilinear form.

Then the dual is the space of all row vectors. So Lax-Milgram tells us that $$\forall r \in (\mathbb{R}^n)^{\ast}$$ there exists a unique $$u \in \mathbb{R}^n$$ such that $$\langle u,v\rangle = r v$$ for all $$v \in \mathbb{R}$$, i.e. the only one is $$r=u^T$$. It just tells us a different way to write the dot product using elements in the dual.

As for the minimisation problem, Lax-Milgram says that the solution to this minimisation problem would be $$\frac{1}{2} \langle u, u \rangle - \langle u,u\rangle = -\frac{1}{2}\|u\|^2$$ I am at a loss for what this represents, but I now have some intuition for the first result! Hopefully someone can comment on this minimisation.

This $$u$$ is what we call a weak solution. The most obvious reason why we ask for weak solutions is because sometimes strong solutions don't exist and we still want to make sense of problems in these cases. The other reason that comes to mind is that even when strong solutions exist, it is still often easier to prove that by showing existence of a weak solution and then a posteriori showing it is a strong solution. This is in part because weak solutions provide a framework for the use of variational methods and so forth.
The finite dimensional analogue of Lax-Milgram is that $$Ax=b$$ has a unique solution if $$A$$ is symmetric positive definite. The "weak formulation" of $$Ax=b$$ is $$\langle v,Ax \rangle = \langle v,b \rangle$$ for all $$v \in \mathbb{R}^n$$, and the bilinear form $$a$$ now is $$a(v,x)=\langle v,Ax \rangle$$. (I use scare quotes on weak formulation because the two formulations are equivalent.)
Relative to the finite dimensional analogy described in the previous paragraph, you are converting from $$Ax=b$$ to $$x=\arg \min f$$ where $$f(x)=\frac{1}{2} \langle x,Ax \rangle - \langle x, b\rangle$$. The fact that $$x=\arg \min f$$ implies $$Ax=b$$ follows from writing out the equation $$\nabla f = 0$$ and applying the symmetry of $$A$$. The positive definiteness is what you need to ensure that the resulting critical point is a minimum.