Lecture 12 - Lax Milgram

#math

Lax Milgram Theorem

Let $(H, ⟨ ., . ⟩, | | | |)$ be a Hilbert space and $a$ continuous, coercive bilinear form on $H \times H$

a : H \times H \to R (or C)

$\exists α, c > 0$ such that

| a (u, v) | \leq c | | u | | v | |

a (u, u) \geq α | | u | |^{2} for any u, v \in H

and $l$ a continuous, linear form on $H$ . Then $\exists! u \in H$ such that $\forall v \in H a (u, v) = l (v)$ . If $a$ is also symmetric, then $u$ is the unique minimum of:

J : v \in H \to \frac{1}{2} a (v, v) - l (v)

Proof

$l$ is a continuous linear form on $H$ , hence by the representation theorem of Riesz, $\exists! f$ such that $\forall v \in H$ ,

l (v) = ⟨ v, f_{l} ⟩

Also, as $a$ is continuous, for $u \in H$ , the map $v \to a (u, v)$ is a continuous linear form on $H : \exists! a_{u} \in H$ such that for any $v \in H$ $a (u, v) = ⟨ v, a_{u} ⟩$
Let $A : u \in H \to a_{u} \in H$ . We have to show that $\exists! u \in H$ such that $A u = f$ .
It suffices to show that $A : H \to H$ is bijective.
Injectivity: For $u, v \in H$ , $λ \in R$ we have:

\begin{aligned} ⟨ w, A (u + λ v) ⟩ & = a (u + λ v, w) \\ = a (u, w) + λ a (v, w) \\ = ⟨ w, A u + λ A v ⟩ \end{aligned}

i.e $A$ is linear. Moreover $\forall u \in H$

\begin{aligned} | | A u | |^{2} & = ⟨ A u, A v ⟩ \\ = a (A u, u) \\ \leq c | | A u | | | | u | | \\ ⟹ & | | A u | | \leq c | | u | | ⟺ A is continuous \end{aligned}

Now, from the coercivity of $a$ , we deduce that $\forall u \in H$ $⟨ u, A u ⟩ = a (u, u) \geq α | | u | |^{2}$ . Where upon if $A u = 0 ⟹ | | u | | = 0 ⟹ u = 0$ i.e $K e r (A) = {0}$ . We deduce that $I m (A)$ is closed. In fact if ${v_{n}}_{n = 1}^{\infty} = {A u_{n}}_{n = 1}^{\infty} i n (I m (A))^{n} \to v \in H$ we have:

\begin{aligned} α | | u_{p} - u_{q} | |^{2} & \leq a (u_{p} - u_{q}, u_{p} - u_{q}) \\ \leq ⟨ u_{p} - u_{q}, A (u_{p} - u_{q}) ⟩ \\ \leq | | v_{p} - v_{q} | | | | u_{p} - u_{q} | | \\ ⟹ & α | | u_{p} - u_{q} | | \leq | | v_{p} - v_{q} | | \end{aligned}

Hence $(u_{n})$ is a Cauchy sequence in $H$ converges to $u \in H$ . Then by continuity $A u = A lim_{n \to \infty} u_{n} = lim_{n \to \infty} A u_{n} = v \in I m (A)$ . That is the image is closed.
Surjectivity: It suffices to verify that

(I m A)^{⊥} = {0}

Let $v \in (I m (A))^{⊥} ⟹ ⟨ v, A u ⟩ = 0 ⟹ a (u, v) = 0 \forall u \in H ⟹ v = 0.$
We proved that $A$ is bijective. In particular, $\exists! u \in H$ such that $\forall v \in H, a (u, v) = l (v)$ . Let $a$ be symmetric then $a (u, v) = a (v, u)$ . Then, for $v \in H$ let $J (u) = \frac{1}{2} a (u, u) - l (u)$

\begin{aligned} J (u + v) & = \frac{1}{2} a (u + v, u + v) - l (u + v) \\ = \frac{1}{2} (a (u, u) + a (u, v) + a (v, u) + a (v, v) - l (u) - l (v)) \\ = \frac{1}{2} a (u, u) + \frac{1}{2} a (v, v) + a (u, v) - l (u) - l (v) \\ = J (u) + \frac{1}{2} a (v, v) \\ \geq J (u) + \frac{α}{2} | | v | |^{2} \end{aligned}

i.e $v = 0$ minimizes our function, that is $u + 0 = u$ is our unique minimizer.