Hilbert Spaces - Applications

The Lax-Milgram Theorem is a cornerstone of the modern theory of partial differential equations, providing existence and uniqueness of weak solutions to elliptic boundary value problems.

TheoremLax-Milgram Theorem

Let $H$ be a Hilbert space and $a : H \times H \to \mathbb{K}$ a bilinear form satisfying:

Boundedness: $|a(x,y)| \leq M \|x\| \|y\|$ for all $x, y \in H$
Coercivity: $|a(x,x)| \geq \alpha \|x\|^2$ for all $x \in H$ and some $\alpha > 0$

Then for every continuous linear functional $\phi \in H^*$ , there exists a unique $u \in H$ such that $a(u, v) = \phi(v)$ for all $v \in H$ . Moreover, $\|u\| \leq \frac{1}{\alpha} \|\phi\|$ .

This theorem transforms the problem of solving $a(u,v) = \phi(v)$ into showing that the bilinear form $a$ satisfies boundedness and coercivity. These conditions are often easy to verify for differential operators.

ExampleApplication to Poisson's Equation

Consider the boundary value problem $-\Delta u = f \text{ in } \Omega, \quad u = 0 \text{ on } \partial\Omega$ where $\Omega \subset \mathbb{R}^n$ is a bounded domain.

Weak Formulation: Find $u \in H_0^1(\Omega)$ such that $\int_\Omega \nabla u \cdot \nabla v \, dx = \int_\Omega f v \, dx$ for all $v \in H_0^1(\Omega)$ .

Verification:

The bilinear form $a(u,v) = \int_\Omega \nabla u \cdot \nabla v \, dx$ is bounded by Cauchy-Schwarz
By Poincaré's inequality, $a(u,u) = \|\nabla u\|_{L^2}^2 \geq C \|u\|_{H^1}^2$ (coercivity)
The functional $\phi(v) = \int_\Omega fv \, dx$ is continuous by Hölder's inequality

Therefore, Lax-Milgram guarantees a unique weak solution.

Proof

Define an operator $A : H \to H$ using Riesz representation: for each $u \in H$ , let $A(u)$ be the unique element satisfying $a(u, v) = \langle A(u), v \rangle$ for all $v \in H$ . The operator $A$ is bounded by the boundedness of $a$ .

By coercivity, $\langle A(u), u \rangle = a(u,u) \geq \alpha \|u\|^2$ , which implies $A$ is injective and $\|A(u)\| \geq \alpha \|u\|$ . This shows $A$ has closed range.

To show $A$ is surjective, use the fact that $\text{Range}(A)^\perp = \ker(A^*) = \{0\}$ by similar coercivity arguments for $A^*$ .

Given $\phi \in H^*$ , let $y \in H$ be its Riesz representer. Since $A$ is surjective, there exists $u \in H$ with $A(u) = y$ , giving $a(u,v) = \langle y, v \rangle = \phi(v)$ for all $v$ .

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Remark

The Lax-Milgram Theorem generalizes the existence theory for linear systems $Ax = b$ to infinite dimensions. The coercivity condition plays the role of positive definiteness, ensuring that the operator $A$ is invertible with bounded inverse.

This theorem has revolutionized the study of PDEs by providing a systematic framework for proving existence and uniqueness of solutions without requiring explicit solution formulas.