---
tag: 85KK79FCN5
kind: statement
role: theorem
name: Lax–Milgram theorem
msc: [35A15, 47A07, 35J20]
uses: [HM2HHE3J51, 0SQZMXB6GP, 86KRHS6X22, 4A82S5FVRT, TTC40N49M6, SJZX81A783, M4CEZ5K9HP, JS2CCR3RNS]
about: [86KRHS6X22]
sources:
  - url: https://en.wikipedia.org/wiki/Babu%C5%A1ka%E2%80%93Lax%E2%80%93Milgram_theorem
---

## Statement

Let $H$ be a real [[Hilbert space|HM2HHE3J51]] with norm $\norm{\cdot}$.
Let $B \colon H \times H \to \R$ be a [[bilinear form|0SQZMXB6GP]] with these two properties:

1. $B$ is bounded: there is $M \ge 0$ with $\abs{B(u, v)} \le M \norm{u} \norm{v}$ for all $u, v \in H$.
2. $B$ is [[coercive|86KRHS6X22]]: there is $\alpha > 0$ with $B(u, u) \ge \alpha \norm{u}^2$ for all $u \in H$.

Then for every bounded linear functional $f$ on $H$, an element of the [[continuous dual space|4A82S5FVRT]] $H'$, there is a unique $u \in H$ such that
$$B(u, v) = f(v) \quad \text{for all } v \in H.$$
Moreover $\norm{u} \le \norm{f}_{H'} / \alpha$.

## Remarks

The form $B$ need not be symmetric.
By the [[Riesz representation theorem|TTC40N49M6]], there are a bounded linear operator $A$ on $H$ with $B(u, v) = \ip{Au}{v}$ and an element $w$ with $f(v) = \ip{w}{v}$.
Coercivity gives $\alpha \norm{u} \le \norm{Au}$, so $A$ is injective with closed range.
If $v$ is orthogonal to the range, then $B(v, v) = \ip{Av}{v} = 0$, so $v = 0$; hence $A$ is onto and $u = A^{-1} w$.
The bound follows from $\alpha \norm{u}^2 \le B(u, u) = f(u) \le \norm{f}_{H'} \norm{u}$.
When $B$ is symmetric, the theorem is the Riesz representation theorem for the inner product $B$.

Example: let $U \subset \R^n$ be a bounded open set and $g \in L^2(U)$.
On $H = H^1_0(U)$, the [[Sobolev space|M4CEZ5K9HP]], take $B(u, v) = \int_U \nabla u \cdot \nabla v \, dx$ and $f(v) = \int_U g v \, dx$.
The [[Poincaré inequality|JS2CCR3RNS]] shows that $B$ is coercive, so there is a unique [[weak solution|SJZX81A783]] $u \in H^1_0(U)$ of $-\Delta u = g$ with $u = 0$ on $\partial U$.

Coercivity cannot be dropped.
On $H = \ell^2$, let $S$ be the right shift and $B(u, v) = \ip{Su}{v}$, which is bounded.
For $f(v) = \ip{e_1}{v}$, a solution would satisfy $S u = e_1$, which is impossible since the first coordinate of $Su$ is $0$.

## Explanations

### L1 human

Many boundary value problems can be written as "find $u$ such that $B(u, v) = f(v)$ for every test vector $v$".
The theorem says that this problem has exactly one solution as soon as $B$ is bounded and positive in a strong sense.
It also says that the solution depends continuously on the data $f$.

### L2 human

This is the standard existence theorem for weak solutions of linear elliptic equations in divergence form.
One checks boundedness from bounds on the coefficients and coercivity from uniform ellipticity together with the Poincaré inequality, possibly after adding a multiple of $u$ to the equation.
The proof is also the basis of finite element error analysis.
Restricting the problem to a finite-dimensional subspace gives the Galerkin approximation, and the constants $M$ and $\alpha$ control its error.
The Babuška version replaces coercivity by an inf-sup condition and allows different trial and test spaces.