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The Lax--Milgram theorem

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a real or complex Hilbert space, let $a$ be a bounded coercive sesquilinear form on $H$ with constants $M, α$ (Bounded, coercive and symmetric sesquilinear forms), and let $F : H \to K$ be a bounded conjugate-linear functional with norm $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ . Then there is a unique $u \in H$ with $a (u, v) = F (v) for every v \in H,$ and it satisfies $α ∥ u ∥ \leq ∥ F ∥$ , that is $∥ u ∥ \leq ∥ F ∥ / α$ . The real bilinear case is the same statement with $a$ symmetric or not, $F$ a bounded linear functional, and the conjugation read as the identity.

Facts & Assumptions

Given: Countable Choice; a real or complex Hilbert space $H$ ; a bounded coercive sesquilinear form $a$ on $H$ with constants $M \geq 0$ and $α > 0$ ; a bounded conjugate-linear functional $F$ on $H$ with $∥ F ∥ = \sup_{∥ v ∥ \leq 1} ∣ F (v) ∣$ ; and the operator $A$ of $a$ , $a (u, v) = (A u, v)$ .

[F1]

$A$ is linear and bounded with $a (u, v) = (A u, v)$ , $∥ A ∥ \leq M$ , and $Re a (u, u) \geq α ∥ u ∥^{2}$ (Bounded, coercive and symmetric sesquilinear forms, A bounded form is represented by a unique bounded operator, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F2]

Riesz representation under Countable Choice: every bounded linear functional $G$ has a unique $w$ with $G (v) = (v, w)$ for all $v$ , and $∥ G ∥ = ∥ w ∥$ (Riesz representation for Hilbert spaces, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

Contractions on complete spaces: a map $T$ of a nonempty complete metric space with $∥ T (u) - T (v) ∥ \leq q ∥ u - v ∥$ , $0 \leq q < 1$ , has exactly one fixed point; and for $H \neq {0}$ and $0 < ρ < 2 α / M^{2}$ the map $u \mapsto u - ρ (A u - w)$ is a strict contraction with constant $q_{ρ} = (1 - 2 ρ α + ρ^{2} M^{2})^{1 / 2} < 1$ (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, Coercivity makes a small form step a strict contraction, Hilbert space).

[F4]

Conjugation: $G (v) : = \overline{F (v)}$ is linear when $F$ is conjugate-linear, $∣ G (v) ∣ = ∣ F (v) ∣$ , and $G$ bounded with the same norm; $Re z \leq ∣ z ∣$ and $∣ z ∣ = ∣ \overline{z} ∣$ (Real and imaginary parts, complex conjugation, and modulus).

[F5]

Linearity of $a$ in the first argument and the estimate for the unique solution follow from $a (u, v) = (A u, v)$ ; the degenerate space $H = {0}$ has the unique solution $u = 0$ and $∥ F ∥ = 0$ (Bounded, coercive and symmetric sesquilinear forms, Hilbert space).

Proof

1.1F1algebra

Assume first $H \neq {0}$ . Then the given bound and coercivity constant satisfy $α \leq M$ and $M > 0$ : choosing $u \neq 0$ and normalising, $α \leq Re a (u, u) \leq ∣ a (u, u) ∣ \leq M ∥ u ∥^{2}$ , so $M \geq α > 0$ ; hence $ρ : = α / M^{2}$ satisfies $0 < ρ < 2 α / M^{2}$ .

1.2F1algebra

Uniqueness: if $u$ satisfies $a (u, v) = 0$ for every $v$ , then testing $v = u$ gives $α ∥ u ∥^{2} \leq Re a (u, u) = 0$ , so $u = 0$ . If $u_{1}, u_{2}$ are two solutions of $a (u, \cdot) = F$ , then first-slot linearity gives $a (u_{1} - u_{2}, v) = 0$ for all $v$ , so $u_{1} = u_{2}$ .

2.1F2F4step 1.1

The conjugate functional: $G (v) : = \overline{F (v)}$ is a bounded linear functional with $∥ G ∥ = ∥ F ∥$ , so by Riesz representation there is a unique $w \in H$ with $G (v) = (v, w)$ for all $v$ , that is $F (v) = (w, v)$ for all $v$ , and $∥ w ∥ = ∥ F ∥$ .

3.1F1F3step 2.1

Existence: fix $ρ$ as in step 1.1 and define $T (u) : = u - ρ (A u - w)$ . By [F3] the map $T$ is a strict contraction of the complete space $H$ with constant $q_{ρ} < 1$ , so by the Banach fixed-point theorem it has a fixed point $u \in H$ ; then $u = u - ρ (A u - w)$ gives $A u = w$ , and hence $a (u, v) = (A u, v) = (w, v) = F (v)$ for every $v \in H$ .

4.1F1F4step 3.1algebra

Estimate: for the solution $u$ of step 3.1, $α ∥ u ∥^{2} \leq Re a (u, u) = Re F (u) \leq ∣ F (u) ∣ \leq ∥ F ∥ ∥ u ∥$ ; if $u \neq 0$ divide by $∥ u ∥$ to get $α ∥ u ∥ \leq ∥ F ∥$ , and if $u = 0$ the inequality holds trivially.

5.1F5given∎

Degenerate space: if $H = {0}$ , then the only element is $0$ , the only functional is $0$ and it is the value at the unique solution $u = 0$ with $α ∥ u ∥ = 0 = ∥ F ∥$ ; uniqueness is immediate. The real bilinear case is the same argument with conjugation read as the identity, and $a$ need not be symmetric.

The Lax--Milgram theorem

Statement

Facts & Assumptions

Proof

Depends on

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Sources