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The Hilbert-space Babuška–Nečas–Babuška criterion

Statement

Assume Countable Choice. Let $X$ and $Y$ be real Hilbert spaces and let $b : X \times Y \to R$ be a bounded bilinear form. Suppose there is $β > 0$ such that $\sup_{∥ y ∥_{Y} \leq 1} ∣ b (x, y) ∣ \geq β ∥ x ∥_{X} (x \in X),$ and the only $y \in Y$ satisfying $b (x, y) = 0$ for every $x \in X$ is $y = 0$ . Then, for every bounded linear functional $F : Y \to R$ , there is a unique $x \in X$ such that $b (x, y) = F (y)$ for every $y \in Y$ , and $∥ x ∥_{X} \leq β^{- 1} ∥ F ∥_{Y^{*}}, ∥ F ∥_{Y^{*}} : = \sup_{∥ y ∥_{Y} \leq 1} ∣ F (y) ∣ .$

Facts & Assumptions

Given: Countable Choice, real Hilbert spaces $X, Y$ , the bounded bilinear form $b$ , its stated inf-sup bound, and the trivial right kernel.

[F1]

Every bounded real linear functional on a Hilbert space has a unique Riesz representative with the same norm (Riesz representation for Hilbert spaces).

[F2]

A closed linear subspace of a Hilbert space and its orthogonal complement give a direct decomposition of the whole space (Orthogonal decomposition by a closed subspace).

[F3]

A Hilbert space is complete in its norm (Hilbert space).

[A1]

Countable Choice is the hypothesis of the Riesz representation and orthogonal-decomposition suppliers (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1A1F1givenconstruct

If $X = {0}$ , the right-kernel hypothesis forces $Y = {0}$ ; the only $F \in Y^{*}$ is zero and $x = 0$ is the unique solution. If $Y = {0}$ and $X \neq {0}$ , the inf-sup hypothesis fails for any nonzero $x$ , so no such instance satisfies the assumptions. Assume henceforth both spaces are nonzero. For each $x \in X$ , the functional $y \mapsto b (x, y)$ is bounded and linear. By [F1], let $S x \in Y$ be the unique vector satisfying $(S x, y)_{Y} = b (x, y)$ for every $y \in Y$ . Bilinearity and uniqueness make $S : X \to Y$ linear, and the Riesz norm identity gives $∥ S x ∥_{Y} = \sup_{∥ y ∥_{Y} \leq 1} ∣ b (x, y) ∣ \leq C_{b} ∥ x ∥_{X}$ , where $C_{b}$ is a continuity bound for $b$ .

2.1F3givenstep 1.1algebra

The inf-sup assumption gives $∥ S x ∥_{Y} \geq β ∥ x ∥_{X}$ . Thus $S$ is injective. If $S x_{n} \to y$ in $Y$ , then $∥ x_{n} - x_{m} ∥_{X} \leq β^{- 1} ∥ S x_{n} - S x_{m} ∥_{Y}$ , so completeness of $X$ gives $x_{n} \to x$ for some $x \in X$ . The bound in step 1.1 gives $S x_{n} \to S x$ , hence $y = S x$ . Therefore $S (X)$ is closed.

3.1A1F2step 2.1given

If $y \in S (X)^{⊥}$ , then $(S x, y)_{Y} = b (x, y) = 0$ for every $x \in X$ ; the right-kernel hypothesis gives $y = 0$ . Apply [F2] to the closed subspace $S (X)$ . Its orthogonal complement is zero, so $Y = S (X) \oplus {0} = S (X)$ ; therefore $S$ is onto.

4.1A1F1step 1.1step 2.1step 3.1algebra∎

For the given $F \in Y^{*}$ , let $g \in Y$ be its Riesz representative, so $F (y) = (y, g)_{Y}$ and $∥ g ∥_{Y} = ∥ F ∥_{Y^{*}}$ . Surjectivity supplies $x \in X$ with $S x = g$ , whence $b (x, y) = (S x, y)_{Y} = (g, y)_{Y} = (y, g)_{Y} = F (y)$ for all $y$ , using symmetry of the real inner product. The lower bound $β ∥ x ∥_{X} \leq ∥ S x ∥_{Y} = ∥ F ∥_{Y^{*}}$ proves the estimate and also uniqueness.

Depends on

Used by

Well-posedness for measurable coercive parabolic forms Theorem

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Christian Clason, Introduction to Finite Element Methods, lecture notes (standard reference, not scraped)