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The elliptic Fredholm range condition is orthogonality to the adjoint kernel

Statement

Assume the Axiom of Choice and Countable Choice. Let $Ω \subseteq R^{n}$ be bounded open, $μ \geq β$ , and let $K_{μ}, K_{μ}^{*}$ be as in The adjoint solution operator solves the adjoint form problem. For $f \in L^{2} (Ω)$ , $K_{μ} f \in ran (I - μ K_{μ}) ⟺ (f, v)_{L^{2}} = 0 for every v \in H_{0}^{1} (Ω) with a^{*} (v, w) = 0 \forall w \in H_{0}^{1} (Ω),$ where $a^{*}$ is the adjoint form of The formal adjoint and the adjoint weak Dirichlet problem and $K_{μ}$ is the shifted solution operator of The shifted elliptic solution operator.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a bounded open set $Ω \subseteq R^{n}$ ; a fixed $μ \geq β$ ; the operators $K_{μ}, K_{μ}^{*}$ on $L^{2} (Ω)$ ; and $f \in L^{2} (Ω)$ .

[F1]

Compactness: $μ K_{μ}$ is a compact operator on the Banach space $L^{2} (Ω)$ , so $A : = I - μ K_{μ}$ is an identity-minus-compact operator (The shifted solution operator is compact on $L^{2}$ , The space $L^{p} (μ)$ as the quotient by null functions, The Axiom of Choice).

[F2]

Fredholm alternative: for a compact operator $K$ on a Banach space and $A = I - K$ , an element $y$ lies in $ran A$ if and only if $φ (y) = 0$ for every $φ$ in $\ker A^{*}$ , where $A^{*} = I - K^{*}$ is the transpose on the dual (Fredholm alternative for identity minus compact, The transpose of a bounded operator).

[F3]

Riesz representation: every bounded linear functional $φ$ on $L^{2} (Ω)$ has the form $φ (y) = (y, g_{φ})_{L^{2}}$ for a unique $g_{φ} \in L^{2} (Ω)$ , and for the Hilbert adjoint $K_{μ}^{*}$ one has $(K_{μ} y, g) = (y, K_{μ}^{*} g)$ (Riesz representation for Hilbert spaces, The Hilbert-space adjoint of a bounded operator, The dual space X^* of a normed space and its dual norm).

[F4]

Kernel of $I - μ K_{μ}^{*}$ : for $v \in L^{2} (Ω)$ , $v \in \ker (I - μ K_{μ}^{*})$ if and only if $v \in H_{0}^{1} (Ω)$ and $a^{*} (v, w) = 0$ for every $w \in H_{0}^{1} (Ω)$ (The adjoint solution operator solves the adjoint form problem).

[F5]

Adjoint identity: $(K_{μ} y, g)_{L^{2}} = (y, K_{μ}^{*} g)_{L^{2}}$ for all $y, g \in L^{2} (Ω)$ , and $μ > 0$ (The adjoint solution operator solves the adjoint form problem, The shifted elliptic solution operator).

Proof

technique · direct

1.1F1F2given

The operator $A = I - μ K_{μ}$ is a bounded linear operator on the Banach space $L^{2} (Ω)$ , and $μ K_{μ}$ is compact by [F1]. By the Fredholm alternative [F2] applied with $K : = μ K_{μ}$ and $y : = K_{μ} f$ , the inclusion $K_{μ} f \in ran A$ is equivalent to the vanishing of $φ (K_{μ} f)$ for every bounded linear functional $φ$ with $A^{*} φ = 0$ .

2.1F3F5step 1.1givenalgebra

Description of $\ker A^{*}$ . For $φ \in (L^{2} (Ω))^{*}$ let $g_{φ} \in L^{2} (Ω)$ be its Riesz vector, $φ (y) = (y, g_{φ})_{L^{2}}$ as in [F3]. Then, using the transpose identity and the Hilbert adjoint, $(A^{*} φ) (y) = φ (A y) = (A y, g_{φ})_{L^{2}} = (y, A^{*} g_{φ})_{L^{2}} = (y, (I - μ K_{μ}^{*}) g_{φ})_{L^{2}}$ for all $y$ , so $φ \in \ker A^{*}$ if and only if $g_{φ} \in \ker (I - μ K_{μ}^{*})$ . For such a vector, [F5] gives $φ (K_{μ} f) = (K_{μ} f, g_{φ})_{L^{2}} = (f, K_{μ}^{*} g_{φ})_{L^{2}} = (f, \frac{1}{μ} g_{φ})_{L^{2}} = \frac{1}{μ} (f, g_{φ})_{L^{2}},$ since $g_{φ} = μ K_{μ}^{*} g_{φ}$ ; because $μ > 0$ this vanishes if and only if $(f, g_{φ})_{L^{2}} = 0$ .

3.1F4step 1.1step 2.1given∎

Weak form of the kernel. By [F4] the condition $g_{φ} \in \ker (I - μ K_{μ}^{*})$ is equivalent to $g_{φ} \in H_{0}^{1} (Ω)$ and $a^{*} (g_{φ}, w) = 0$ for every $w \in H_{0}^{1} (Ω)$ . Substituting into step 2.1, $K_{μ} f \in ran (I - μ K_{μ})$ holds if and only if $(f, v)_{L^{2}} = 0$ for every $v \in H_{0}^{1} (Ω)$ with $a^{*} (v, w) = 0$ for all $w \in H_{0}^{1} (Ω)$ , as claimed.

Depends on

Used by

The Fredholm alternative for weak elliptic Dirichlet problems Theorem

Dependency tree · two levels

66 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

John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page notes) (standard reference, not scraped)
Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer Universitext, 2011, complete 614-page text) (standard reference, not scraped)