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✓ 25 results · all verified · 16 also independently AI-judged

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Fredholm Elliptic Problems and the Elliptic Spectrum

1 · Prerequisites

2 · Summary

This page builds the elliptic Fredholm theory and the spectral theory of a symmetric divergence-form Dirichlet operator. Gårding's inequality gives an explicit lower bound for the form in terms of the ellipticity and coefficient constants, and a sufficiently large shift makes the form coercive with constant $θ / 2$ . The shifted solution operator $K_{μ}$ is then bounded from $L^{2}$ to $H_{0}^{1}$ , and on a bounded open set it is compact after composition with the Rellich embedding. The unshifted weak equation is algebraically equivalent to the identity-minus-compact equation $(I - μ K_{μ}) u = K_{μ} f$ ; reading the abstract Fredholm alternative through this reduction yields the two-alternative theorem for weak Dirichlet problems with $L^{2}$ data, the finite dimension and equality of the dimensions of the homogeneous and adjoint homogeneous spaces, and the uniqueness-implies-existence corollary with a bounded solution map. The formal adjoint is defined form-first, and its solution operator is the Hilbert-space adjoint of $K_{μ}$ , which turns the range condition into orthogonality to the weak adjoint kernel.

In the symmetric case the associated $L^{2}$ operator $L$ has dense domain, is symmetric and lower bounded. Surjectivity of the shifted operators establishes self-adjointness, and compactness of the shifted inverse gives compact resolvent on bounded domains; $K_{μ}$ restricted to the symmetric case is positive and self-adjoint. The compact self-adjoint spectral theorem then produces a nondecreasing eigenvalue list with finite multiplicities and an orthonormal eigenbasis of $L^{2}$ , with the eigenbasis expanding every $H_{0}^{1}$ element in the form norm. The Rayleigh and Courant--Fischer variational principles identify the eigenvalues as min-max values of the Rayleigh quotient, the first Dirichlet eigenvalue is monotone under domain inclusion, and the reciprocal square root of the first eigenvalue of the Dirichlet Laplacian is the optimal zero-trace Poincaré constant. The resolvent is a spectral series with norm the reciprocal distance to the spectrum, and the non-invertible shifts form a closed discrete set.

Conventions: $Ω \subseteq R^{n}$ is open, $n \geq 1$ , with boundedness and connectedness imposed only where stated; coefficients are measurable, essentially bounded and uniformly elliptic, with the sesquilinear convention linear in the first argument and conjugate-linear in the second; all spaces are almost-everywhere classes, and $H_{0}^{1}$ is the closure of $C_{c}^{\infty}$ . Countable Choice is declared for the Sobolev, Lebesgue and Hilbert-space interfaces, and the Axiom of Choice is carried exactly where compactness of the Rellich embedding and the abstract Fredholm alternative are used. No boundary regularity of $\partial Ω$ is assumed anywhere on this page, and the Fredholm alternative is stated for $L^{2}$ data; $H^{- 1}$ data are not treated.

3 · Logical flowchart

4 · Definitions, theorems and proofs

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Garding's inequality for a divergence-form elliptic operator

Statement

Assume Countable Choice (CC) (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the Sobolev and Lebesgue interfaces used by Uniformly elliptic divergence-form operators and their sesquilinear forms and The elliptic form is well defined and bounded on $H^{1}$ . Let $Ω \subseteq R^{n}$ be open, $n \geq 1$ , $K \in {R, C}$ , and let $L$ and its sesquilinear form $a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms, with ellipticity constant $θ > 0$ and coefficient bounds $M_{a}, M_{b}, M_{c}$ . Then every $u \in H^{1} (Ω)$ satisfies $Re a (u, u) \geq \frac{θ}{2} ∥ D u ∥_{L^{2} (Ω)}^{2} - (\frac{n M_{b}^{2}}{2 θ} + M_{c}) ∥ u ∥_{L^{2} (Ω)}^{2},$ and consequently, with the explicit constants $α : = θ / 2$ and $β : = θ / 2 + n M_{b}^{2} / (2 θ) + M_{c}$ , $Re a (u, u) \geq α ∥ u ∥_{H^{1} (Ω)}^{2} - β ∥ u ∥_{L^{2} (Ω)}^{2} .$ Both inequalities restrict to $u \in H_{0}^{1} (Ω)$ . No Poincare inequality, no boundedness of $Ω$ and no symmetry of $a$ is used; the constants are explicit and are not claimed to be optimal.

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ , $n \geq 1$ ; $K \in {R, C}$ ; a uniformly elliptic divergence-form operator $L$ and its form $a$ with ellipticity constant $θ > 0$ and coefficient bounds $M_{a}, M_{b}, M_{c}$ ; and $u \in H^{1} (Ω)$ (or $u \in H_{0}^{1} (Ω)$ ).

[F1]

Coefficients and form: $a^{i j}, b^{i}, c$ are measurable and essentially bounded with $∣ a^{i j} ∣ \leq M_{a}$ , $∣ b^{i} ∣ \leq M_{b}$ , $∣ c ∣ \leq M_{c}$ almost everywhere, the uniform ellipticity condition $Re (\sum_{i, j} a^{i j} (x) ξ_{j} \overline{ξ_{i}}) \geq θ ∣ ξ ∣^{2}$ holds for almost every $x$ and all $ξ \in C^{n}$ , and $a (u, u) = \int_{Ω} (a^{i j} D_{j} u \overline{D_{i} u} + b^{i} D_{i} u \overline{u} + c u \overline{u}) d x$ (Uniformly elliptic divergence-form operators and their sesquilinear forms, The essential supremum of a measurable function with respect to a measure, The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

[F2]

The three integrals in [F1] are absolutely convergent for $u \in H^{1} (Ω)$ , so the real part of $a (u, u)$ is the sum of the real parts of the three integrals (The elliptic form is well defined and bounded on $H^{1}$ , Complex Lp classes and Euclidean test-function conventions).

[F3]

Holder and the coefficient bounds: for measurable functions with $∣ b^{i} ∣ \leq M_{b}$ , $∣ c ∣ \leq M_{c}$ almost everywhere, $∣ \int_{Ω} b^{i} D_{i} u \overline{u} d x ∣ \leq M_{b} ∥ D_{i} u ∥_{L^{2}} ∥ u ∥_{L^{2}}$ and $∣ \int_{Ω} c ∣ u ∣^{2} d x ∣ \leq M_{c} ∥ u ∥_{L^{2}}^{2}$ (Holder's inequality for integrals, including the endpoint cases, The space $L^{p} (μ)$ as the quotient by null functions).

[F4]

For real $r, s \geq 0$ and $θ > 0$ , Young's inequality with $p = q = 2$ gives $\sqrt{n} M_{b} r s \leq \frac{θ}{2} r^{2} + \frac{n M_{b}^{2}}{2 θ} s^{2}$ (Young's inequality for conjugate real exponents).

[F5]

Norm identity: on $H^{1} (Ω)$ and on its subspace $H_{0}^{1} (Ω)$ the norm satisfies $∥ u ∥_{H^{1}}^{2} = ∥ u ∥_{L^{2}}^{2} + ∥ D u ∥_{L^{2}}^{2}$ , where $∣ D u ∣^{2} = \sum_{i = 1}^{n} ∣ D_{i} u ∣^{2}$ (Integer-order Sobolev spaces and their norms, The notation $H^{k}$ and the reserved zero-boundary symbol, Zero-boundary Sobolev space as a norm closure).

[F6]

Finite-index Cauchy--Schwarz is Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs applied to $(∥ D_{i} u ∥_{2})_{i = 1}^{n}$ and $(1)_{i = 1}^{n}$ in Euclidean space. Elementary inequalities $∣ z ∣ \geq Re z$ and $∣ z ∣ \geq ∣ Re z ∣$ for complex $z$ , and $∥ D u ∥_{L^{2}}^{2} = \int_{Ω} ∣ D u ∣^{2}$ (Real and imaginary parts, complex conjugation, and modulus, Complex Lp classes and Euclidean test-function conventions).

Proof

technique · direct

1.1F1F6algebra

Principal part. Since $D u (x) \in C^{n}$ , applying the ellipticity hypothesis with $ξ = D u (x)$ gives $Re (a^{i j} (x) D_{j} u (x) \overline{D_{i} u (x)}) \geq θ ∣ D u (x) ∣^{2}$ for almost every $x \in Ω$ , and integration over $Ω$ yields $\int_{Ω} Re (a^{i j} D_{j} u \overline{D_{i} u}) d x \geq θ ∥ D u ∥_{L^{2}}^{2} .$

1.2F1F3F6algebra

Drift term. Pointwise $∣ b^{i} D_{i} u \overline{u} ∣ \leq M_{b} ∣ D_{i} u ∣ ∣ u ∣$ almost everywhere, so [F3] gives $∣ \int_{Ω} b^{i} D_{i} u \overline{u} d x ∣ \leq M_{b} ∥ D_{i} u ∥_{L^{2}} ∥ u ∥_{L^{2}}$ for each $i$ . Summing the $n$ terms and applying Cauchy--Schwarz in the index $i$ , $\sum_{i = 1}^{n} ∥ D_{i} u ∥_{L^{2}} \leq \sqrt{n} ∥ D u ∥_{L^{2}}$ , hence $∣ \int_{Ω} b^{i} D_{i} u \overline{u} d x ∣ \leq \sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}},$ and in particular $Re \int_{Ω} b^{i} D_{i} u \overline{u} d x \geq - ∣ \int_{Ω} b^{i} D_{i} u \overline{u} d x ∣ \geq - \sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}}$ .

1.3F1F3F6algebra

Reaction term. Since $∣ c u \overline{u} ∣ = ∣ c ∣ ∣ u ∣^{2} \leq M_{c} ∣ u ∣^{2}$ almost everywhere, [F3] gives $∣ \int_{Ω} c ∣ u ∣^{2} d x ∣ \leq M_{c} ∥ u ∥_{L^{2}}^{2}$ , so $Re \int_{Ω} c u \overline{u} d x \geq - M_{c} ∥ u ∥_{L^{2}}^{2}$ .

2.1F2step 1.1step 1.2step 1.3algebra

Combine the three terms. By [F2] the real part of $a (u, u)$ is the sum of the three real parts estimated in steps 1.1, 1.2 and 1.3: $Re a (u, u) \geq θ ∥ D u ∥_{L^{2}}^{2} - \sqrt{n} M_{b} ∥ D u ∥_{L^{2}} ∥ u ∥_{L^{2}} - M_{c} ∥ u ∥_{L^{2}}^{2} .$

3.1F4step 2.1algebra

Absorb the drift term. With $r = ∥ D u ∥_{L^{2}}$ and $s = ∥ u ∥_{L^{2}}$ , [F4] gives $\sqrt{n} M_{b} r s \leq \frac{θ}{2} r^{2} + \frac{n M_{b}^{2}}{2 θ} s^{2}$ , so step 2.1 yields the first displayed inequality $Re a (u, u) \geq \frac{θ}{2} ∥ D u ∥_{L^{2}}^{2} - (\frac{n M_{b}^{2}}{2 θ} + M_{c}) ∥ u ∥_{L^{2}}^{2} .$

4.1F5step 3.1givenalgebra∎

Replace the gradient norm using [F5]: $\frac{θ}{2} ∥ D u ∥_{L^{2}}^{2} = \frac{θ}{2} ∥ u ∥_{H^{1}}^{2} - \frac{θ}{2} ∥ u ∥_{L^{2}}^{2}$ , so the inequality of step 3.1 becomes $Re a (u, u) \geq α ∥ u ∥_{H^{1}}^{2} - β ∥ u ∥_{L^{2}}^{2}$ with $α = θ / 2$ and $β = θ / 2 + n M_{b}^{2} / (2 θ) + M_{c}$ ; both estimates descend to $u \in H_{0}^{1} (Ω)$ because $H_{0}^{1} (Ω) \subseteq H^{1} (Ω)$ and the norms agree, and no Poincare inequality, boundedness of $Ω$ or symmetry of $a$ entered any step.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

A sufficiently large shift is coercive

Statement

Assume Countable Choice. In the setting of Garding's inequality for a divergence-form elliptic operator put $a_{μ} (u, v) : = a (u, v) + μ (u, v)_{L^{2}}$ for $μ \in R$ . If $μ \geq β$ with $β : = θ / 2 + n M_{b}^{2} / (2 θ) + M_{c}$ , then $a_{μ}$ is a bounded sesquilinear form on $H^{1} (Ω)$ satisfying $Re a_{μ} (u, u) \geq \frac{θ}{2} ∥ u ∥_{H^{1} (Ω)}^{2} for every u \in H^{1} (Ω),$ and the same inequality holds for the restriction of $a_{μ}$ to $H_{0}^{1} (Ω)$ , so $a_{μ}$ is coercive with constant $θ / 2$ , independent of $μ$ once $μ \geq β$ (Bounded, coercive and symmetric sesquilinear forms). No boundedness of $Ω$ is used and the shift $μ$ is fixed.

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ ; a uniformly elliptic operator $L$ and form $a$ with constants $θ, M_{a}, M_{b}, M_{c}$ ; a real $μ \geq β = θ / 2 + n M_{b}^{2} / (2 θ) + M_{c}$ ; and the form $a_{μ} = a + μ (\cdot, \cdot)_{L^{2}}$ .

[F1]

Garding's inequality: for every $u \in H^{1} (Ω)$ , $Re a (u, u) \geq \frac{θ}{2} ∥ D u ∥_{L^{2}}^{2} - (\frac{n M_{b}^{2}}{2 θ} + M_{c}) ∥ u ∥_{L^{2}}^{2}$ and $Re a (u, u) \geq \frac{θ}{2} ∥ u ∥_{H^{1}}^{2} - β ∥ u ∥_{L^{2}}^{2}$ (Garding's inequality for a divergence-form elliptic operator).

[F2]

The form $a$ is sesquilinear on $H^{1} (Ω)$ and bounded: $∣ a (u, v) ∣ \leq (n M_{a} + n M_{b} + M_{c}) ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}}$ (The elliptic form is well defined and bounded on $H^{1}$ , Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F3]

The $L^{2}$ pairing $(\cdot, \cdot)_{L^{2}}$ is sesquilinear and $∣ (u, v)_{L^{2}} ∣ \leq ∥ u ∥_{L^{2}} ∥ v ∥_{L^{2}} \leq ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}}$ , since on $H^{1}$ the norm satisfies $∥ w ∥_{H^{1}}^{2} = ∥ w ∥_{L^{2}}^{2} + ∥ D w ∥_{L^{2}}^{2}$ (Integer-order Sobolev spaces and their norms, The notation $H^{k}$ and the reserved zero-boundary symbol, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F4]

$H^{1} (Ω)$ and its closed subspace $H_{0}^{1} (Ω)$ are Hilbert spaces for the Sobolev inner product (The Sobolev space $H^{1}$ is a Hilbert space). Coercivity on a Hilbert space means $Re a (u, u) \geq α ∥ u ∥^{2}$ with a constant $α > 0$ , and restriction of a form to the closed subspace $H_{0}^{1} (Ω) \subseteq H^{1} (Ω)$ preserves sesquilinearity and estimates (Bounded, coercive and symmetric sesquilinear forms, Zero-boundary Sobolev space as a norm closure).

Proof

technique · direct

1.1F2F3givenalgebra

Sesquilinearity and boundedness. The sum of the sesquilinear forms $a$ and $μ (\cdot, \cdot)_{L^{2}}$ is sesquilinear, and [F2] with [F3] gives for all $u, v \in H^{1} (Ω)$ $∣ a_{μ} (u, v) ∣ \leq ∣ a (u, v) ∣ + ∣ μ ∣ ∣ (u, v)_{L^{2}} ∣ \leq (n M_{a} + n M_{b} + M_{c} + ∣ μ ∣) ∥ u ∥_{H^{1}} ∥ v ∥_{H^{1}},$ so $a_{μ}$ is a bounded sesquilinear form on $H^{1} (Ω)$ .

1.2F1givenalgebra

Coercivity. For $u \in H^{1} (Ω)$ , $a_{μ} (u, u) = a (u, u) + μ ∥ u ∥_{L^{2}}^{2}$ has real part $Re a_{μ} (u, u) = Re a (u, u) + μ ∥ u ∥_{L^{2}}^{2} \geq \frac{θ}{2} ∥ u ∥_{H^{1}}^{2} - (β - μ) ∥ u ∥_{L^{2}}^{2} \geq \frac{θ}{2} ∥ u ∥_{H^{1}}^{2},$ by [F1] and $β - μ \leq 0$ . Hence $a_{μ}$ is coercive on $H^{1} (Ω)$ with constant $θ / 2$ .

2.1F4step 1.1step 1.2given∎

Restriction and conclusion. For $u \in H_{0}^{1} (Ω) \subseteq H^{1} (Ω)$ the same computation applies verbatim because the $H^{1}$ norm on the subspace is the restricted norm, so $a_{μ} ∣_{H_{0}^{1}}$ is bounded and satisfies $Re a_{μ} (u, u) \geq \frac{θ}{2} ∥ u ∥_{H^{1}}^{2}$ with the same constant $θ / 2$ ; the constant does not depend on $μ$ once $μ \geq β$ , and no boundedness of $Ω$ or Poincare inequality entered steps 1.1 and 1.2.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The shifted elliptic solution operator

Definition

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open, let $L, a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms with ellipticity constant $θ$ and coefficient bounds $M_{a}, M_{b}, M_{c}$ , and let $μ \geq β : = θ / 2 + n M_{b}^{2} / (2 θ) + M_{c}$ , so that $a_{μ} = a + μ (\cdot, \cdot)_{L^{2}}$ is bounded and coercive on $H_{0}^{1} (Ω)$ with constant $α : = θ / 2$ (A sufficiently large shift is coercive). For $f \in L^{2} (Ω)$ the functional $F_{f} (v) : = (f, v)_{L^{2} (Ω)}$ is conjugate-linear and bounded on $H_{0}^{1} (Ω)$ with $∥ F_{f} ∥ \leq ∥ f ∥_{L^{2} (Ω)}$ : Cauchy--Schwarz gives $∣ F_{f} (v) ∣ \leq ∥ f ∥_{L^{2}} ∥ v ∥_{L^{2}}$ , and the standard $H_{0}^{1}$ norm satisfies $∥ v ∥_{L^{2}} \leq ∥ v ∥_{H_{0}^{1}}$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Integer-order Sobolev spaces and their norms). The shifted elliptic solution operator $K_{μ}$ assigns to $f \in L^{2} (Ω)$ the unique $u \in H_{0}^{1} (Ω)$ with $a_{μ} (u, v) = (f, v)_{L^{2} (Ω)} for every v \in H_{0}^{1} (Ω),$ whose existence and uniqueness are The Lax--Milgram theorem; it is linear in $f$ with $∥ K_{μ} f ∥_{H_{0}^{1}} \leq ∥ f ∥_{L^{2}} / α$ (The Lax--Milgram solution operator has norm at most $1 / α$ ). $K_{μ}$ is first a map $L^{2} (Ω) \to H_{0}^{1} (Ω)$ ; it is regarded on $L^{2} (Ω)$ through the inclusion $H_{0}^{1} (Ω) \subset L^{2} (Ω)$ , and the two maps are distinguished throughout. No boundedness or boundary regularity of $Ω$ is used, and the shift is fixed and never silently changed.

Well-definedness, recorded with the definition. The form $a_{μ}$ is bounded and coercive on the Hilbert space $H_{0}^{1} (Ω)$ with the restricted Sobolev inner product and completeness supplied by The Sobolev space $H^{1}$ is a Hilbert space; the $L^{2}$ integral pairing is a Hilbert inner product by $L^{2}$ with the integral pairing is a Hilbert space: boundedness is the shift corollary, and coercivity holds with constant $α = θ / 2$ independent of $μ$ once $μ \geq β$ . The datum functional $F_{f}$ is conjugate-linear in $v$ in the convention of Bounded, coercive and symmetric sesquilinear forms and bounded by $∥ f ∥_{L^{2}}$ , so The Lax--Milgram theorem applies and produces a unique $u \in H_{0}^{1} (Ω)$ ; for the norm estimate, testing the defining identity at $v = u$ gives $α ∥ u ∥_{H_{0}^{1}}^{2} \leq Re a_{μ} (u, u) = Re (f, u)_{L^{2}} \leq ∥ f ∥_{L^{2}} ∥ u ∥_{H_{0}^{1}}$ . Linearity of $f \mapsto K_{μ} f$ follows from uniqueness, and the same uniqueness makes $K_{μ}$ independent of any choice of representative of $f$ ; the map $K_{μ}$ is defined for the fixed $μ$ and is never applied at any other shift.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The shifted solution operator is compact on $L^{2}$

Statement

Assume the Axiom of Choice, inherited through the compact-embedding supplier named below, together with Countable Choice. Let $Ω \subseteq R^{n}$ be open and bounded, and let $K_{μ}$ be the shifted solution operator of The shifted elliptic solution operator for a fixed $μ \geq β$ . Then $K_{μ}$ , regarded as an operator on $L^{2} (Ω)$ , is compact: $∥ K_{μ} f ∥_{H_{0}^{1}} \leq α^{- 1} ∥ f ∥_{L^{2}}$ (The Lax--Milgram solution operator has norm at most $1 / α$ ) and $H_{0}^{1} (Ω) ↪ L^{2} (Ω)$ is compact (Compactness of $W_{0}^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded open sets at $p = 2$ ), so $K_{μ}$ maps bounded subsets of $L^{2} (Ω)$ to relatively compact subsets of $L^{2} (Ω)$ . No regularity of $\partial Ω$ is assumed.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a bounded open set $Ω \subseteq R^{n}$ ; a fixed $μ \geq β$ ; the shifted solution operator $K_{μ} : L^{2} (Ω) \to H_{0}^{1} (Ω)$ with coercivity constant $α = θ / 2$ .

[F1]

$K_{μ}$ is well defined and linear, and for every $f \in L^{2} (Ω)$ one has $∥ K_{μ} f ∥_{H_{0}^{1}} \leq α^{- 1} ∥ f ∥_{L^{2}}$ (The shifted elliptic solution operator, The Lax--Milgram solution operator has norm at most $1 / α$ ).

[F2]

$H_{0}^{1} (Ω) = W_{0}^{1, 2} (Ω)$ , and for bounded open $Ω$ the inclusion $W_{0}^{1, 2} (Ω) ↪ L^{2} (Ω)$ is compact: every sequence bounded in $H_{0}^{1} (Ω)$ has a subsequence converging in $L^{2} (Ω)$ (Compactness of $W_{0}^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded open sets, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

Compositions: if $T : X \to Y$ is compact and $B : Z \to X$ is bounded linear, then $T \circ B : Z \to Y$ is compact (Compositions with a compact operator are compact, Compact linear operator).

Proof

technique · direct

1.1F1given

The map $K_{μ} : L^{2} (Ω) \to H_{0}^{1} (Ω)$ is linear and bounded with operator bound $∥ K_{μ} ∥ \leq α^{- 1}$ by [F1], so it maps bounded subsets of $L^{2} (Ω)$ to bounded subsets of $H_{0}^{1} (Ω)$ .

1.2F2given

The inclusion $ι : H_{0}^{1} (Ω) \to L^{2} (Ω)$ is compact by [F2], because $Ω$ is bounded and open and $p = 2$ is admissible.

2.1F3step 1.1step 1.2given∎

The $L^{2}$ realization of $K_{μ}$ is the composite $ι \circ K_{μ}$ . By step 1.1 the first factor is bounded linear and by step 1.2 the second is compact, so [F3] makes the composite compact; hence bounded subsets of $L^{2} (Ω)$ are mapped into relatively compact subsets of $L^{2} (Ω)$ . No boundary regularity was used, and the Axiom of Choice is inherited through the Rellich supplier [F2].

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

On bounded domains, the unshifted equation is an identity-minus-compact equation

Statement

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open, let $μ \geq β$ and let $K_{μ}$ be the shifted solution operator of The shifted elliptic solution operator for the form $a$ of Uniformly elliptic divergence-form operators and their sesquilinear forms. For $f \in L^{2} (Ω)$ and $u \in H_{0}^{1} (Ω)$ the following are equivalent:

$a (u, v) = (f, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , i.e. $u$ is a weak Dirichlet solution in the sense of Weak Dirichlet solutions for a divergence-form operator;
$(I - μ K_{μ}) u = K_{μ} f$ as elements of $L^{2} (Ω)$ , where $I$ is the identity of $L^{2} (Ω)$ .

Both sides of (2) lie in $H_{0}^{1} (Ω)$ . Thus for arbitrary open $Ω$ the weak equation is equivalent to this identity-minus-bounded-operator equation. If $Ω$ is bounded, also assume the Axiom of Choice; then $K_{μ}$ on $L^{2} (Ω)$ , and hence $μ K_{μ}$ , is compact by The shifted solution operator is compact on $L^{2}$ , so (2) is an identity-minus-compact Fredholm equation. The algebraic equivalence applies to the general divergence-form operator, including nonsymmetric lower-order terms.

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ ; a fixed $μ \geq β$ ; the shifted solution operator $K_{μ}$ of the general divergence-form operator; $f \in L^{2} (Ω)$ and $u \in H_{0}^{1} (Ω)$ .

[F1]

Definition of $K_{μ}$ : for $g \in L^{2} (Ω)$ , $K_{μ} g$ is the unique class in $H_{0}^{1} (Ω)$ with $a_{μ} (K_{μ} g, v) = (g, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ , where $a_{μ} = a + μ (\cdot, \cdot)_{L^{2}}$ ; $K_{μ}$ is linear and maps $L^{2} (Ω)$ into $H_{0}^{1} (Ω)$ (The shifted elliptic solution operator, Zero-boundary Sobolev space as a norm closure).

[F2]

Weak Dirichlet solutions: $u$ is a weak solution of $a (u, v) = (f, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ exactly when the identity holds for all test classes $v \in H_{0}^{1} (Ω)$ with datum $f \in L^{2} (Ω)$ (Weak Dirichlet solutions for a divergence-form operator, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F3]

Compactness: if $Ω$ is bounded and the Axiom of Choice holds, the $L^{2}$ realization of $K_{μ}$ is compact, hence so is $μ K_{μ}$ , and $I - μ K_{μ}$ is an identity-minus-compact operator on $L^{2} (Ω)$ (The shifted solution operator is compact on $L^{2}$ , The space $L^{p} (μ)$ as the quotient by null functions, The Axiom of Choice).

Proof

technique · direct

1.1F1F2givenalgebra

Equivalence. Condition (1) says $a (u, v) = (f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ . Adding $μ (u, v)_{L^{2}}$ to both sides, this is equivalent to $a_{μ} (u, v) = (f + μ u, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ , where $f + μ u \in L^{2} (Ω)$ . By the defining uniqueness clause of [F1] for the datum $f + μ u$ , this holds exactly when $u = K_{μ} (f + μ u)$ in $H_{0}^{1} (Ω)$ . Rearranging the linear identity gives $u - μ K_{μ} u = K_{μ} f$ , that is $(I - μ K_{μ}) u = K_{μ} f$ in $L^{2} (Ω)$ , and conversely the same rearrangement recovers the defining identity for $f + μ u$ and hence condition (1). No symmetry of $a$ and no sign condition on the lower-order coefficients is used.

1.2F1givenalgebra

Location of the two sides. Since $u \in H_{0}^{1} (Ω)$ by hypothesis and $K_{μ}$ maps $L^{2} (Ω)$ into $H_{0}^{1} (Ω)$ , both $u$ and $μ K_{μ} u$ , hence both sides of (2), lie in $H_{0}^{1} (Ω)$ ( $μ K_{μ} u \in H_{0}^{1}$ because $H_{0}^{1}$ is a linear subspace); the equality itself is an equality of $L^{2}$ classes.

2.1F3step 1.1step 1.2given∎

Compact case. If $Ω$ is bounded and the Axiom of Choice is assumed, [F3] makes the $L^{2}$ realization of $μ K_{μ}$ compact, so (2) is the equation $(I - C) u = K_{μ} f$ with $C : = μ K_{μ}$ compact, an identity-minus-compact equation; for unbounded $Ω$ the equivalence of step 1.1 remains valid as an identity-minus-bounded-operator equation and no compactness or Fredholm claim is made.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

The formal adjoint and the adjoint weak Dirichlet problem

Definition

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open and let $L, a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms. The adjoint form is $a^{*} (u, v) : = \overline{a (v, u)},$ a bounded sesquilinear form on $H^{1} (Ω)$ with the same bound as $a$ (Bounded, coercive and symmetric sesquilinear forms, The elliptic form is well defined and bounded on $H^{1}$ ); in coefficients $a^{*} (u, v) = \int_{Ω} (\overline{a^{j i}} D_{j} u \overline{D_{i} v} + \overline{b^{i}} u \overline{D_{i} v} + \overline{c} u \overline{v}) d x .$ The formal adjoint is the expression $L^{*} u : = - D_{i} (\overline{a^{j i}} D_{j} u) - D_{i} (\overline{b^{i}} u) + \overline{c} u$ , understood as a distribution: for $w, v \in C_{c}^{\infty} (Ω)$ , $⟨ L^{*} w, \overline{v} ⟩ = a^{*} (w, v)$ . Indeed the coefficient products are locally integrable, hence define regular distributions by Locally integrable functions embed in distributions, and the signed derivative rule of Distributional derivative gives exactly the displayed form. Even for smooth $w$ , $L^{*} w$ need not be a locally integrable function when the coefficients are merely measurable; an integral $\int (L^{*} w) \overline{v}$ is used only when it is represented by such a function; the form $a^{*}$ is the primary object and is defined before any integration by parts. The adjoint weak Dirichlet problem with datum $f \in L^{2} (Ω)$ asks for $v \in H_{0}^{1} (Ω)$ with $a^{*} (v, w) = (f, w)_{L^{2}} for every w \in H_{0}^{1} (Ω),$ and its homogeneous version is $a^{*} (v, w) = 0$ for all $w$ . No orthogonality is invoked in this definition. Since $Re a^{*} (u, u) = Re a (u, u)$ , the Garding constants of Garding's inequality for a divergence-form elliptic operator also apply to $a^{*}$ , and $a_{μ}^{*} : = a^{*} + μ (\cdot, \cdot)_{L^{2}}$ is coercive for $μ \geq β$ (A sufficiently large shift is coercive).

Conventions recorded with the definition. All pairings are the $L^{2}$ or $H_{0}^{1}$ pairings of the cited items, with conjugation in the second slot; the datum $f \in L^{2} (Ω)$ acts through the conjugate-linear functional $w \mapsto (f, w)_{L^{2}}$ , which is an element of $H^{- 1} (Ω)$ by the Cauchy--Schwarz estimate $∥ w ∥_{L^{2}} \leq ∥ w ∥_{H_{0}^{1}}$ (The negative Sobolev space $H^{- 1} (Ω)$ , Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure, Real and imaginary parts, complex conjugation, and modulus). The formal expression $L^{*}$ is recorded as the operator whose weak pairing reproduces $a^{*}$ on smooth compactly supported functions; $L^{*}$ is not claimed to be of the same divergence form as $L$ , and no boundary condition is attached to it beyond the test class $H_{0}^{1} (Ω)$ . The adjoint weak problem is stated for $H_{0}^{1}$ test functions exactly as in Weak Dirichlet solutions for a divergence-form operator, and no existence or uniqueness is asserted here.

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The adjoint solution operator solves the adjoint form problem

Statement

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open, $μ \geq β$ , and let $a^{*}$ be the adjoint form of The formal adjoint and the adjoint weak Dirichlet problem. Define $K_{μ}^{*} f$ , for $f \in L^{2} (Ω)$ , to be the unique $v \in H_{0}^{1} (Ω)$ with $a_{μ}^{*} (v, w) = (f, w)_{L^{2}}$ for every $w \in H_{0}^{1} (Ω)$ (existence and uniqueness by The Lax--Milgram theorem). Then $K_{μ}^{*}$ is well defined and linear on $L^{2} (Ω)$ , and it is the Hilbert-space adjoint of the shifted solution operator $K_{μ}$ of The shifted elliptic solution operator: $(K_{μ} f, g)_{L^{2}} = (f, K_{μ}^{*} g)_{L^{2}} for all f, g \in L^{2} (Ω) .$ If $Ω$ is bounded and the Axiom of Choice is also assumed, then $K_{μ}^{*}$ is compact on $L^{2} (Ω)$ (The shifted solution operator is compact on $L^{2}$ is proved by the same bounded-map/Rellich composition for $a^{*}$ ). Moreover, for $v \in L^{2} (Ω)$ one has $v \in \ker (I - μ K_{μ}^{*})$ if and only if $v \in H_{0}^{1} (Ω)$ and $a^{*} (v, w) = 0$ for every $w \in H_{0}^{1} (Ω)$ .

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ ; the divergence-form operator $L$ , its form $a$ , the adjoint form $a^{*}$ , a fixed $μ \geq β$ , and the operators $K_{μ}, K_{μ}^{*}$ on $L^{2} (Ω)$ .

[F1]

The adjoint form $a^{*}$ and its shift: $a^{*} (v, w) = \overline{a (w, v)}$ , $a_{μ}^{*} = a^{*} + μ (\cdot, \cdot)_{L^{2}}$ , and $Re a_{μ}^{*} (u, u) = Re a_{μ} (u, u) \geq \frac{θ}{2} ∥ u ∥_{H_{0}^{1}}^{2}$ , so $a_{μ}^{*}$ is bounded and coercive on $H_{0}^{1} (Ω)$ with the same constants as $a_{μ}$ ; the datum $w \mapsto (f, w)_{L^{2}}$ is a bounded conjugate-linear functional on $H_{0}^{1} (Ω)$ (The formal adjoint and the adjoint weak Dirichlet problem, The shifted elliptic solution operator, The space $L^{p} (μ)$ as the quotient by null functions, Zero-boundary Sobolev space as a norm closure).

[F2]

Lax--Milgram: a bounded coercive sesquilinear form on a Hilbert space and a bounded conjugate-linear functional have a unique solution, and the solution map is linear with norm at most $1 / α$ (The Lax--Milgram theorem, A bounded linear operator between normed spaces, Integer-order Sobolev spaces and their norms).

[F3]

Hilbert-space adjoints: $S^{*}$ is the operator with $(S f, g) = (f, S^{*} g)$ for all $f, g$ , and it is unique (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F4]

The $L^{2}$ realization of $K_{μ}$ is compact when $Ω$ is bounded: a bounded linear map $L^{2} (Ω) \to H_{0}^{1} (Ω)$ followed by the compact Rellich inclusion $H_{0}^{1} (Ω) ↪ L^{2} (Ω)$ is compact (The shifted solution operator is compact on $L^{2}$ , Compositions with a compact operator are compact, Compactness of $W_{0}^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded open sets, Compact linear operator, The Axiom of Choice).

Proof

technique · direct

1.1F1F2given

Well-definedness and linearity. By [F1] and [F2] applied to $a_{μ}^{*}$ , for each $f \in L^{2} (Ω)$ there is a unique $v \in H_{0}^{1} (Ω)$ with $a_{μ}^{*} (v, w) = (f, w)_{L^{2}}$ for all $w \in H_{0}^{1} (Ω)$ ; uniqueness makes $K_{μ}^{*}$ independent of any choice, and linearity of $f \mapsto K_{μ}^{*} f$ follows from uniqueness exactly as for $K_{μ}$ , since $a_{μ}^{*}$ and the datum functional are linear in that slot.

2.1F1F3step 1.1givenalgebra

Adjoint identity. For $f, g \in L^{2} (Ω)$ put $v : = K_{μ}^{*} g$ , so that $a_{μ}^{*} (v, w) = (g, w)_{L^{2}}$ for every $w \in H_{0}^{1} (Ω)$ . Testing at $w = K_{μ} f$ and conjugating, and using $a_{μ}^{*} (v, u) = \overline{a_{μ} (u, v)}$ , $(K_{μ} f, g)_{L^{2}} = \overline{(g, K_{μ} f)_{L^{2}}} = \overline{a_{μ}^{*} (v, K_{μ} f)} = a_{μ} (K_{μ} f, v) = (f, v)_{L^{2}} = (f, K_{μ}^{*} g)_{L^{2}},$ where the penultimate identity is the defining equation of $K_{μ}$ . Hence $K_{μ}^{*}$ is the Hilbert-space adjoint of $K_{μ}$ by [F3].

2.2F4step 1.1given

Compactness on bounded $Ω$ . If $Ω$ is bounded, [F1] and [F2] make $K_{μ}^{*} : L^{2} (Ω) \to H_{0}^{1} (Ω)$ bounded linear, and composing with the compact Rellich inclusion $H_{0}^{1} (Ω) ↪ L^{2} (Ω)$ expresses $K_{μ}^{*}$ on $L^{2} (Ω)$ as a bounded map followed by a compact one, hence compact by [F4]; the Axiom of Choice is inherited through the Rellich supplier.

3.1F1F2step 1.1givenalgebra∎

Kernel at $μ$ . For $v \in L^{2} (Ω)$ one has $v \in \ker (I - μ K_{μ}^{*})$ if and only if $v = μ K_{μ}^{*} v$ , and since $ran K_{μ}^{*} \subseteq H_{0}^{1} (Ω)$ this forces $v \in H_{0}^{1} (Ω)$ and, by the defining equation of $K_{μ}^{*}$ with datum $μ v$ , $a_{μ}^{*} (v, w) = μ (v, w)_{L^{2}} for every w \in H_{0}^{1} (Ω),$ which is exactly $a^{*} (v, w) = 0$ for every $w$ . Conversely, if $v \in H_{0}^{1} (Ω)$ satisfies $a^{*} (v, w) = 0$ for all $w$ , then $a_{μ}^{*} (v, w) = μ (v, w)_{L^{2}}$ for all $w$ , so uniqueness in [F2] gives $K_{μ}^{*} (μ v) = v$ , that is $μ K_{μ}^{*} v = v$ .

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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.

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The Fredholm alternative for weak elliptic Dirichlet problems

Statement

Assume the Axiom of Choice and Countable Choice. Let $Ω \subseteq R^{n}$ be bounded open, $K \in {R, C}$ , and let $L, a$ be as in Uniformly elliptic divergence-form operators and their sesquilinear forms with ellipticity constant $θ$ and coefficient bounds $M_{a}, M_{b}, M_{c}$ . Let $a^{*}$ be the adjoint form of The formal adjoint and the adjoint weak Dirichlet problem. Consider the weak Dirichlet problem $a (u, v) = (f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ , with datum $f \in L^{2} (Ω)$ (Weak Dirichlet solutions for a divergence-form operator). Then exactly one of the following alternatives holds. (1) The homogeneous problem $a (u, v) = 0$ for all $v \in H_{0}^{1} (Ω)$ has only the solution $u = 0$ . Then for every $f \in L^{2} (Ω)$ the problem has exactly one weak solution $u \in H_{0}^{1} (Ω)$ . (2) The homogeneous problem has a nonzero solution. Then both homogeneous solution spaces $N : = {u \in H_{0}^{1} (Ω) : a (u, v) = 0 \forall v}, N^{*} : = {v \in H_{0}^{1} (Ω) : a^{*} (v, w) = 0 \forall w}$ are finite-dimensional and nontrivial with $\dim N = \dim N^{*}$ ; for $f \in L^{2} (Ω)$ the problem has a solution if and only if $(f, v)_{L^{2}} = 0$ for every $v \in N^{*}$ ; and whenever a solution exists the solution set is an affine translate of $N$ , so uniqueness fails. The data class is $L^{2} (Ω)$ ; the weaker class $H^{- 1} (Ω)$ is deliberately not treated here.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a bounded open set $Ω \subseteq R^{n}$ ; the divergence-form operator $L$ and form $a$ with constants $θ, M_{a}, M_{b}, M_{c}$ ; the adjoint form $a^{*}$ ; a fixed $μ \geq β$ ; the shifted solution operator $K_{μ}$ and its adjoint $K_{μ}^{*}$ ; and $f \in L^{2} (Ω)$ .

[F1]

Operator reduction: for $u \in H_{0}^{1} (Ω)$ the weak equation $a (u, v) = (f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ is equivalent to $(I - μ K_{μ}) u = K_{μ} f$ in $L^{2} (Ω)$ , and both sides of that equation lie in $H_{0}^{1} (Ω)$ (On bounded domains, the unshifted equation is an identity-minus-compact equation).

[F2]

Compactness and the abstract alternative: $A : = I - μ K_{μ}$ is an identity-minus-compact operator on the Banach space $L^{2} (Ω)$ , the homogeneous spaces satisfy $N = \ker A$ and, under the Riesz identification, $N^{*} = \ker A^{*}$ , and $K_{μ} f \in ran A$ if and only if $(f, v)_{L^{2}} = 0$ for every $v \in N^{*}$ (The adjoint solution operator solves the adjoint form problem, The elliptic Fredholm range condition is orthogonality to the adjoint kernel, Fredholm alternative for identity minus compact, Kernel of identity minus compact is finite dimensional, Compact linear operator, The Axiom of Choice).

[F3]

Abstract Fredholm alternatives: for a compact $K$ on a Banach space and $A = I - K$ , either $A$ is injective, in which case it is bijective with bounded inverse, or $\ker A$ and the cokernel are finite dimensional and nontrivial with equal dimensions (Fredholm alternative for identity minus compact). The range of $A = I - μ K_{μ}$ is closed by Range of identity minus compact is closed; AC supplies its DC premise by AC supplies the countable and dependent choices used in Banach integration. By Orthogonal decomposition by a closed subspace, $L^{2} = ran A \oplus (ran A)^{⊥}$ . The adjoint identity gives $(ran A)^{⊥} = \ker (I - μ K_{μ}^{*})$ , and $v \mapsto v + ran A$ restricts to a linear bijection from this orthogonal kernel onto the cokernel.

[F4]

Data conventions: weak solutions are $H_{0}^{1}$ classes, the datum $f \in L^{2} (Ω)$ acts through the conjugate-linear pairing $v \mapsto (f, v)_{L^{2}}$ , and the adjoint form is $a^{*} (v, w) = \overline{a (w, v)}$ (Weak Dirichlet solutions for a divergence-form operator, The formal adjoint and the adjoint weak Dirichlet problem, The space $L^{p} (μ)$ as the quotient by null functions, Zero-boundary Sobolev space as a norm closure, The shifted elliptic solution operator, Uniformly elliptic divergence-form operators and their sesquilinear forms, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1F1F2F4given

Identification of the spaces. By [F1] with $f = 0$ , a class $u \in H_{0}^{1} (Ω)$ solves the homogeneous problem $a (u, v) = 0$ for all $v$ if and only if $(I - μ K_{μ}) u = 0$ ; since $ran K_{μ} \subseteq H_{0}^{1} (Ω)$ , this identifies $N$ with $\ker A$ , where $A = I - μ K_{μ}$ is bounded on $L^{2} (Ω)$ . By [F2] the adjoint homogeneous space $N^{*}$ is, under the Riesz identification of $L^{2} (Ω)$ with its dual, exactly the kernel of the transpose $A^{*}$ , and for $f \in L^{2} (Ω)$ the solvability of the weak problem is equivalent to $K_{μ} f \in ran A$ , hence to $(f, v)_{L^{2}} = 0$ for every $v \in N^{*}$ .

2.1F2F3step 1.1given

The two alternatives. Since $μ K_{μ}$ is compact, [F3] gives exactly the following dichotomy for $A = I - μ K_{μ}$ : either $A$ is injective, hence bijective with bounded inverse, or $\ker A \neq {0}$ and both $\ker A$ and the cokernel are finite dimensional with equal dimensions. In the first case $\ker A = N = {0}$ by step 1.1.

3.1F1step 2.1given

Case (1). Assume $N = {0}$ . Then $A$ is injective, so by step 2.1 it is bijective and boundedly invertible; for every $f \in L^{2} (Ω)$ the equation $A u = K_{μ} f$ has the unique solution $u = A^{- 1} K_{μ} f \in L^{2} (Ω)$ ; the equation gives $u = K_{μ} f + μ K_{μ} u \in H_{0}^{1} (Ω)$ , and then [F1] shows that it solves the weak problem, and by the equivalence [F1] any weak solution gives a solution of $A u = K_{μ} f$ , so the weak solution is unique. This proves alternative (1).

3.2F1F2step 1.1step 2.1givenalgebra

Case (2). Assume $N \neq {0}$ . Then $\ker A = N$ is nontrivial and finite dimensional, and by step 2.1 its dimension equals that of the cokernel, which under the Riesz identification is $\dim \ker A^{*} = \dim N^{*}$ ; so $N$ and $N^{*}$ are finite-dimensional and nontrivial with $\dim N = \dim N^{*}$ . By step 1.1 the weak problem is solvable exactly when $(f, v)_{L^{2}} = 0$ for every $v \in N^{*}$ . If $u_{0}$ is one solution, then for any $u$ the class $u - u_{0}$ satisfies the homogeneous problem, i.e. lies in $N$ , and conversely $u_{0} + n$ with $n \in N$ is a solution; hence the solution set is the affine translate $u_{0} + N$ , which is not a singleton because $N \neq {0}$ , so uniqueness fails. This proves alternative (2).

4.1F2F3step 3.1step 3.2given∎

Exhaustiveness. Steps 3.1 and 3.2 cover the two mutually exclusive possibilities of step 2.1, so exactly one of the alternatives holds; the datum class is $L^{2} (Ω)$ throughout, no $H^{- 1} (Ω)$ data are used, and the Axiom of Choice is inherited only through the compactness of $K_{μ}$ and the abstract Fredholm alternative.

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The elliptic kernel and cokernel are finite dimensional

Statement

Assume the Axiom of Choice and Countable Choice. In the setting of The Fredholm alternative for weak elliptic Dirichlet problems the weak homogeneous space $N = {u \in H_{0}^{1} (Ω) : a (u, v) = 0 \forall v}$ and the weak adjoint space $N^{*} = {v \in H_{0}^{1} (Ω) : a^{*} (v, w) = 0 \forall w}$ are finite-dimensional over $K$ with $\dim N = \dim N^{*}$ . Explicitly $N = \ker (I - μ K_{μ})$ and $N^{*} = \ker (I - μ K_{μ}^{*})$ as subspaces of $L^{2} (Ω)$ (On bounded domains, the unshifted equation is an identity-minus-compact equation, The adjoint solution operator solves the adjoint form problem), so this common dimension is the dimension of $E_{1} (T) = \ker (T - I)$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism), for each compact operator $T = μ K_{μ}$ and $T = μ K_{μ}^{*}$ . When the common dimension is positive, $1$ is an eigenvalue and this dimension is its geometric multiplicity; when it is zero, $1$ is not an eigenvalue. No equality of algebraic multiplicities is asserted. The common geometric multiplicity is independent of the admissible shift $μ \geq β$ .

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a bounded open set $Ω \subseteq R^{n}$ ; the divergence-form operator with form $a$ and adjoint $a^{*}$ ; a fixed $μ \geq β$ ; the shifted solution operators $K_{μ}, K_{μ}^{*}$ .

[F1]

Identifications: $N = {u \in H_{0}^{1} (Ω) : (I - μ K_{μ}) u = 0}$ and $N^{*} = {v \in H_{0}^{1} (Ω) : (I - μ K_{μ}^{*}) v = 0}$ ; moreover $\ker (I - μ K_{μ}) \subseteq H_{0}^{1} (Ω)$ and $\ker (I - μ K_{μ}^{*}) \subseteq H_{0}^{1} (Ω)$ , so the two sets are exactly the indicated kernels in $L^{2} (Ω)$ (On bounded domains, the unshifted equation is an identity-minus-compact equation, The adjoint solution operator solves the adjoint form problem, The shifted elliptic solution operator, Zero-boundary Sobolev space as a norm closure).

[F2]

Abstract Fredholm dimension: for a compact operator $μ K_{μ}$ on $L^{2} (Ω)$ and $A = I - μ K_{μ}$ , the kernel and the cokernel of $A$ are finite dimensional with equal dimensions; the kernel of the transpose $A^{*}$ is finite dimensional (Fredholm alternative for identity minus compact, Kernel of identity minus compact is finite dimensional, Compact linear operator, The Axiom of Choice). The range of $A = I - μ K_{μ}$ is closed by Range of identity minus compact is closed; AC supplies its DC premise by AC supplies the countable and dependent choices used in Banach integration. By Orthogonal decomposition by a closed subspace, $L^{2} = ran A \oplus (ran A)^{⊥}$ . The adjoint identity gives $(ran A)^{⊥} = \ker (I - μ K_{μ}^{*})$ , and $v \mapsto v + ran A$ restricts to a linear bijection from this orthogonal kernel onto the cokernel.

[F3]

The weak spaces $N, N^{*}$ are the homogeneous and adjoint homogeneous solution spaces of the weak elliptic problem, and the abstract identities of [F1] hold for every admissible shift $μ \geq β$ (The Fredholm alternative for weak elliptic Dirichlet problems).

[F4]

For a linear endomorphism $T$ , its eigenspace for eigenvalue $1$ is $E_{1} (T) = \ker (T - I)$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism). Its dimension is the geometric multiplicity used here; algebraic multiplicity is not part of this conclusion.

Proof

technique · direct

1.1F1given

Identifications. By [F1], $N = \ker (I - μ K_{μ})$ and $N^{*} = \ker (I - μ K_{μ}^{*})$ as subspaces of $L^{2} (Ω)$ : an element of either kernel lies in $H_{0}^{1} (Ω)$ because the range of the corresponding solution operator is contained in $H_{0}^{1} (Ω)$ .

2.1F2step 1.1given

Finite dimension and equality. Since $μ K_{μ}$ is compact on the Banach space $L^{2} (Ω)$ , [F2] gives $\dim \ker (I - μ K_{μ}) < \infty$ and $\dim \ker (I - μ K_{μ}) = \dim coker (I - μ K_{μ})$ . Under the Riesz identification of $L^{2} (Ω)$ with its dual, the annihilator of the range is $\ker (I - μ K_{μ}^{*})$ , so the cokernel dimension equals $\dim \ker (I - μ K_{μ}^{*})$ ; by step 1.1 these are $\dim N$ and $\dim N^{*}$ , both finite, with $\dim N = \dim N^{*}$ .

3.1F1F3F4step 2.1givenalgebra∎

Independence of the shift and geometric multiplicity. For a further admissible shift $μ^{'} \geq β$ the same argument with $μ^{'}$ gives $N = \ker (I - μ^{'} K_{μ^{'}})$ and $\dim N = \dim N^{*}$ ; the spaces $N, N^{*}$ themselves are defined by the weak equations alone and do not mention any shift, so the common dimension is independent of the choice of admissible $μ$ . By [F4], when positive, these dimensions are the geometric multiplicities of eigenvalue $1$ of $μ K_{μ}$ and $μ K_{μ}^{*}$ , respectively, because $N$ and $N^{*}$ are exactly their eigenspaces. This identifies no algebraic multiplicity.

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Uniqueness implies existence for the elliptic Dirichlet problem

Statement

Assume the Axiom of Choice and Countable Choice. In the setting of The Fredholm alternative for weak elliptic Dirichlet problems suppose both homogeneous problems are trivial: $a (u, v) = 0$ for all $v \in H_{0}^{1} (Ω)$ implies $u = 0$ , and $a^{*} (v, w) = 0$ for all $w \in H_{0}^{1} (Ω)$ implies $v = 0$ (the two conditions are equivalent by the finite dimension and equality of dimensions in The elliptic kernel and cokernel are finite dimensional). Then for every $f \in L^{2} (Ω)$ there is exactly one $u \in H_{0}^{1} (Ω)$ with $a (u, v) = (f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ . Moreover the solution map $f \mapsto u$ is a bounded linear operator from $L^{2} (Ω)$ to $H_{0}^{1} (Ω)$ : explicitly $u = K_{μ} (I - μ K_{μ})^{- 1} f$ for every admissible $μ \geq β$ , with $(I - μ K_{μ})^{- 1}$ bounded on $L^{2} (Ω)$ (The shifted elliptic solution operator, Neumann series and small perturbations of bounded inverses).

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a bounded open set $Ω \subseteq R^{n}$ ; the weak Dirichlet problem for form $a$ and adjoint $a^{*}$ ; a fixed $μ \geq β$ ; the compact operator $μ K_{μ}$ on $L^{2} (Ω)$ ; and the assumption that both homogeneous problems are trivial.

[F1]

Fredholm alternative: exactly one of the alternatives of The Fredholm alternative for weak elliptic Dirichlet problems holds; alternative (2) holds exactly when the homogeneous problem has a nonzero solution; alternative (1) gives existence and uniqueness for every $f \in L^{2} (Ω)$ .

[F2]

The homogeneous space $N$ of [ $F 1$ ] equals $\ker (I - μ K_{μ})$ , and $u$ is a weak solution with datum $f$ exactly when $(I - μ K_{μ}) u = K_{μ} f$ (On bounded domains, the unshifted equation is an identity-minus-compact equation, The elliptic kernel and cokernel are finite dimensional, The shifted elliptic solution operator).

[F3]

Abstract Fredholm alternative: for a compact $K$ on a Banach space, $I - K$ is injective if and only if it is surjective, and then it is boundedly invertible (Fredholm alternative for identity minus compact, Bounded inverse theorem, Neumann series and small perturbations of bounded inverses).

[F4]

Bounded linear operators compose to bounded linear operators, and $K_{μ} : L^{2} (Ω) \to H_{0}^{1} (Ω)$ is bounded linear (A bounded linear operator between normed spaces, The shifted elliptic solution operator).

Proof

technique · direct

1.1F1F2given

Existence and uniqueness. If the homogeneous problem is trivial then $N = {0} = \ker (I - μ K_{μ})$ by [F2], so alternative (2) of [F1] is excluded; by the dichotomy of [F1] alternative (1) holds. Hence for every $f \in L^{2} (Ω)$ the weak problem has exactly one solution $u \in H_{0}^{1} (Ω)$ . The triviality of the adjoint homogeneous problem is not needed for this conclusion, and the two triviality hypotheses are equivalent by The elliptic kernel and cokernel are finite dimensional.

2.1F2F3F4step 1.1givenalgebra

Bounded solution map. Under the hypothesis $N = {0}$ the operator $A : = I - μ K_{μ}$ is injective on $L^{2} (Ω)$ , so [F3] makes it boundedly invertible there; since $u$ is a weak solution with datum $f$ exactly when $A u = K_{μ} f$ by [F2], the unique solution is $u = A^{- 1} K_{μ} f$ . The operator $K_{μ}$ commutes with $A = I - μ K_{μ}$ , hence also with $A^{- 1}$ , so $u = K_{μ} A^{- 1} f$ . Both factors in this expression are bounded linear operators, with $K_{μ}$ mapping into $H_{0}^{1} (Ω)$ by [F4], so $f \mapsto u$ is a bounded linear operator from $L^{2} (Ω)$ to $H_{0}^{1} (Ω)$ ; the expression is independent of the admissible shift because $u$ is.

3.1F1F3step 1.1step 2.1given∎

Conclusion. Steps 1.1 and 2.1 give existence, uniqueness and the bounded solution map for every $f \in L^{2} (Ω)$ , with the explicit representation $u = K_{μ} (I - μ K_{μ})^{- 1} f$ ; no compactness is used beyond the Fredholm alternative inherited from $μ K_{μ}$ , and the Axiom of Choice supplies the hypotheses of the Rellich compactness and abstract Fredholm suppliers.

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Smooth compactly supported functions of an open set are dense in $L^{2}$

Statement

Assume Countable Choice. Let $Ω \subseteq R^{n}$ be open with $n \geq 1$ and $K \in {R, C}$ . Then $C_{c}^{\infty} (Ω; K)$ is dense in $L^{2} (Ω; K)$ : for every $f \in L^{2} (Ω; K)$ and every $δ > 0$ there is $φ \in C_{c}^{\infty} (Ω; K)$ with $∥ φ - f ∥_{L^{2} (Ω)} < δ$ . Consequently $H_{0}^{1} (Ω)$ is dense in $L^{2} (Ω)$ , and if a class $h \in L^{2} (Ω)$ satisfies $(h, v)_{L^{2}} = 0$ for all $v \in H_{0}^{1} (Ω)$ , then $h = 0$ .

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ with $n \geq 1$ ; $K \in {R, C}$ ; a class $f \in L^{2} (Ω; K)$ ; and a tolerance $δ > 0$ .

[F1]

$L^{2}$ classes and zero extension: $L^{2} (Ω; K)$ is a space of almost-everywhere classes with norm $∥ u ∥_{L^{2} (Ω)} = (\int_{Ω} ∣ u ∣^{2})^{1 / 2}$ and pairing $(u, v)_{L^{2}} = \int_{Ω} u \overline{v}$ ; the zero extension $F = 1_{Ω} f$ of $f$ , equal to $f$ on $Ω$ and $0$ off $Ω$ , is a well-defined class in $L^{2} (R^{n}; K)$ with $∥ F ∥_{L^{2} (R^{n})} = ∥ f ∥_{L^{2} (Ω)}$ (The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions, Integral over a measurable subset).

[F2]

A continuous real function on a nonempty compact metric space attains its minimum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value). Exhaustion tools: a closed and bounded subset of $R^{n}$ is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), and for nonempty $A$ the function $x \mapsto dist (x, A)$ is $1$ -Lipschitz, hence continuous ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz).

[F3]

Mollifier existence: A smooth bump between concentric Euclidean balls gives a smooth $0 \leq q \leq 1$ equal to $1$ on ${\overline{B}}_{1 / 4} (0)$ and supported in $B_{1 / 2} (0)$ . Its support has finite measure by Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, and its inner ball contains a positive-volume box by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, so $0 < \int q < \infty$ . Thus $ρ = q / \int q$ is a real unit-mass smooth bump supported in $B_{1} (0)$ ; radiality is unnecessary.

[F4]

Monotone and dominated convergence: a nondecreasing sequence of nonnegative measurable functions has integral limit equal to the integral of its pointwise limit, and a sequence dominated by one integrable function has integrals converging to the integral of its pointwise limit (Monotone convergence for the integral, Dominated convergence).

[F5]

Global smoothing: if $G \in L^{2} (R^{n})$ , Hölder on each finite-measure compact set makes $G$ locally integrable (Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure). Its convolution with the real unit-mass bump of [F3] is smooth on all of $R^{n}$ by Convolution with a mollifier is smooth, and derivatives pass under the integral sign. The rescaled family is that of The mollifier family generated by a unit-mass smooth bump.

[F6]

Approximate identity convergence: a mollifier family is an $L^{1}$ approximate identity (A unit-mass smooth bump generates an $L^{1}$ approximate identity), and for every $1 \leq q < \infty$ and $g \in L^{q} (R^{n})$ one has $∥ ρ_{ε} * g - g ∥_{L^{q}} \to 0$ as $ε \to 0^{+}$ (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ ).

[F7]

Zero-boundary Sobolev space: every $φ \in C_{c}^{\infty} (Ω; K)$ lies in $W^{k, p} (Ω; K)$ because its classical derivatives are weak derivatives (Classical derivatives agree with weak derivatives, Integer-order Sobolev spaces and their norms), and $H_{0}^{1} (Ω) = W_{0}^{1, 2} (Ω)$ is by definition the closure of $C_{c}^{\infty} (Ω; K)$ in the $H^{1}$ norm (Zero-boundary Sobolev space as a norm closure).

[F8]

Inner product: on $L^{2} (Ω; K)$ the pairing of [F1] is an inner product inducing the $L^{2}$ norm ( $L^{2}$ with the integral pairing is a Hilbert space), and Cauchy--Schwarz gives $∣ (u, v)_{L^{2}} ∣ \leq ∥ u ∥_{L^{2}} ∥ v ∥_{L^{2}}$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

Proof

technique · direct

1.1F2given

If $Ω = \emptyset$ , use $φ = 0$ . Otherwise, for $m = 0, 1, 2, \dots$ , when $Ω \neq R^{n}$ , set $K_{m} = {x : ∣ x ∣ \leq m + 1, dist (x, R^{n} ∖ Ω) \geq 1 / (m + 1)}$ ; when $Ω = R^{n}$ , set $K_{m} = {\overline{B}}_{m + 1} (0)$ . By [F2] each $K_{m}$ is compact and contained in $Ω$ , the sets increase, and $⋃_{m} K_{m} = Ω$ : in the proper-open-set case every point has positive distance from the complement by openness.

2.1F1F4step 1.1choose

Let $F : = 1_{Ω} f \in L^{2} (R^{n})$ be the zero extension of [F1] and let $F_{m} : = 1_{K_{m}} F$ . Since $K_{m} ↑ Ω$ , the sequence $∣ F - F_{m} ∣^{2} = ∣ F ∣^{2} 1_{Ω ∖ K_{m}}$ decreases pointwise to $0$ and is bounded by the integrable function $∣ F ∣^{2}$ ; hence $∥ F - F_{m} ∥_{L^{2}}^{2} = \int_{R^{n}} ∣ F - F_{m} ∣^{2} \to 0$ by dominated convergence, and $F_{m} \to F$ pointwise. Equivalently, the integrals $\int_{K_{m}} ∣ F ∣^{2}$ increase to $∥ F ∥_{L^{2}}^{2}$ by monotone convergence, so the same limit follows. Choose $m$ with $∥ F - F_{m} ∥_{L^{2}} < δ / 2$ .

3.1F1F2step 2.1choosealgebra

Put $G = F_{m}$ for the $m$ selected in step 2.1, with a representative zero outside $K_{m}$ . Then $∥ F - G ∥_{2} < δ / 2$ . If $K_{m} = \emptyset$ , $G = 0$ and $φ = 0$ already has error less than $δ$ ; hence assume $K_{m} \neq \emptyset$ . Consider every pair $(y, r)$ with $y \in K_{m}$ , $0 < r \leq 1$ and $B_{r} (y) \subseteq Ω$ . The balls $B_{r / 2} (y)$ cover $K_{m}$ , so compactness gives a finite nonempty subcover $B_{r_{j} / 2} (y_{j})$ ; put $ε_{0} = \min_{j} r_{j} / 4 > 0$ . If $y \in K_{m}$ and $∣ h ∣ \leq ε_{0}$ , some covering ball gives $∣ y + h - y_{j} ∣ < r_{j} / 2 + ε_{0} < r_{j}$ , so $y + h \in Ω$ . Thus $K_{m} + {\overline{B}}_{ε_{0}} (0) \subseteq Ω$ . This sumset is compact: it is bounded, and for a point $x$ outside it the continuous function $y \mapsto ∣ x - y ∣$ attains on $K_{m}$ a minimum greater than $ε_{0}$ , so its complement is open.

4.1F3F5step 3.1construct

Choose the unit-mass smooth bump $ρ$ constructed in [F3], and put $G_{ε} = G * ρ_{ε}$ for $0 < ε < ε_{0}$ . By [F5] this is smooth globally. If $x \notin K_{m} + {\overline{B}}_{ε} (0)$ , every $y \in K_{m}$ has $ρ_{ε} (x - y) = 0$ , while $G (y) = 0$ off $K_{m}$ ; hence the defining integral is zero. Its support therefore lies in the compact set $K_{m} + {\overline{B}}_{ε} (0) \subseteq Ω$ , so $G_{ε} ∣_{Ω} \in C_{c}^{\infty} (Ω)$ .

5.1F1F6step 3.1step 4.1choosealgebra

By [F6], $∥ G_{ε} - G ∥_{2} \to 0$ . Choose $0 < ε < ε_{0}$ with this norm less than $δ / 2$ and set $φ = G_{ε} ∣_{Ω}$ . Since $F, G, G_{ε}$ vanish outside $Ω$ , $∥ φ - f ∥_{L^{2} (Ω)} = ∥ G_{ε} - F ∥_{2} \leq ∥ G_{ε} - G ∥_{2} + ∥ G - F ∥_{2} < δ$ .

6.1F1F7F8step 5.1∎

Density of $C_{c}^{\infty} (Ω; K)$ in $L^{2} (Ω)$ follows because $f$ and $δ$ were arbitrary in step 5.1. Consequently $H_{0}^{1} (Ω)$ is dense in $L^{2} (Ω)$ : given $g \in L^{2} (Ω)$ and $η > 0$ , step 5.1 supplies $φ \in C_{c}^{\infty} (Ω; K)$ with $∥ φ - g ∥_{L^{2}} < η$ , and $φ \in H_{0}^{1} (Ω)$ by [F7]. Finally let $h \in L^{2} (Ω)$ satisfy $(h, v)_{L^{2}} = 0$ for every $v \in H_{0}^{1} (Ω)$ . By density choose classes $v_{k} \in H_{0}^{1} (Ω)$ with $∥ v_{k} - h ∥_{L^{2}} \to 0$ ; then Cauchy--Schwarz [F8] gives $∣ (h, h) ∣ = ∣ (h, h - v_{k}) ∣ \leq ∥ h ∥_{L^{2}} ∥ h - v_{k} ∥_{L^{2}} \to 0$ , so $∥ h ∥_{L^{2}}^{2} = (h, h) = 0$ and $h = 0$ .

DefinitionDefinition: Literature-sourcedProof: AI-adaptedOpen item page →

The $L^{2}$ operator associated with a symmetric elliptic form

Definition

Assume Countable Choice. Symmetric case. Let $Ω \subseteq R^{n}$ be open and let $a$ be the divergence-form sesquilinear form of Uniformly elliptic divergence-form operators and their sesquilinear forms with $b^{i} \equiv 0$ , coefficients satisfying $a^{i j} = \overline{a^{j i}}$ a.e. and real $c$ , all measurable and essentially bounded, and with uniform ellipticity constant $θ$ . Thus $a (u, v) = \int_{Ω} (a^{i j} D_{j} u \overline{D_{i} v} + c u \overline{v}) d x$ is a bounded symmetric form, $a (u, v) = \overline{a (v, u)}$ (Bounded, coercive and symmetric sesquilinear forms, The formal adjoint and the adjoint weak Dirichlet problem). Define $D (L) : = {u \in H_{0}^{1} (Ω) : \exists f \in L^{2} (Ω) with a (u, v) = (f, v)_{L^{2}} \forall v \in H_{0}^{1} (Ω)}, L u : = f .$ This is well defined: if $f, g$ both satisfy the defining identity then $(f - g, v)_{L^{2}} = 0$ for every $v \in H_{0}^{1} (Ω)$ , and $H_{0}^{1} (Ω)$ is dense in $L^{2} (Ω)$ (Smooth compactly supported functions of an open set are dense in $L^{2}$ ), so $f = g$ in $L^{2} (Ω)$ . The space $D (L)$ is a linear subspace of $H_{0}^{1} (Ω)$ containing the range of every shifted solution operator (The shifted elliptic solution operator), and $L : D (L) \to L^{2} (Ω)$ is linear. With only bounded measurable coefficient hypotheses, $D (L)$ may be a proper subspace of the form domain $H_{0}^{1} (Ω)$ ; those hypotheses alone do not assert $C_{c}^{\infty} (Ω) \subseteq D (L)$ . Membership $u \in D (L)$ with $L u = f$ is exactly the weak statement of $L u = f$ with zero boundary values in $L^{2}$ data (Weak Dirichlet solutions for a divergence-form operator).

Well-definedness and symmetry, recorded with the definition. Boundedness of $a$ on $H^{1} (Ω)$ is The elliptic form is well defined and bounded on $H^{1}$ with $b = 0$ , and symmetry follows by conjugating the defining integrand: with $a^{i j} = \overline{a^{j i}}$ and $c$ real, $\overline{a (v, u)} = \int_{Ω} (\overline{a^{j i}} D_{j} u \overline{D_{i} v} + \overline{c} \overline{v} u) d x = a (u, v)$ after re-indexing. Hence the pair $a, ⟨ \cdot, \cdot ⟩_{L^{2}}$ is the symmetric sesquilinear pair whose weak identity defines $D (L)$ . The representing datum is unique by the density argument above, so $L u$ is a well-defined class; linearity of $L$ follows from linearity of $a$ and of the $L^{2}$ pairing. The range inclusion $ran K_{μ} \subseteq D (L)$ holds because $K_{μ} g$ satisfies $a (K_{μ} g, v) = a_{μ} (K_{μ} g, v) - μ (K_{μ} g, v)_{L^{2}} = (g - μ K_{μ} g, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ , with datum $g - μ K_{μ} g \in L^{2} (Ω)$ (The shifted elliptic solution operator, The space $L^{p} (μ)$ as the quotient by null functions, Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure, Real and imaginary parts, complex conjugation, and modulus, The Axiom of Countable Choice ( ${A C}_{ω}$ )). No claim of self-adjointness, closedness, density of $D (L)$ , or identification with a classical differential expression is made here; those belong to the following items.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The associated elliptic operator is densely defined, symmetric and lower bounded

Statement

Assume Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form, let $Ω \subseteq R^{n}$ be open, fix $μ \geq β$ with $β$ as in Garding's inequality for a divergence-form elliptic operator, and let $K_{μ}$ be the shifted solution operator of The shifted elliptic solution operator. Then:

$D (L)$ is dense in $L^{2} (Ω)$ ;
$L$ is symmetric, i.e. $(L u, v)_{L^{2}} = (u, L v)_{L^{2}}$ for all $u, v \in D (L)$ ;
$L$ is lower bounded, i.e. $(L u, u)_{L^{2}} = a (u, u) \geq - β ∥ u ∥_{L^{2}}^{2}$ for every $u \in D (L)$ (and also $a (u, u) \geq - μ ∥ u ∥_{L^{2}}^{2}$ ). No boundary regularity of $Ω$ is used.

Facts & Assumptions

Given: Countable Choice; an open set $Ω \subseteq R^{n}$ ; the symmetric divergence-form form $a$ with constants $θ, M_{a}, M_{c}$ and $b = 0$ ; a fixed $μ \geq β$ ; the shifted solution operator $K_{μ}$ ; the operator $L : D (L) \to L^{2} (Ω)$ of The $L^{2}$ operator associated with a symmetric elliptic form.

[F1]

Range identity: for every $f \in L^{2} (Ω)$ the class $K_{μ} f$ lies in $D (L)$ with $L (K_{μ} f) = f - μ K_{μ} f$ , because $a (K_{μ} f, v) = a_{μ} (K_{μ} f, v) - μ (K_{μ} f, v)_{L^{2}} = (f - μ K_{μ} f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ (The shifted elliptic solution operator, The $L^{2}$ operator associated with a symmetric elliptic form).

[F2]

Coercivity: $Re a_{μ} (u, u) = a_{μ} (u, u) \geq \frac{θ}{2} ∥ u ∥_{H_{0}^{1}}^{2}$ for $u \in H_{0}^{1} (Ω)$ , since $μ \geq β$ (A sufficiently large shift is coercive, Bounded, coercive and symmetric sesquilinear forms, Integer-order Sobolev spaces and their norms).

[F3]

Density: $H_{0}^{1} (Ω)$ is dense in $L^{2} (Ω)$ , and the closure of a linear subspace equals its double orthogonal complement, so a subspace is dense in the Hilbert space $L^{2} (Ω)$ exactly when its orthogonal complement is trivial (Smooth compactly supported functions of an open set are dense in $L^{2}$ , The double orthogonal complement of a subspace is its closure, Orthogonality and the orthogonal complement, Hilbert space, The space $L^{p} (μ)$ as the quotient by null functions).

[F4]

Garding's inequality: $a (u, u) \geq \frac{θ}{2} ∥ u ∥_{H^{1}}^{2} - β ∥ u ∥_{L^{2}}^{2}$ and $a (u, u) \geq - β ∥ u ∥_{L^{2}}^{2}$ (Garding's inequality for a divergence-form elliptic operator).

[F5]

Weak representer: for $u \in D (L)$ and $f = L u$ one has $a (u, v) = (f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ (The $L^{2}$ operator associated with a symmetric elliptic form, Zero-boundary Sobolev space as a norm closure).

Proof

technique · direct

1.1F1F2F3given

Range inclusion. By [F1] every element of $ran K_{μ}$ lies in $D (L)$ , so it suffices to show that $ran K_{μ}$ is dense in $L^{2} (Ω)$ . Let $y \in L^{2} (Ω)$ be orthogonal to $ran K_{μ}$ . Then $(y, K_{μ} y)_{L^{2}} = 0$ , while the defining equation of $K_{μ}$ at $v = K_{μ} y$ gives $a_{μ} (K_{μ} y, K_{μ} y) = (y, K_{μ} y)_{L^{2}} = 0$ ; coercivity [F2] yields $∥ K_{μ} y ∥_{H_{0}^{1}}^{2} \leq \frac{2}{θ} a_{μ} (K_{μ} y, K_{μ} y) = 0$ , so $K_{μ} y = 0$ . For every $v \in H_{0}^{1} (Ω)$ the defining equation then gives $(y, v)_{L^{2}} = a_{μ} (K_{μ} y, v) = 0$ , and density of $H_{0}^{1} (Ω)$ in $L^{2} (Ω)$ ([F3]) forces $y = 0$ . Since $K_{μ}$ is linear, its range is a linear subspace; by [F3] its orthogonal complement is trivial, so $ran K_{μ}$ is dense; hence its superset $D (L)$ is dense in $L^{2} (Ω)$ .

1.2F5givenalgebra

Symmetry. Let $u, v \in D (L)$ and write $f = L u$ , $g = L v$ . By [F5], $(L u, v)_{L^{2}} = (f, v)_{L^{2}} = a (u, v)$ and $(u, L v)_{L^{2}} = (u, g)_{L^{2}} = \overline{(g, u)_{L^{2}}} = \overline{a (v, u)}$ . Since the coefficients are Hermitian and $b = 0$ with real $c$ , the form $a$ is symmetric, $a (u, v) = \overline{a (v, u)}$ (The $L^{2}$ operator associated with a symmetric elliptic form, Bounded, coercive and symmetric sesquilinear forms), so $(L u, v)_{L^{2}} = (u, L v)_{L^{2}}$ .

2.1F2F4F5step 1.1givenalgebra∎

Lower bound. For $u \in D (L)$ , [F5] with $v = u$ gives $(L u, u)_{L^{2}} = a (u, u)$ , which is real by symmetry; Garding's inequality [F4] yields $a (u, u) \geq - β ∥ u ∥_{L^{2}}^{2}$ , and a second application with the positive shift gives $a (u, u) = a_{μ} (u, u) - μ ∥ u ∥_{L^{2}}^{2} \geq - μ ∥ u ∥_{L^{2}}^{2}$ because $a_{μ} (u, u) \geq 0$ by [F2]. No boundary regularity of $Ω$ was used.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The symmetric elliptic form operator is self-adjoint with compact resolvent

Statement

Assume Countable Choice, together with the Axiom of Choice where the compactness clause is used. Let $Ω \subseteq R^{n}$ be open and let $L, D (L)$ be the symmetric-case operator of The $L^{2}$ operator associated with a symmetric elliptic form, with $D (L)$ dense and $L$ symmetric and lower bounded (The associated elliptic operator is densely defined, symmetric and lower bounded); let its scalar field be $K \in {R, C}$ and fix $μ \geq β$ . Then $L$ is self-adjoint: $L = L^{*}$ , the adjoint being taken in $L^{2} (Ω; K)$ for the densely defined operator $L$ (Adjoint of a densely defined operator, Symmetric, self-adjoint and essentially self-adjoint operators). Moreover $L + μ : D (L) \to L^{2} (Ω; K)$ is a bijection with inverse $K_{μ}$ . If $Ω$ is bounded and the Axiom of Choice holds, then $K_{μ}$ is compact (The shifted solution operator is compact on $L^{2}$ ). Under these boundedness and AC hypotheses, for $K = C$ , $L$ has compact resolvent: $(L - λ)^{- 1}$ is compact on $L^{2} (Ω; C)$ for every $λ$ in the resolvent set of $L$ (Resolvent and spectrum of an unbounded operator). For $K = R$ , identify $L^{2} (Ω; C)$ canonically with $L^{2} (Ω; R)_{C}$ by $u + i v \leftrightarrow (u, v)$ , and let $L_{C} (u + i v) : = L u + i L v$ on $D (L_{C}) = D (L) + i D (L)$ (Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding, Complexification of a real-linear map, Complex Lp classes and Euclidean test-function conventions). Then $L_{C}$ is self-adjoint and has compact resolvent: $(L_{C} - λ)^{- 1}$ is compact on $L^{2} (Ω; C)$ for every $λ$ in its resolvent set (Resolvent and spectrum of an unbounded operator).

Facts & Assumptions

Given: Countable Choice; the symmetric divergence-form case with form $a$ and operator $L$ ; a fixed $μ \geq β$ ; the shifted solution operator $K_{μ}$ ; and the field $K \in {R, C}$ .

[F1]

$D (L)$ is dense in $L^{2} (Ω)$ and $L$ is symmetric; $L + μ$ is defined on $D (L)$ and is symmetric as well (The associated elliptic operator is densely defined, symmetric and lower bounded, The $L^{2}$ operator associated with a symmetric elliptic form, Symmetric, self-adjoint and essentially self-adjoint operators).

[F2]

Solution operator: $a_{μ} (K_{μ} f, v) = (f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ and $f \in L^{2} (Ω)$ , with $a_{μ} = a + μ (\cdot, \cdot)_{L^{2}}$ bounded and coercive on $H_{0}^{1} (Ω)$ with constant $α = θ / 2$ (The shifted elliptic solution operator, A sufficiently large shift is coercive, Zero-boundary Sobolev space as a norm closure).

[F3]

Range description: $a (K_{μ} f, v) = (f - μ K_{μ} f, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ , so $K_{μ} f \in D (L)$ and $L (K_{μ} f) = f - μ K_{μ} f$ (The $L^{2}$ operator associated with a symmetric elliptic form).

[F4]

Lax--Milgram applies to bounded coercive sesquilinear forms on $H_{0}^{1} (Ω)$ and bounded conjugate-linear data; the shifted forms $a_{μ} \pm i (\cdot, \cdot)_{L^{2}}$ in the complex case have the same real part as $a_{μ}$ (The Lax--Milgram theorem, Bounded, coercive and symmetric sesquilinear forms).

[F5]

Range criterion: a densely defined symmetric operator on a complex Hilbert space is self-adjoint if and only if $ran (T \pm i) = H$ (Range criterion for self-adjointness, Adjoint of a densely defined operator).

[F6]

Compactness: if $Ω$ is bounded and the Axiom of Choice holds, the $L^{2}$ realization of $K_{μ}$ is compact, and a bounded operator times a compact operator is compact (The shifted solution operator is compact on $L^{2}$ , Compositions with a compact operator are compact, Compact linear operator, A bounded linear operator between normed spaces, The Axiom of Choice).

[F7]

Complex resolvent: for a densely defined operator $\tilde{L}$ on a complex Hilbert space and $λ$ in its resolvent set, $\tilde{L} - λ : D (\tilde{L}) \to H_{C}$ is a bijection with bounded inverse $(\tilde{L} - λ)^{- 1}$ (Resolvent and spectrum of an unbounded operator).

[F8]

Canonical Hilbert-space complexification. By the componentwise convention for complex $L^{2}$ , every complex class has a unique decomposition $u + i v$ with real $u, v \in L^{2} (Ω; R)$ . The map $1 \otimes u + i \otimes v \mapsto u + i v$ identifies $(L^{2} (Ω; R))_{C}$ with $L^{2} (Ω; C)$ ; expanding the complex integral pairing gives $⟨ u + i v, p + i q ⟩_{C} = (u, p)_{R} + (v, q)_{R} + i ((v, p)_{R} - (u, q)_{R}), ∥ u + i v ∥_{L^{2} (C)}^{2} = ∥ u ∥_{L^{2} (R)}^{2} + ∥ v ∥_{L^{2} (R)}^{2} .$ Thus the identification is a complex-linear Hilbert isometry. For a real densely defined operator $T$ , its complexification is $T_{C} (u + i v) = T u + i T v$ on $D (T) + i D (T)$ (Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding, Complexification of a real-linear map, The complex $L^{2}$ pairing on equivalence classes, Complex Lp classes and Euclidean test-function conventions, $L^{2}$ with the integral pairing is a Hilbert space, The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz, Real and complex inner-product spaces and their induced length, Hilbert space).

[F9]

If a real bounded operator $S$ is compact, then its componentwise complexification is compact: for any bounded sequence $u_{j} + i v_{j}$ , the real and imaginary sequences are bounded; compactness of $S$ and the metric compactness equivalences give a subsequence on which $S u_{j}$ converges, then a further subsequence on which $S v_{j}$ converges. Boundedness follows from $∥ S_{C} (u + i v) ∥^{2} = ∥ S u ∥^{2} + ∥ S v ∥^{2} \leq ∥ S ∥^{2} ∥ u + i v ∥^{2}$ . AC supplies the Countable and Dependent Choice hypotheses of the metric compactness equivalences. This applies to the compact real shifted inverse $K_{μ}$ (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice, AC supplies the countable and dependent choices used in Banach integration).

Proof

technique · direct

1.1F2F3given

Surjectivity of $L + μ$ . Let $f \in L^{2} (Ω)$ and put $u : = K_{μ} f \in H_{0}^{1} (Ω)$ ; by [F3] $u \in D (L)$ and $L u = f - μ u$ , that is $(L + μ) u = f$ . Hence $ran (L + μ) = L^{2} (Ω)$ , and $(L + μ) u = 0$ forces $u = K_{μ} 0 = 0$ by [F2], so $L + μ$ is a bijection of $D (L)$ onto $L^{2} (Ω)$ with inverse $K_{μ}$ .

1.2F2F3F4given

The complex case: $ran (L + μ \pm i) = L^{2} (Ω)$ . Assume $K = C$ and let $f \in L^{2} (Ω)$ . The forms $a_{μ}^{\pm} (u, v) : = a_{μ} (u, v) \pm i (u, v)_{L^{2}}$ are bounded and coercive on $H_{0}^{1} (Ω)$ , because $Re a_{μ}^{\pm} (u, u) = Re a_{μ} (u, u) \geq α ∥ u ∥_{H_{0}^{1}}^{2}$ and boundedness is inherited from $a_{μ}$ and the $L^{2}$ pairing; applying [F4] to the bounded conjugate-linear datum $v \mapsto (f, v)$ gives a unique $u_{\pm} \in H_{0}^{1} (Ω)$ with $a_{μ} (u_{\pm}, v) \pm i (u_{\pm}, v)_{L^{2}} = (f, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , which rearranges to $a (u_{\pm}, v) = (f - μ u_{\pm} \mp i u_{\pm}, v)_{L^{2}}$ . Hence $u_{\pm} \in D (L)$ with $(L + μ \pm i) u_{\pm} = f$ , so both ranges are all of $L^{2} (Ω)$ .

2.1F1F5step 1.2given

Complex case: self-adjointness. Assume $K = C$ and put $T : = L + μ$ on the dense domain $D (L)$ . By [F1] the operator $T$ is densely defined and symmetric, and by step 1.2 $ran (T \pm i) = L^{2} (Ω)$ ; the range criterion [F5] then makes $T$ self-adjoint, and $L = T - μ$ is self-adjoint because subtracting the real scalar $μ$ preserves the adjoint relation $D (L^{*}) = D (T^{*}) = D (L)$ and $L^{*} = T^{*} - μ$ .

2.2F1step 1.1given

Real case: self-adjointness. Assume $K = R$ . By step 1.1 the densely defined symmetric operator $T = L + μ$ has full range $L^{2} (Ω; R)$ . Let $v \in D (T^{*})$ and put $w : = T^{*} v \in L^{2} (Ω; R)$ ; by surjectivity choose $u \in D (T)$ with $T u = w$ . Then for every $z \in D (T)$ , symmetry gives $(T z, v) = (z, T^{*} v) = (z, T u) = (T z, u)$ , so $v - u$ is orthogonal to $ran T = L^{2} (Ω; R)$ and hence $v = u \in D (T)$ with $T v = T^{*} v$ . Therefore $D (T^{*}) = D (T)$ and $T$ is self-adjoint, and so is $L = T - μ$ .

3.1F8step 2.2givenalgebra

Complexification of the real branch. Write $H_{C} = L^{2} (Ω; C) = H_{R} + i H_{R}$ using [F8], and define $L_{C} (u + i v) = L u + i L v$ on $D (L) + i D (L)$ . Its domain is dense because $D (L)$ is dense in the real space and each real and imaginary component can be approximated there. To check self-adjointness, let $y = x + i z \in D (L_{C}^{*})$ and write $L_{C}^{*} y = p + i q$ . Testing the adjoint identity at real $h \in D (L)$ gives $(L h, x)_{R} - i (L h, z)_{R} = ⟨ L_{C} h, y ⟩_{C} = ⟨ h, p + i q ⟩_{C} = (h, p)_{R} - i (h, q)_{R} .$ Thus $x, z \in D (L^{*}) = D (L)$ and $p = L x$ , $q = L z$ . Hence $y \in D (L_{C})$ and $L_{C}^{*} y = L_{C} y$ . Conversely, the real self-adjoint identities give $L_{C} \subseteq L_{C}^{*}$ , so equality holds.

4.1F6F7F8F9step 1.1step 2.1step 2.2step 3.1givenalgebra∎

Compact resolvent. If $K = C$ , put $\tilde{L} = L$ and $\tilde{K} = K_{μ}$ ; [F6] gives compactness of $\tilde{K}$ when $Ω$ is bounded. If $K = R$ , put $\tilde{L} = L_{C}$ and $\tilde{K} (u + i v) = K_{μ} u + i K_{μ} v$ ; [F6] makes $K_{μ}$ compact on the real $L^{2}$ space, and [F9] gives compactness of $\tilde{K}$ . By step 1.1 (componentwise in the real case), $\tilde{K} = (\tilde{L} + μ)^{- 1}$ . In either case let $λ$ lie in the resolvent set of $\tilde{L}$ , write $R = (\tilde{L} - λ)^{- 1}$ and $c = λ + μ$ . Using the inverse relations on their domains gives $R - \tilde{K} = R [(\tilde{L} + μ) - (\tilde{L} - λ)] \tilde{K} = c R \tilde{K},$ and also $R - \tilde{K} = \tilde{K} [(\tilde{L} + μ) - (\tilde{L} - λ)] R = c \tilde{K} R .$ Therefore $Q : = I - c \tilde{K}$ is boundedly invertible, with $Q^{- 1} = I + c R$ , since $(I - c \tilde{K}) (I + c R) = I = (I + c R) (I - c \tilde{K})$ by these identities. On $D (\tilde{L})$ one has $\tilde{L} - λ = (\tilde{L} + μ) Q$ , whence $(\tilde{L} - λ)^{- 1} = Q^{- 1} \tilde{K} = (I + c R) \tilde{K} .$ This is a bounded operator composed with the compact $\tilde{K}$ , so it is compact. The complex resolvent definition applies to $\tilde{L}$ in both scalar-field cases.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Symmetric elliptic weak eigenpairs

Definition

Assume Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form, a weak eigenpair of the Dirichlet problem for $L$ is a pair $(λ, u)$ with $λ \in R$ , $u \in H_{0}^{1} (Ω) ∖ {0}$ and $a (u, v) = λ (u, v)_{L^{2}} for every v \in H_{0}^{1} (Ω);$ $λ$ is a weak eigenvalue and $u$ a weak eigenfunction. The eigenspace $E_{λ} : = {u \in H_{0}^{1} (Ω) : a (u, v) = λ (u, v)_{L^{2}} \forall v}$ is a closed linear subspace. Weak eigenpairs are exactly operator eigenpairs: $(λ, u)$ is a weak eigenpair if and only if $u \in D (L) ∖ {0}$ and $L u = λ u$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism, The $L^{2}$ operator associated with a symmetric elliptic form). Eigenfunctions are $L^{2}$ equivalence classes; no canonical representative or canonical vector in a multiple eigenspace is selected. The multiplicity of $λ$ is $\dim E_{λ}$ , with no finiteness assertion for general open $Ω$ .

Well-definedness, recorded with the definition. $E_{λ}$ is a linear subspace because $a$ and the $L^{2}$ pairing are linear in the first argument; it is closed in $H_{0}^{1} (Ω)$ because both $u \mapsto a (u, v)$ and $u \mapsto (u, v)_{L^{2}}$ are continuous on $H_{0}^{1} (Ω)$ for each fixed $v$ : boundedness of $a$ and the estimate $∣ (u, v)_{L^{2}} ∣ \leq ∥ u ∥_{L^{2}} ∥ v ∥_{L^{2}} \leq ∥ u ∥_{H_{0}^{1}} ∥ v ∥_{H_{0}^{1}}$ show that weak limits of vectors in $E_{λ}$ remain in $E_{λ}$ . The equivalence with operator eigenpairs is the definition of $D (L)$ and $L u$ : the weak identity for $(λ, u)$ says exactly that the datum $f = λ u$ represents $a (u, \cdot)$ on $H_{0}^{1} (Ω)$ , which, together with $u \neq 0$ , is the pair of conditions $u \in D (L) ∖ {0}$ and $L u = λ u$ . The eigenvalue $λ$ is required to be real, as is forced for the symmetric form once $u \neq 0$ : taking $v = u$ gives $λ ∥ u ∥_{L^{2}}^{2} = a (u, u) = \overline{a (u, u)}$ . All equalities are equalities of $L^{2}$ and $H_{0}^{1}$ classes (The space $L^{p} (μ)$ as the quotient by null functions, Zero-boundary Sobolev space as a norm closure, Uniformly elliptic divergence-form operators and their sesquilinear forms, The Axiom of Countable Choice ( ${A C}_{ω}$ )); no regularity of eigenfunctions and no boundary values beyond membership in $H_{0}^{1}$ are asserted.

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The symmetric shifted solution operator is positive and self-adjoint

Statement

Assume Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form, with $Ω$ open and $μ \geq β$ , the shifted solution operator $K_{μ}$ of The shifted elliptic solution operator, regarded on $L^{2} (Ω)$ , is self-adjoint and positive: $(K_{μ} f, g)_{L^{2}} = (f, K_{μ} g)_{L^{2}}, (K_{μ} f, f)_{L^{2}} = a_{μ} (K_{μ} f, K_{μ} f) \geq α ∥ K_{μ} f ∥_{H_{0}^{1}}^{2} \geq 0$ for all $f, g \in L^{2} (Ω)$ , with $(K_{μ} f, f)_{L^{2}} = 0$ if and only if $f = 0$ ; in particular $K_{μ}$ is injective. Moreover $a (K_{μ} f, v) = (f - μ K_{μ} f, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , so $K_{μ} f \in D (L)$ and $L (K_{μ} f) = f - μ K_{μ} f$ . Symmetry of $K_{μ}$ is verified from the form; no self-adjointness of the differential expression is assumed.

Facts & Assumptions

Given: Countable Choice; the symmetric divergence-form case with form $a$ and $μ \geq β$ ; the shifted solution operator $K_{μ}$ and the shifted form $a_{μ} = a + μ (\cdot, \cdot)_{L^{2}}$ ; $f, g \in L^{2} (Ω)$ .

[F1]

Defining identity and coercivity: $a_{μ} (K_{μ} h, v) = (h, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ and all $h \in L^{2} (Ω)$ , and $a_{μ} (u, u) \geq α ∥ u ∥_{H_{0}^{1}}^{2}$ with $α = θ / 2$ for $u \in H_{0}^{1} (Ω)$ (The shifted elliptic solution operator, A sufficiently large shift is coercive, Bounded, coercive and symmetric sesquilinear forms).

[F2]

Symmetry: $a (u, v) = \overline{a (v, u)}$ and $(v, u)_{L^{2}} = \overline{(u, v)_{L^{2}}}$ , so $a_{μ}$ is symmetric as well; in particular $a_{μ} (u, u)$ is real (The $L^{2}$ operator associated with a symmetric elliptic form, Bounded, coercive and symmetric sesquilinear forms, The formal adjoint and the adjoint weak Dirichlet problem).

[F3]

Density: $H_{0}^{1} (Ω)$ is dense in $L^{2} (Ω)$ (Smooth compactly supported functions of an open set are dense in $L^{2}$ ).

[F4]

Hilbert adjoints and positivity: an operator $T$ on a Hilbert space is self-adjoint when $(T f, g) = (f, T g)$ for all $f, g$ , and positive when $(T f, f) \geq 0$ (The Hilbert-space adjoint of a bounded operator, Self-adjoint, positive, unitary and normal operators).

[F5]

The operator $L$ and its domain: $u \in D (L)$ with $L u = h$ means $u \in H_{0}^{1} (Ω)$ and $a (u, v) = (h, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ (The $L^{2}$ operator associated with a symmetric elliptic form, The associated elliptic operator is densely defined, symmetric and lower bounded).

Proof

technique · direct

1.1F1F2F4givenalgebra

Self-adjointness. By [F1] applied to $g$ and [F2], and then to $f$ , $(K_{μ} f, g)_{L^{2}} = \overline{(g, K_{μ} f)_{L^{2}}} = \overline{a_{μ} (K_{μ} g, K_{μ} f)} = a_{μ} (K_{μ} f, K_{μ} g) = (f, K_{μ} g)_{L^{2}},$ for all $f, g \in L^{2} (Ω)$ ; hence $K_{μ}$ is self-adjoint by [F4].

2.1F1F3F4step 1.1givenalgebra

Positivity and injectivity. Taking $g = f$ in the computation of step 1.1 and using [F1], $(K_{μ} f, f)_{L^{2}} = a_{μ} (K_{μ} f, K_{μ} f) \geq α ∥ K_{μ} f ∥_{H_{0}^{1}}^{2} \geq 0$ . If $(K_{μ} f, f)_{L^{2}} = 0$ , then $α ∥ K_{μ} f ∥_{H_{0}^{1}}^{2} \leq 0$ , so $K_{μ} f = 0$ ; then for every $v \in H_{0}^{1} (Ω)$ the defining identity gives $(f, v)_{L^{2}} = a_{μ} (K_{μ} f, v) = 0$ , and density of $H_{0}^{1} (Ω)$ in $L^{2} (Ω)$ ([F3]) gives $f = 0$ . Conversely $f = 0$ gives $K_{μ} f = 0$ and hence $(K_{μ} f, f)_{L^{2}} = 0$ ; thus $K_{μ}$ is positive and injective.

3.1F1F5givenalgebra∎

Range description. For $f \in L^{2} (Ω)$ and every $v \in H_{0}^{1} (Ω)$ , $a (K_{μ} f, v) = a_{μ} (K_{μ} f, v) - μ (K_{μ} f, v)_{L^{2}} = (f - μ K_{μ} f, v)_{L^{2}},$ because the datum $f - μ K_{μ} f$ lies in $L^{2} (Ω)$ . By [F5] this says $K_{μ} f \in D (L)$ and $L (K_{μ} f) = f - μ K_{μ} f$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Discrete spectrum of a symmetric elliptic Dirichlet operator

Statement

Assume the Axiom of Choice and Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form, let $Ω \subseteq R^{n}$ be nonempty, open and bounded, fix $μ \geq β$ , and let $K_{μ}$ be the shifted solution operator of The shifted elliptic solution operator (The shifted solution operator is compact on $L^{2}$ ). Then the following hold.

There are real numbers $λ_{1} \leq λ_{2} \leq \dots$ with $λ_{j} \to + \infty$ , each eigenvalue repeated according to its finite multiplicity, and an orthonormal basis ${e_{j}}_{j \geq 1}$ of $L^{2} (Ω)$ with $e_{j} \in H_{0}^{1} (Ω)$ and $a (e_{j}, v) = λ_{j} (e_{j}, v)_{L^{2}} for every v \in H_{0}^{1} (Ω),$ equivalently $e_{j} \in D (L)$ and $L e_{j} = λ_{j} e_{j}$ in the sense of Symmetric elliptic weak eigenpairs. Explicitly $λ_{j} = ν_{j}^{- 1} - μ$ , where $ν_{j} > 0$ , $ν_{j} ↓ 0$ , are the nonzero eigenvalues of the compact self-adjoint positive operator $K_{μ}$ with $K_{μ} e_{j} = ν_{j} e_{j}$ (The symmetric shifted solution operator is positive and self-adjoint).
Every weak eigenvalue of the Dirichlet problem occurs in the list, and each listed $λ_{j}$ is a weak eigenvalue with finite-dimensional eigenspace; eigenspaces belonging to distinct eigenvalues are $L^{2}$ -orthogonal.
$λ_{j} > - μ$ for every $j$ , and $a (u, u) \geq - β ∥ u ∥_{L^{2}}^{2}$ for every $u \in H_{0}^{1} (Ω)$ . The proof applies the compact self-adjoint spectral theorem once to $K_{μ}$ ; since $\ker K_{μ} = {0}$ , no Hilbert basis of the kernel is ever selected, and the union of orthonormal bases of the nonzero eigenspaces of $K_{μ}$ is already a Hilbert basis of $L^{2} (Ω)$ .

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the symmetric divergence-form case with constants $θ, M_{a}, M_{c}$ ; a fixed $μ \geq β$ ; the shifted solution operator $K_{μ}$ and the operator $L$ of The $L^{2}$ operator associated with a symmetric elliptic form.

[F1]

Spectral data: $K_{μ}$ , regarded on $L^{2} (Ω)$ , is compact, self-adjoint, positive and injective, with $\ker K_{μ} = {0}$ (The symmetric shifted solution operator is positive and self-adjoint, The shifted solution operator is compact on $L^{2}$ , Compact linear operator, The Axiom of Choice). Since $Ω$ contains a box of positive finite measure (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included), it contains countably many disjoint positive-measure subboxes whose indicators give an infinite orthogonal family in $L^{2} (Ω)$ .

[F2]

Compact self-adjoint spectral theorem: the nonzero eigenvalues of $K_{μ}$ are real, of finite multiplicity and accumulate only at $0$ ; the eigenspaces for distinct eigenvalues are orthogonal; the closed span of their union is $(\ker K_{μ})^{⊥}$ , so it is all of $L^{2} (Ω)$ because $K_{μ}$ is injective; and $K_{μ} x = \sum_{ν} ν P_{ν} x$ in norm, where $P_{ν}$ is the orthogonal projection onto the eigenspace of $ν$ (Spectral theorem for compact self adjoint operators, Eigenspaces of a self adjoint operator are orthogonal, Orthonormal families, complete orthonormal systems and Hilbert bases, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

Translation of eigenvectors: if $K_{μ} e = ν e$ with $e \neq 0$ and $ν > 0$ , then $e = ν^{- 1} K_{μ} e \in H_{0}^{1} (Ω)$ because $K_{μ}$ maps $L^{2}$ into $H_{0}^{1}$ ; hence for every $v \in H_{0}^{1} (Ω)$ one has $a (e, v) = a_{μ} (e, v) - μ (e, v)_{L^{2}} = (ν^{- 1} - μ) (e, v)_{L^{2}}$ . Conversely if $(λ, u)$ is a weak eigenpair of the symmetric case, then $a_{μ} (u, v) = (λ + μ) (u, v)_{L^{2}}$ for all $v$ , so $K_{μ} ((λ + μ) u) = u$ by uniqueness, and $λ + μ > 0$ because $0 < a_{μ} (u, u) = (λ + μ) ∥ u ∥_{L^{2}}^{2}$ ; hence $K_{μ} u = ν u$ with $ν = (λ + μ)^{- 1}$ (The shifted elliptic solution operator, Symmetric elliptic weak eigenpairs, The associated elliptic operator is densely defined, symmetric and lower bounded, The symmetric elliptic form operator is self-adjoint with compact resolvent).

[F4]

Finite-dimensional Hilbert spaces have orthonormal bases (Every finite-dimensional real or complex inner product space has an orthonormal basis), and Garding's inequality gives $a (u, u) \geq - β ∥ u ∥_{L^{2}}^{2}$ for all $u \in H_{0}^{1} (Ω)$ , with $β$ the constant of Garding's inequality for a divergence-form elliptic operator.

Proof

technique · direct

1.1F1F2given

Spectral data of $K_{μ}$ . By [F1] and [F2] the nonzero eigenvalues $ν$ of $K_{μ}$ are real and of finite multiplicity; positivity gives $ν > 0$ for each of them, and they accumulate only at $0$ . The eigenspaces $E_{ν}$ are finite dimensional and pairwise orthogonal, and their union spans a dense subspace: its closed span is $(\ker K_{μ})^{⊥} = L^{2} (Ω)$ .

2.1F3step 1.1givenalgebra

Translation. Let $ν > 0$ be an eigenvalue of $K_{μ}$ with eigenvector $e$ . Since $K_{μ} e \in H_{0}^{1} (Ω)$ and $K_{μ} e = ν e$ , we have $e \in H_{0}^{1} (Ω)$ ; then [F3] shows $a (e, v) = (ν^{- 1} - μ) (e, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , so $(ν^{- 1} - μ, e)$ is a weak eigenpair with finite-dimensional eigenspace equal to the $ν$ -eigenspace of $K_{μ}$ ; conversely every weak eigenpair $(λ, u)$ arises this way from $ν = (λ + μ)^{- 1}$ , and $λ = ν^{- 1} - μ$ . In particular the two eigenvalue lists correspond bijectively, and each weak eigenvalue is real and of finite multiplicity.

3.1F1F2F4step 1.1step 2.1givenchoose

Enumeration. By [F2] the closed span of the nonzero eigenspaces is all of $L^{2} (Ω)$ , which is infinite dimensional by [F1]. Since each eigenspace is finite dimensional, there must be infinitely many nonzero eigenvalues; compactness gives at most countably many. By [F4], together with Countable Choice, choose an orthonormal basis of each eigenspace $E_{ν}$ ; their union ${e_{j}}_{j \geq 1}$ is an orthonormal family whose closed span is $L^{2} (Ω)$ by step 1.1, hence a Hilbert basis of $L^{2} (Ω)$ with $e_{j} \in H_{0}^{1} (Ω)$ and $K_{μ} e_{j} = ν_{j} e_{j}$ , where the eigenvalues $ν_{j} > 0$ are listed in decreasing order with multiplicity, so that $ν_{j} ↓ 0$ . Set $λ_{j} : = ν_{j}^{- 1} - μ$ ; then $λ_{j}$ is nondecreasing and tends to $+ \infty$ , and step 2.1 gives $a (e_{j}, v) = λ_{j} (e_{j}, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , equivalently $L e_{j} = λ_{j} e_{j}$ by Symmetric elliptic weak eigenpairs.

4.1F2F4step 2.1step 3.1givenalgebra

Claim 2 and the lower bounds. Every weak eigenvalue occurs in the list ${λ_{j}}$ by step 2.1, and each listed $λ_{j}$ is a weak eigenvalue; eigenspaces for distinct eigenvalues are $L^{2}$ -orthogonal by [F2], since they are eigenspaces of $K_{μ}$ for distinct $ν$ . Finally $λ_{j} = ν_{j}^{- 1} - μ > - μ$ because $ν_{j} > 0$ , and Garding's inequality gives $a (u, u) \geq - β ∥ u ∥_{L^{2}}^{2}$ for every $u \in H_{0}^{1} (Ω)$ .

5.1F1step 3.1step 4.1given∎

Conclusion. Steps 3.1 and 4.1 establish all three assertions: the list ${λ_{j}}$ , the orthonormal basis ${e_{j}}$ with the weak eigenrelations and the operator form $L e_{j} = λ_{j} e_{j}$ , the completeness of the eigenvalue list with finite multiplicities and orthogonality of distinct eigenspaces, and the lower bounds $λ_{j} > - μ$ and $a (u, u) \geq - β ∥ u ∥_{L^{2}}^{2}$ . No Hilbert basis of $\ker K_{μ}$ was selected, because $K_{μ}$ is injective by [F1]; only orthonormal bases of the finite-dimensional nonzero eigenspaces were chosen.

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Eigenbasis expansion in the form norm

Statement

Assume the Axiom of Choice and Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form with $Ω$ nonempty bounded open, let ${e_{j}}$ and ${λ_{j}}$ be the eigenbasis and eigenvalues of Discrete spectrum of a symmetric elliptic Dirichlet operator and fix $μ \geq β$ . Then:

for every $u \in H_{0}^{1} (Ω)$ the series $\sum_{j} (u, e_{j})_{L^{2}} e_{j}$ converges to $u$ in the $H_{0}^{1} (Ω)$ norm (equivalently in the inner-product norm $a_{μ}$ ), and $a_{μ} (u, u) = \sum_{j \geq 1} (λ_{j} + μ) ∣ (u, e_{j})_{L^{2}} ∣^{2};$
for every $f \in L^{2} (Ω)$ the series $\sum_{j} (f, e_{j})_{L^{2}} e_{j}$ converges to $f$ in $L^{2} (Ω)$ and $∥ f ∥_{L^{2}}^{2} = \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2}$ (Parseval);
consequently $a (u, u) = \sum_{j} λ_{j} ∣ (u, e_{j})_{L^{2}} ∣^{2}$ for every $u \in H_{0}^{1} (Ω)$ , the series being absolutely convergent. This expansion is the form-domain companion of the $L^{2}$ eigenbasis of the spectral theorem.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the symmetric divergence-form case with form $a$ and operator $L$ ; the eigenbasis ${e_{j}}$ and eigenvalues $λ_{j} \to + \infty$ of the discrete spectral theorem, orthonormal in $L^{2}$ ; a fixed $μ \geq β$ .

[F1]

Eigenrelations: $e_{j} \in H_{0}^{1} (Ω)$ , $a (e_{j}, v) = λ_{j} (e_{j}, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , the $λ_{j}$ are real, and ${e_{j}}$ is a Hilbert basis of $L^{2} (Ω)$ (Discrete spectrum of a symmetric elliptic Dirichlet operator, Symmetric elliptic weak eigenpairs, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F2]

Shifted positivity: $a_{μ} = a + μ (\cdot, \cdot)_{L^{2}}$ is symmetric, and $a_{μ} (u, u) \geq α ∥ u ∥_{H_{0}^{1}}^{2}$ with $α = θ / 2$ ; in particular $a_{μ}$ is an inner product on $H_{0}^{1} (Ω)$ . Boundedness of the shifted form gives $a_{μ} (u, u) \leq M_{μ} ∥ u ∥_{H_{0}^{1}}^{2}$ with $M_{μ} = n M_{a} + M_{c} + ∣ μ ∣$ , so together with the coercive lower bound its norm is equivalent to the Sobolev norm and is complete by The Sobolev space $H^{1}$ is a Hilbert space. It defines the norm $∥ u ∥_{a_{μ}} = a_{μ} (u, u)^{1 / 2}$ , and $λ_{j} + μ > 0$ for every $j$ (A sufficiently large shift is coercive, The shifted elliptic solution operator, The symmetric shifted solution operator is positive and self-adjoint, Hilbert space).

[F3]

Fourier expansion and Parseval: in a Hilbert space with complete orthonormal family the finite-subset net of the coefficients converges in norm and the squared norm is computed by the sum of the squared coefficient moduli; Bessel's inequality and square-summability control the partial sums (Fourier expansion in a Hilbert space, Parseval equivalences for an orthonormal family, The finite Bessel inequality and best approximation by a finite orthonormal family, Square-summable orthogonal families have norm-convergent finite sums, Orthogonality and the orthogonal complement, The $L^{2}$ operator associated with a symmetric elliptic form, The Axiom of Choice).

Proof

technique · direct

1.1F1F2givenalgebra

Orthonormality in the form. For $j, k$ use symmetry of $a$ and the eigenrelation [F1] with $v = e_{k}$ : $a (e_{j}, e_{k}) = λ_{j} (e_{j}, e_{k})_{L^{2}} = λ_{j} δ_{j k}$ . Hence $a_{μ} (e_{j}, e_{k}) = (λ_{j} + μ) (e_{j}, e_{k})_{L^{2}} = (λ_{j} + μ) δ_{j k},$ so the family $f_{j} : = (λ_{j} + μ)^{- 1 / 2} e_{j}$ (well defined by [F2]) is orthonormal in the inner product $a_{μ}$ on $H_{0}^{1} (Ω)$ .

1.2F1F3given

Parseval in $L^{2}$ . Since ${e_{j}}$ is a Hilbert basis of $L^{2} (Ω)$ , [F3] gives $f = \sum_{j} (f, e_{j})_{L^{2}} e_{j}$ in $L^{2} (Ω)$ and $∥ f ∥_{L^{2}}^{2} = \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2}$ for every $f \in L^{2} (Ω)$ , which is claim 2.

2.1F1F2F3step 1.1givenalgebra

Completeness in the form. Let $u \in H_{0}^{1} (Ω)$ satisfy $a_{μ} (u, f_{j}) = 0$ for every $j$ . Then $a_{μ} (u, e_{j}) = (λ_{j} + μ) (u, e_{j})_{L^{2}} = 0$ , so $(u, e_{j})_{L^{2}} = 0$ for every $j$ ; since ${e_{j}}$ is a Hilbert basis of $L^{2} (Ω)$ , $u = 0$ as an $L^{2}$ class, hence $u = 0$ . Thus $(f_{j})$ is a complete orthonormal family in the Hilbert space $(H_{0}^{1} (Ω), a_{μ})$ , and by [F3] for every $u \in H_{0}^{1} (Ω)$ the net of finite partial sums of $\sum_{j} a_{μ} (u, f_{j}) f_{j}$ converges to $u$ in the $a_{μ}$ norm, with $a_{μ} (u, u) = \sum_{j} ∣ a_{μ} (u, f_{j}) ∣^{2} .$ Since $a_{μ} (u, f_{j}) = \overline{a_{μ} (f_{j}, u)} = \overline{(λ_{j} + μ)^{- 1 / 2} a_{μ} (e_{j}, u)} = \overline{(λ_{j} + μ)^{1 / 2} (e_{j}, u)_{L^{2}}} = (λ_{j} + μ)^{1 / 2} (u, e_{j})_{L^{2}}$ , the partial sums are $\sum_{j} (u, e_{j})_{L^{2}} e_{j}$ and claim 1 follows; the $H_{0}^{1}$ norm and the $a_{μ}$ norm are equivalent by [F2].

3.1F2step 1.2step 2.1givenalgebra∎

Claim 3. Let $u \in H_{0}^{1} (Ω) \subseteq L^{2} (Ω)$ . By claim 2 applied to $u$ , $∥ u ∥_{L^{2}}^{2} = \sum_{j} ∣ (u, e_{j})_{L^{2}} ∣^{2}$ , and by claim 1 $a_{μ} (u, u) = \sum_{j} (λ_{j} + μ) ∣ (u, e_{j})_{L^{2}} ∣^{2}$ with both sides finite. Subtracting $μ$ times the first identity from the second gives $a (u, u) = a_{μ} (u, u) - μ ∥ u ∥_{L^{2}}^{2} = \sum_{j} λ_{j} ∣ (u, e_{j})_{L^{2}} ∣^{2} .$ This series is absolutely convergent: since $λ_{j} \to + \infty$ , only finitely many $λ_{j}$ are negative, and for all remaining indices $0 \leq λ_{j} ∣ (u, e_{j})_{L^{2}} ∣^{2} \leq (λ_{j} + μ) ∣ (u, e_{j})_{L^{2}} ∣^{2}$ , whose sum is finite by claim 1.

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The Rayleigh principle for the first Dirichlet eigenvalue

Statement

Assume the Axiom of Choice and Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form with $Ω$ nonempty bounded open, let ${λ_{j}}$ be the eigenvalues of Discrete spectrum of a symmetric elliptic Dirichlet operator. Then $λ_{1} = \min_{u \in H_{0}^{1} (Ω) ∖ {0}} \frac{a (u, u)}{∥ u ∥_{L^{2}}^{2}},$ the minimum is attained exactly at the nonzero elements of the eigenspace $E_{λ_{1}}$ , and $λ_{1}$ is the smallest weak eigenvalue. If in addition the form $a$ is coercive on $H_{0}^{1} (Ω)$ with constant $α^{'} > 0$ (for instance when the hypotheses of Lax--Milgram solvability for coercive divergence-form equations hold, or $a$ is the principal Dirichlet form), then $λ_{1} \geq α^{'} > 0$ ; in general only $λ_{1} > - μ$ and the Garding bound $λ_{1} \geq - β$ are asserted.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the symmetric divergence-form case with form $a$ ; the eigenbasis ${e_{j}}$ and nondecreasing eigenvalue list $λ_{j} \to + \infty$ of the discrete spectral theorem; and $u \in H_{0}^{1} (Ω) ∖ {0}$ .

[F1]

Eigenbasis expansion: $a (u, u) = \sum_{j} λ_{j} ∣ (u, e_{j})_{L^{2}} ∣^{2}$ absolutely convergent and $∥ u ∥_{L^{2}}^{2} = \sum_{j} ∣ (u, e_{j})_{L^{2}} ∣^{2}$ for every $u \in H_{0}^{1} (Ω)$ ; each $e_{j}$ is a weak eigenfunction with eigenvalue $λ_{j}$ (Eigenbasis expansion in the form norm, Discrete spectrum of a symmetric elliptic Dirichlet operator, Symmetric elliptic weak eigenpairs).

[F2]

The list is nondecreasing with $λ_{j} \to + \infty$ , the eigenvalue $0$ is not in the list unless it is an eigenvalue, and $λ_{1}$ is the smallest weak eigenvalue; the nonzero elements $E_{λ_{1}} ∖ {0}$ are exactly the weak eigenfunctions for $λ_{1}$ (Discrete spectrum of a symmetric elliptic Dirichlet operator).

[F3]

Coercivity: if $a (u, u) \geq α^{'} ∥ u ∥_{H_{0}^{1}}^{2}$ for all $u$ , then in particular $a (u, u) \geq α^{'} ∥ u ∥_{L^{2}}^{2}$ , and $λ_{1} \geq α^{'} > 0$ whenever the quotient is bounded below by $α^{'}$ (Bounded, coercive and symmetric sesquilinear forms, Lax--Milgram solvability for coercive divergence-form equations, The space $L^{p} (μ)$ as the quotient by null functions).

Proof

technique · direct

1.1F1F2givenalgebra

Weighted average. Let $u \in H_{0}^{1} (Ω) ∖ {0}$ and put $c_{j} : = (u, e_{j})_{L^{2}}$ . By [F1], $\sum_{j} ∣ c_{j} ∣^{2} = ∥ u ∥_{L^{2}}^{2} > 0$ and $a (u, u) = \sum_{j} λ_{j} ∣ c_{j} ∣^{2}$ , so $\frac{a (u, u)}{∥ u ∥_{L^{2}}^{2}} = \frac{\sum_{j} λ_{j} ∣ c_{j} ∣^{2}}{\sum_{j} ∣ c_{j} ∣^{2}} .$ Since $λ_{j} \geq λ_{1}$ for every $j$ by [F2], the quotient is at least $λ_{1}$ , with equality if and only if $c_{j} = 0$ for every $j$ with $λ_{j} > λ_{1}$ ; that is, if and only if $u$ lies in the closed span of the $e_{j}$ with $λ_{j} = λ_{1}$ , which is exactly $E_{λ_{1}}$ .

2.1F1F2step 1.1given

Attainment. The vector $e_{1}$ is a nonzero weak eigenfunction with $a (e_{1}, e_{1}) = λ_{1} ∥ e_{1} ∥_{L^{2}}^{2}$ and $∥ e_{1} ∥_{L^{2}} = 1$ , so the quotient at $u = e_{1}$ equals $λ_{1}$ ; combined with step 1.1, the infimum is the minimum $λ_{1}$ , attained exactly on $E_{λ_{1}} ∖ {0}$ , and $λ_{1}$ is the smallest weak eigenvalue by [F2].

3.1F3step 2.1givenalgebra∎

Lower bounds. If $a$ is coercive with constant $α^{'} > 0$ then [F3] gives $a (u, u) / ∥ u ∥_{L^{2}}^{2} \geq α^{'}$ for every nonzero $u$ , hence $λ_{1} \geq α^{'} > 0$ by step 2.1. In the general case only the bounds $λ_{1} > - μ$ and $λ_{1} \geq - β$ of the discrete spectral theorem and Garding's inequality are asserted.

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The Courant-Fischer min-max principle for elliptic eigenvalues

Statement

Assume the Axiom of Choice and Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form with $Ω$ nonempty bounded open, let $λ_{1} \leq λ_{2} \leq \dots$ be the eigenvalues of Discrete spectrum of a symmetric elliptic Dirichlet operator, repeated according to multiplicity. Then for every $k \geq 1$ , $λ_{k} = \min {\max_{u \in S ∖ {0}} \frac{a (u, u)}{∥ u ∥_{L^{2}}^{2}} : S \subseteq H_{0}^{1} (Ω), \dim S = k} = \max {\inf_{u \in (H_{0}^{1} (Ω) \cap T^{⊥_{L^{2}}}) ∖ {0}} \frac{a (u, u)}{∥ u ∥_{L^{2}}^{2}} : T \subseteq H_{0}^{1} (Ω), \dim T = k - 1},$ where $T^{⊥_{L^{2}}}$ is the orthogonal complement in $L^{2} (Ω)$ , and for $k = 1$ the maximum is over $T = {0}$ , so $H_{0}^{1} (Ω) \cap T^{⊥_{L^{2}}} = H_{0}^{1} (Ω)$ . Both outer extrema are attained: the first at $S = span {e_{1}, \dots, e_{k}}$ , and the second at $T = span {e_{1}, \dots, e_{k - 1}}$ , where the inner infimum is attained at $e_{k}$ . No smoothness of $\partial Ω$ is required.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the symmetric divergence-form case with form $a$ ; the orthonormal eigenbasis ${e_{j}}$ and nondecreasing eigenvalue list $λ_{j} \to + \infty$ ; and $k \geq 1$ .

[F1]

Weighted average: for every $u \in H_{0}^{1} (Ω) ∖ {0}$ with $c_{j} : = (u, e_{j})_{L^{2}}$ one has $a (u, u) = \sum_{j} λ_{j} ∣ c_{j} ∣^{2}, ∥ u ∥_{L^{2}}^{2} = \sum_{j} ∣ c_{j} ∣^{2},$ both series converging; the expansion is unconditional over the Hilbert basis (Eigenbasis expansion in the form norm, Discrete spectrum of a symmetric elliptic Dirichlet operator, Orthogonality and the orthogonal complement).

[F2]

Finite-dimensional intersection: a linear map from a $k$ -dimensional space into a $(k - 1)$ -dimensional space has nonzero kernel (Rank-nullity: $\dim_{F} V = nullity T + rank T$ , Hilbert space).

Proof

technique · direct

1.1F1givenalgebra

For a finite-dimensional nonzero $S \subseteq H_{0}^{1}$ , choose an $L^{2}$ -orthonormal basis using Every finite-dimensional real or complex inner product space has an orthonormal basis. In its real coordinates (real and imaginary coordinates when the field is complex), the $L^{2}$ unit sphere is a nonempty compact Euclidean sphere by Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, and $a (u, u)$ is a continuous quadratic polynomial there. It has a maximum by A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value. Homogeneity identifies that maximum with $\max_{S ∖ {0}} Q$ , so every inner maximum in the statement exists. Weighted averages. For $u \neq 0$ the quotient is the weighted average $Q (u) : = a (u, u) / ∥ u ∥_{L^{2}}^{2} = \sum_{j} λ_{j} ∣ c_{j} ∣^{2} / \sum_{j} ∣ c_{j} ∣^{2}$ with $\sum_{j} ∣ c_{j} ∣^{2} > 0$ ; if $u$ lies in the span of finitely many eigenvectors $e_{j_{1}}, \dots, e_{j_{m}}$ then the quotient is the corresponding finite convex combination of the $λ_{j_{i}}$ .

2.1F1F2step 1.1givenalgebra

The min-max identity. Let $k \geq 1$ and put $S_{k} : = span {e_{1}, \dots, e_{k}}$ , a $k$ -dimensional subspace on which $Q (u)$ is a weighted average of $λ_{1}, \dots, λ_{k}$ , hence at most $λ_{k}$ , with value $λ_{k}$ at $u = e_{k}$ ; therefore the min over $k$ -dimensional $S$ of $\max_{S ∖ {0}} Q$ is at most $λ_{k}$ . Conversely, for any $k$ -dimensional $S \subseteq H_{0}^{1} (Ω)$ the $(k - 1)$ linear functionals $u \mapsto (u, e_{j})_{L^{2}}$ , $j < k$ , have a nonzero common zero $u \in S ∖ {0}$ by [F2]; then $c_{j} = 0$ for $j < k$ , so $Q (u)$ is a weighted average of $λ_{k}, λ_{k + 1}, \dots$ and is at least $λ_{k}$ . Hence every $k$ -dimensional $S$ contains a direction of quotient at least $λ_{k}$ , so the minimum is exactly $λ_{k}$ , attained at $S_{k}$ ; this proves the first displayed identity.

2.2F1F2step 1.1givenalgebra

The max-inf identity. Let $T_{k - 1} : = span {e_{1}, \dots, e_{k - 1}}$ (the zero subspace for $k = 1$ ); on $H_{0}^{1} (Ω) \cap T_{k - 1}^{⊥_{L^{2}}}$ the quotient is a weighted average of $λ_{k}, λ_{k + 1}, \dots$ by [F1], so its infimum equals $λ_{k}$ , attained at $u = e_{k}$ ; hence the outer maximum is at least $λ_{k}$ . Conversely, let $T$ be any $(k - 1)$ -dimensional subspace and choose a basis $t_{1}, \dots, t_{k - 1}$ (the empty basis when $k = 1$ ). The linear map $J : span {e_{1}, \dots, e_{k}} \to K^{k - 1}$ given by $J (u) = ((u, t_{1})_{L^{2}}, \dots, (u, t_{k - 1})_{L^{2}})$ has a nonzero kernel by [F2], since its domain has dimension $k$ and its codomain has dimension $k - 1$ . A nonzero $u$ in that kernel lies in $H_{0}^{1} (Ω) \cap T^{⊥_{L^{2}}} \cap span {e_{1}, \dots, e_{k}}$ ; its quotient $Q (u)$ is a weighted average of $λ_{1}, \dots, λ_{k}$ and is therefore at most $λ_{k}$ . Thus the infimum over $H_{0}^{1} (Ω) \cap T^{⊥_{L^{2}}}$ is at most $λ_{k}$ for every $T$ . Hence the outer maximum is exactly $λ_{k}$ , attained at $T = T_{k - 1}$ .

3.1F1step 2.1step 2.2given∎

Attainment. The extreme subspaces $S_{k}$ and $T_{k - 1}$ are explicit finite-dimensional spans of the eigenbasis, and the values $λ_{k}$ are attained at $e_{k}$ ; no smoothness of $\partial Ω$ entered the argument, which uses only the eigenbasis expansion and linear algebra.

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The Poincare constant is the reciprocal square root of the first Dirichlet eigenvalue

Statement

Assume the Axiom of Choice and Countable Choice. Let $Ω \subseteq R^{n}$ be nonempty bounded open and consider the Dirichlet Laplacian, i.e. the symmetric case with $a^{i j} = δ^{i j}$ , $b = 0$ , $c = 0$ (Uniformly elliptic divergence-form operators and their sesquilinear forms). Then its first eigenvalue satisfies $λ_{1} > 0$ and $λ_{1} = \min_{u \in H_{0}^{1} (Ω) ∖ {0}} \frac{∥ D u ∥_{L^{2}}^{2}}{∥ u ∥_{L^{2}}^{2}}, ∥ u ∥_{L^{2} (Ω)} \leq λ_{1}^{- 1 / 2} ∥ D u ∥_{L^{2} (Ω)} (u \in H_{0}^{1} (Ω)),$ with equality for nonzero $u$ exactly at the nonzero first eigenfunctions; $u = 0$ is also the trivial equality case. Hence $λ_{1}^{- 1 / 2}$ is the optimal (smallest) constant in the $L^{2}$ zero-trace Poincare inequality on $Ω$ : every constant $C$ with $∥ u ∥_{L^{2}} \leq C ∥ D u ∥_{L^{2}}$ for all $u \in H_{0}^{1} (Ω)$ satisfies $C \geq λ_{1}^{- 1 / 2}$ , and the positive admissible constant $C_{P}$ of The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction at $p = 2$ therefore satisfies $C_{P} \geq λ_{1}^{- 1 / 2}$ . No numerical value or domain formula for $λ_{1}$ is asserted.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the Dirichlet Laplacian form $a (u, v) = \int_{Ω} D u \cdot \overline{D v} d x$ on $H_{0}^{1} (Ω)$ with eigenvalues $λ_{j}$ and eigenbasis ${e_{j}}$ .

[F1]

Rayleigh principle: for the symmetric case, $λ_{1} = \min_{u \neq 0} a (u, u) / ∥ u ∥_{L^{2}}^{2}$ , attained exactly on the nonzero elements of the first eigenspace, and $λ_{1}$ is the smallest weak eigenvalue; for the Dirichlet Laplacian $a (u, u) = ∥ D u ∥_{L^{2}}^{2}$ (The Rayleigh principle for the first Dirichlet eigenvalue, The $L^{2}$ operator associated with a symmetric elliptic form, Discrete spectrum of a symmetric elliptic Dirichlet operator, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F2]

Since $Ω$ is bounded, it lies in a finite-width slab. The supplier at $p = 2$ gives a finite Poincare constant for every $u \in W_{0}^{1, 2} (Ω; C) = H_{0}^{1} (Ω)$ ; enlarge it if necessary and fix a positive admissible $C_{P}$ , so $∥ u ∥_{L^{2} (Ω)} \leq C_{P} ∥ D u ∥_{L^{2} (Ω)}$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions, The Axiom of Choice).

Proof

technique · direct

1.1F1F2givenalgebra

Positivity and the minimum. With $a (u, u) = ∥ D u ∥_{L^{2}}^{2}$ , the Rayleigh principle [F1] identifies $λ_{1}$ with the displayed minimum, attained exactly on the nonzero first eigenfunctions. Fix the positive admissible $C_{P}$ of [F2]. For every $u \neq 0$ , Poincare gives $∥ D u ∥_{L^{2}}^{2} / ∥ u ∥_{L^{2}}^{2} \geq C_{P}^{- 2} > 0$ , so $λ_{1} \geq C_{P}^{- 2} > 0$ .

2.1F1step 1.1givenalgebra

The inequality and its equality cases. For nonzero $u \in H_{0}^{1} (Ω)$ , the identity $λ_{1} = \min_{v \neq 0} ∥ D v ∥^{2} / ∥ v ∥^{2}$ gives $∥ D u ∥_{L^{2}}^{2} \geq λ_{1} ∥ u ∥_{L^{2}}^{2}$ , hence $∥ u ∥_{L^{2}} \leq λ_{1}^{- 1 / 2} ∥ D u ∥_{L^{2}}$ . Equality for nonzero $u$ holds exactly when its Rayleigh quotient equals $λ_{1}$ , which by [F1] is exactly at the nonzero first eigenfunctions. At $u = 0$ both sides are zero.

3.1F2step 2.1givenalgebra∎

Optimality. Let $C$ be any constant with $∥ u ∥_{L^{2}} \leq C ∥ D u ∥_{L^{2}}$ for all $u \in H_{0}^{1} (Ω)$ . Testing at a nonzero first eigenfunction $e_{1}$ , step 2.1 gives $∥ e_{1} ∥_{L^{2}} = λ_{1}^{- 1 / 2} ∥ D e_{1} ∥_{L^{2}} \leq C ∥ D e_{1} ∥_{L^{2}}$ . Moreover $∥ D e_{1} ∥_{L^{2}} > 0$ : if it were zero, [F2] would imply $∥ e_{1} ∥_{L^{2}} \leq C_{P} ∥ D e_{1} ∥_{L^{2}} = 0$ , contradicting $e_{1} \neq 0$ . Thus $C \geq λ_{1}^{- 1 / 2}$ . Hence $λ_{1}^{- 1 / 2}$ is the smallest admissible constant, and the particular constant $C_{P}$ of the zero-trace Poincare inequality satisfies $C_{P} \geq λ_{1}^{- 1 / 2}$ .

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Non-invertible elliptic shifts form a discrete set in the self-adjoint case

Statement

Assume the Axiom of Choice and Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form, let the scalar field be $K \in {R, C}$ and let $Ω$ be nonempty, bounded and open. Write $L, D (L)$ for the symmetric-case operator. Define the complex Hilbert space $\tilde{H} : = L^{2} (Ω; C)$ and the complex operator $\tilde{L}$ as follows: if $K = C$ , set $\tilde{L} = L$ ; if $K = R$ , use the canonical isometric identification $L^{2} (Ω; R)_{C} ≅ L^{2} (Ω; C)$ and set $\tilde{L} = L_{C}$ , the complexification $L_{C} (u + i v) = L u + i L v$ on $D (L) + i D (L)$ (The complex $L^{2}$ pairing on equivalence classes, Complex Lp classes and Euclidean test-function conventions, Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding, Complexification of a real-linear map, The symmetric elliptic form operator is self-adjoint with compact resolvent). Let ${λ_{j}}$ be the eigenvalues of Discrete spectrum of a symmetric elliptic Dirichlet operator, repeated according to multiplicity. For every real $λ$ , the base-field operator $L - λ : D (L) \to L^{2} (Ω; K)$ is bijective with bounded inverse if and only if $λ \notin {λ_{j}}$ . For such $λ$ the inverse $R_{λ} : = (L - λ)^{- 1}$ , in the adopted $L - λ$ convention, is given by the convergent series $R_{λ} f = \sum_{j \geq 1} \frac{(f, e_{j})_{L^{2}}}{λ_{j} - λ} e_{j} (f \in L^{2} (Ω; K)),$ which converges in $L^{2} (Ω; K)$ and in $H_{0}^{1} (Ω; K)$ , and $∥ R_{λ} ∥ = 1 / dist (λ, {λ_{j}})$ . In the real case this inverse complexifies to $(\tilde{L} - λ)^{- 1}$ with the same operator norm, and conversely the complex resolvent at a real $λ$ restricts to the real inverse. Finally, the complex spectrum is $σ (\tilde{L}) = {λ_{j}}$ : a closed discrete subset of $R$ , bounded below, unbounded above, with no finite accumulation point; the nonreal resolvent exclusion follows from Resolvent of a self-adjoint operator: nonreal resolvents and the estimate.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the symmetric divergence-form case with operator $L$ and form $a$ ; the orthonormal eigenbasis ${e_{j}}$ and nondecreasing eigenvalue list $λ_{j} \to + \infty$ of the discrete spectral theorem; a real $λ$ ; and $f \in L^{2} (Ω)$ .

[F1]

Eigenbasis expansion: for $u \in H_{0}^{1} (Ω)$ and $g \in L^{2} (Ω)$ , $u = \sum_{j} (u, e_{j})_{L^{2}} e_{j}$ in $H_{0}^{1}$ and $g = \sum_{j} (g, e_{j})_{L^{2}} e_{j}$ in $L^{2}$ , with $∥ g ∥_{L^{2}}^{2} = \sum_{j} ∣ (g, e_{j})_{L^{2}} ∣^{2}$ and $a_{μ} (u, u) = \sum_{j} (λ_{j} + μ) ∣ (u, e_{j})_{L^{2}} ∣^{2}$ (Eigenbasis expansion in the form norm, Discrete spectrum of a symmetric elliptic Dirichlet operator).

[F2]

The distinct eigenvalues of $L$ are exactly the list ${λ_{j}}$ , the list is nondecreasing with $λ_{j} \to + \infty$ , and every weak eigenpair occurs there (Discrete spectrum of a symmetric elliptic Dirichlet operator, Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

[F3]

Form norm: $a_{μ}$ is a complete inner product on $H_{0}^{1} (Ω)$ equivalent to the standard Sobolev norm, with $a_{μ} (w, w) \geq α ∥ w ∥_{H_{0}^{1}}^{2}$ for $α = θ / 2$ ; also $a$ is bounded on $H_{0}^{1} (Ω)$ and $a_{μ} (e_{j}, e_{j}) = λ_{j} + μ > 0$ for each normalized eigenfunction (A sufficiently large shift is coercive, Hilbert space, The $L^{2}$ operator associated with a symmetric elliptic form, The shifted elliptic solution operator).

[F4]

Resolvent convention: for a complex operator $\tilde{L}$ , membership in the resolvent set means bijectivity of $\tilde{L} - λ$ with an everywhere-defined bounded inverse, whose negative is the library resolvent $(λ - \tilde{L})^{- 1}$ (Resolvent and spectrum of an unbounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). In the real case the canonical Hilbert complexification has $∥ u + i v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2}$ , so a real bounded inverse complexifies to a bounded inverse with the same norm (The symmetric elliptic form operator is self-adjoint with compact resolvent).

[F5]

Nonreal resolvent exclusion: $\tilde{L}$ is self-adjoint on the complex Hilbert space $\tilde{H}$ , so every nonreal $z$ belongs to its resolvent set and $σ (\tilde{L}) \subseteq R$ (The symmetric elliptic form operator is self-adjoint with compact resolvent, Resolvent of a self-adjoint operator: nonreal resolvents and the estimate).

Proof

technique · direct

1.1F1F2F4givenalgebra

The candidate series. Suppose $λ \notin {λ_{j}}$ . Since $λ_{j} \to + \infty$ and the distinct eigenvalues have no finite accumulation point, the set of distinct eigenvalues is closed and its distance $δ : = dist (λ, {λ_{j}})$ to $λ$ is positive. Set $c_{j} : = (f, e_{j})_{L^{2}} (λ_{j} - λ)^{- 1}$ . Then $∣ c_{j} ∣ \leq δ^{- 1} ∣ (f, e_{j})_{L^{2}} ∣$ , and [F1] gives $\sum_{j} ∣ c_{j} ∣^{2} \leq δ^{- 2} \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2} = δ^{- 2} ∥ f ∥_{L^{2}}^{2} < \infty$ ; hence $u : = \sum_{j} c_{j} e_{j}$ converges in $L^{2} (Ω)$ with $∥ u ∥_{L^{2}} \leq δ^{- 1} ∥ f ∥_{L^{2}}$ .

2.1F1F2F3step 1.1givenalgebra

Strong form convergence and the equation. For $M < N$ , form orthogonality of the eigenfunctions gives $a_{μ} (u_{N} - u_{M}, u_{N} - u_{M}) = \sum_{M < j \leq N} (λ_{j} + μ) ∣ c_{j} ∣^{2} .$ The ratio $(λ_{j} + μ) (λ_{j} - λ)^{- 2}$ is bounded over $j$ (the denominator is nonzero and quadratic growth dominates the linear numerator), so Parseval [F1] gives $\sum_{j} (λ_{j} + μ) ∣ c_{j} ∣^{2} \leq C \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2} = C ∥ f ∥_{L^{2}}^{2} < \infty .$ Its tails tend to zero, so $(u_{N})$ is Cauchy in the $a_{μ}$ norm; by [F3] this norm is complete and equivalent to $H_{0}^{1}$ , hence $u_{N}$ converges strongly in $H_{0}^{1}$ to some $\tilde{u}$ . The continuous inclusion $H_{0}^{1} ↪ L^{2}$ and the $L^{2}$ convergence of step 1.1 identify $\tilde{u} = u$ , so $u \in H_{0}^{1}$ and $u_{N} \to u$ strongly there. For each $v \in H_{0}^{1} (Ω)$ , boundedness of $a$ and the eigenrelations give $a (u, v) = \lim_{N} a (u_{N}, v) = \lim_{N} \sum_{j \leq N} [(f, e_{j})_{L^{2}} + λ c_{j}] (e_{j}, v)_{L^{2}} = (f, v)_{L^{2}} + λ (u, v)_{L^{2}},$ where the last equality uses the $L^{2}$ basis expansions of $f$ , $u$ and $v$ . Thus $a (u, v) = (f + λ u, v)_{L^{2}}$ for all $v$ , so $u \in D (L)$ and $(L - λ) u = f$ .

3.1F2step 1.1step 2.1given

Bijectivity. If $λ = λ_{k}$ for some $k$ , the eigenfunction $e_{k} \neq 0$ satisfies $(L - λ_{k}) e_{k} = 0$ , so $L - λ$ is not injective and hence not bijective. If $λ \notin {λ_{j}}$ , step 2.1 produces a solution of $(L - λ) u = f$ for every $f \in L^{2} (Ω)$ , so $L - λ$ is surjective; it is injective, because $(L - λ) u = 0$ makes $u$ a weak eigenfunction with eigenvalue $λ$ , forcing $λ \in {λ_{j}}$ by [F2] or $u = 0$ . The solution estimate of step 1.1 gives $∥ (L - λ)^{- 1} f ∥_{2} \leq δ^{- 1} ∥ f ∥_{2}$ , so $L - λ$ is bijective with bounded inverse exactly for $λ \notin {λ_{j}}$ .

4.1F1F4step 1.1step 3.1givenalgebra

The inverse series and its norm. For $λ \notin {λ_{j}}$ the series of step 1.1 has coefficients $(f, e_{j})_{L^{2}} (λ_{j} - λ)^{- 1}$ , so the solution is $R_{λ} f = \sum_{j} (f, e_{j})_{L^{2}} (λ_{j} - λ)^{- 1} e_{j}$ , converging in $L^{2}$ and, by step 2.1, with $H_{0}^{1}$ membership; its $L^{2}$ norm satisfies $∥ R_{λ} f ∥_{L^{2}}^{2} = \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2} ∣ λ_{j} - λ ∣^{- 2} \leq δ^{- 2} ∥ f ∥_{L^{2}}^{2}$ . Choose $k$ with $∣ λ_{k} - λ ∣ = δ$ (attained because the eigenvalue set is closed); testing at $f = e_{k}$ gives $∥ R_{λ} e_{k} ∥_{L^{2}} = 1 / δ$ , so the operator norm is exactly $∥ R_{λ} ∥ = 1 / δ = 1 / dist (λ, {λ_{j}})$ .

5.1F2F4F5step 3.1step 4.1given∎

Complex spectrum. By [F5], every nonreal scalar is in $ρ (\tilde{L})$ . For a real $λ \notin {λ_{j}}$ , step 3.1 gives a bounded inverse for $L - λ$ over the base field; if $K = C$ this is directly the complex resolvent, while if $K = R$ its complexification is a bounded inverse of $\tilde{L} - λ$ . Conversely, each $λ_{j}$ is an eigenvalue, so $\tilde{L} - λ_{j}$ is not injective (in the real case, complexify its nonzero real eigenfunction). Therefore $σ (\tilde{L}) = {λ_{j}}$ , which is discrete with no finite accumulation point because $λ_{j} \to + \infty$ , bounded below by $λ_{1} \geq - β$ from the discrete spectral theorem, and unbounded above because $λ_{j} \to + \infty$ .

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The elliptic resolvent identity

Statement

Assume the Axiom of Choice and Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form, let the scalar field be $K \in {R, C}$ and let $Ω$ be bounded and open. Define $\tilde{H} : = L^{2} (Ω; C)$ and $\tilde{L} = L$ if $K = C$ ; if $K = R$ , use the canonical isometric identification $L^{2} (Ω; R)_{C} ≅ L^{2} (Ω; C)$ and set $\tilde{L} = L_{C}$ , the complexification $L_{C} (u + i v) = L u + i L v$ on $D (L) + i D (L)$ (The complex $L^{2}$ pairing on equivalence classes, Complex Lp classes and Euclidean test-function conventions, Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding, Complexification of a real-linear map, The symmetric elliptic form operator is self-adjoint with compact resolvent). Write $σ (\tilde{L})$ for its complex spectrum as in Resolvent and spectrum of an unbounded operator. For $z \notin σ (\tilde{L})$ put $R_{z} : = (\tilde{L} - z)^{- 1}$ in the adopted $\tilde{L} - z$ convention, so that $R_{z} : \tilde{H} \to D (\tilde{L})$ is bijective onto $D (\tilde{L})$ with $(\tilde{L} - z) R_{z} = I$ on $\tilde{H}$ and $R_{z} (\tilde{L} - z) = I$ on $D (\tilde{L})$ . Then for all $z, w \notin σ (\tilde{L})$ $R_{z} - R_{w} = (z - w) R_{z} R_{w} = (z - w) R_{w} R_{z},$ the identities holding on all of $\tilde{H}$ ; in particular $R_{z} R_{w} = R_{w} R_{z}$ . Each $R_{z}$ is compact on $\tilde{H}$ .

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a bounded open set $Ω \subseteq R^{n}$ ; the symmetric-case operator $L$ and its complex realization $\tilde{L}$ with its complex spectrum $σ (\tilde{L})$ ; complex numbers $z, w \notin σ (\tilde{L})$ ; and the resolvents $R_{z}, R_{w}$ .

[F1]

Resolvent data: for $z \notin σ (\tilde{L})$ the operator $\tilde{L} - z : D (\tilde{L}) \to \tilde{H}$ is bijective with bounded inverse $R_{z}$ , $R_{z}$ maps $\tilde{H}$ into $D (\tilde{L})$ , $(\tilde{L} - z) R_{z} = I$ on $\tilde{H}$ and $R_{z} (\tilde{L} - z) = I$ on $D (\tilde{L})$ (Resolvent and spectrum of an unbounded operator).

[F2]

Compact resolvent: each $R_{z}$ is compact on $\tilde{H}$ (The symmetric elliptic form operator is self-adjoint with compact resolvent, Compact linear operator, The Axiom of Choice).

[F3]

Composition conventions: products of the resolvents in either order are defined on all of $\tilde{H}$ because $R_{z}, R_{w}$ map $\tilde{H}$ into $D (\tilde{L})$ , on which the other resolvent is defined; a general second-resolvent identity for closed operators is available for comparison (Second resolvent identity for a closed perturbation, The $L^{2}$ operator associated with a symmetric elliptic form, Resolvent and spectrum of an unbounded operator).

Proof

technique · direct

1.1F1F3givenalgebra

The identity. On $\tilde{H}$ insert the two inverse relations of [F1]: $R_{z} - R_{w} = R_{z} (\tilde{L} - w) R_{w} - R_{z} (\tilde{L} - z) R_{w} = R_{z} [(\tilde{L} - w) - (\tilde{L} - z)] R_{w} = (z - w) R_{z} R_{w},$ where the first equality uses $(\tilde{L} - w) R_{w} = I$ and $R_{z} (\tilde{L} - z) = I$ ; all products are everywhere defined by [F3]. Exchanging $z, w$ gives $R_{z} - R_{w} = (z - w) R_{w} R_{z}$ ; comparing the two expressions gives $(z - w) R_{z} R_{w} = (z - w) R_{w} R_{z}$ . If $z \neq w$ , divide by $z - w$ ; if $z = w$ , the products are identical.

2.1F2step 1.1given∎

Compactness. Each $R_{z}$ is compact on $\tilde{H}$ by [F2]; the resolvent identity itself is an operator identity on all of $\tilde{H}$ and involves no compactness, and no choice beyond [F1] and [F2] is used.

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Eigenfunctions for distinct symmetric elliptic eigenvalues are $L^{2}$ -orthogonal

Statement

Assume Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form, let $(λ, u)$ and $(μ, v)$ be weak eigenpairs as in Symmetric elliptic weak eigenpairs with $λ \neq μ$ . Then $(u, v)_{L^{2}} = 0$ . If the scalar field is $C$ and the coefficients are real, conjugation preserves weak eigenpairs at the same eigenvalue; each nonzero real or imaginary part of an eigenfunction is then a real-valued weak eigenfunction, so an eigenfunction can be chosen real.

Facts & Assumptions

Given: Countable Choice; the symmetric divergence-form case of The $L^{2}$ operator associated with a symmetric elliptic form; weak eigenpairs $(λ, u)$ and $(μ, v)$ with $λ \neq μ$ and $u, v \neq 0$ .

[F1]

Weak eigenpair equations: $a (u, w) = λ (u, w)_{L^{2}}$ and $a (v, w) = μ (v, w)_{L^{2}}$ for every $w \in H_{0}^{1} (Ω)$ , with real $λ, μ$ (Symmetric elliptic weak eigenpairs).

[F2]

Symmetry: $a (u, v) = \overline{a (v, u)}$ for all arguments, and $a (w, w)$ is real (Bounded, coercive and symmetric sesquilinear forms, The $L^{2}$ operator associated with a symmetric elliptic form).

[F3]

Conjugation: for real coefficients the form satisfies $a (\overline{u}, \overline{v}) = \overline{a (u, v)}$ . With the inner product linear in its first argument, $(\overline{u}, w)_{L^{2}} = \overline{(u, \overline{w})_{L^{2}}}$ ; conjugation also preserves $H_{0}^{1} (Ω)$ (Real and imaginary parts, complex conjugation, and modulus, The $L^{2}$ operator associated with a symmetric elliptic form, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1F1F2givenalgebra

Test the eigenequation of $(λ, u)$ at $w = v$ and that of $(μ, v)$ at $w = u$ : [F1] gives $λ (u, v)_{L^{2}} = a (u, v)$ and $μ (v, u)_{L^{2}} = a (v, u)$ . Conjugating the second identity and using symmetry [F2], $μ \overline{(v, u)_{L^{2}}} = \overline{a (v, u)} = a (u, v)$ ; since $\overline{(v, u)_{L^{2}}} = (u, v)_{L^{2}}$ and $λ, μ$ are real, comparison gives $λ (u, v)_{L^{2}} = μ (u, v)_{L^{2}}$ , that is $(λ - μ) (u, v)_{L^{2}} = 0$ . As $λ \neq μ$ and the scalar field is $R$ or $C$ , $(u, v)_{L^{2}} = 0$ .

2.1F1F3step 1.1givenalgebra∎

Real coefficients. Suppose the scalar field is $C$ and the coefficients $a^{i j}, c$ are real (with $b = 0$ ). For $w \in H_{0}^{1} (Ω)$ , [F3] gives $a (\overline{u}, w) = \overline{a (u, \overline{w})} = \overline{λ (u, \overline{w})_{L^{2}}} = λ (\overline{u}, w)_{L^{2}}$ , so $\overline{u}$ is a weak eigenfunction with eigenvalue $λ$ . By linearity, each nonzero one of $Re u = \frac{1}{2} (u + \overline{u})$ and $Im u = \frac{1}{2 i} (u - \overline{u})$ is a real-valued weak eigenfunction at $λ$ ; since $u \neq 0$ , at least one is nonzero, so an eigenfunction can be chosen real. The orthogonality conclusion of step 1.1 is independent of this representative remark.

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Spectral series solution of an invertible symmetric elliptic problem

Statement

Assume the Axiom of Choice and Countable Choice. In the symmetric case of The $L^{2}$ operator associated with a symmetric elliptic form with $Ω$ nonempty bounded open, suppose $0$ is not an eigenvalue of $L$ (equivalently, by Discrete spectrum of a symmetric elliptic Dirichlet operator, no nonzero $u \in H_{0}^{1} (Ω)$ satisfies $a (u, v) = 0$ for all $v$ ; this holds in particular when $a$ is coercive on $H_{0}^{1} (Ω)$ ). Then $L : D (L) \to L^{2} (Ω)$ is bijective, and for every $f \in L^{2} (Ω)$ the unique weak solution $u \in H_{0}^{1} (Ω)$ of $a (u, v) = (f, v)_{L^{2}}$ for all $v$ is $u = L^{- 1} f = \sum_{j \geq 1} \frac{(f, e_{j})_{L^{2}}}{λ_{j}} e_{j},$ the series converging in $L^{2} (Ω)$ and in $H_{0}^{1} (Ω)$ ; moreover $u \in D (L)$ with $L u = f$ and $∥ u ∥_{L^{2}} \leq (\min_{j} ∣ λ_{j} ∣)^{- 1} ∥ f ∥_{L^{2}}$ (and $∥ u ∥_{L^{2}} \leq λ_{1}^{- 1} ∥ f ∥_{L^{2}}$ when $λ_{1} > 0$ , in particular under coercivity of $a$ ).

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the symmetric divergence-form case with form $a$ and operator $L$ ; the eigenbasis ${e_{j}}$ and eigenvalue list $λ_{j} \to + \infty$ of the discrete spectral theorem; the hypothesis that $0$ is not an eigenvalue; and $f \in L^{2} (Ω)$ .

[F1]

Spectral series for the inverse at $λ = 0$ : the complex spectrum of the relevant complex realization is $σ (\tilde{L}) = {λ_{j}}$ , and the corollary gives the base-field inverse series for every real parameter outside this list. Since $0$ is not an eigenvalue, $L$ is bijective with bounded inverse $L^{- 1}$ , and $L^{- 1} f = \sum_{j} (f, e_{j})_{L^{2}} λ_{j}^{- 1} e_{j}$ with convergence in $L^{2} (Ω)$ and in $H_{0}^{1} (Ω)$ (Non-invertible elliptic shifts form a discrete set in the self-adjoint case, Discrete spectrum of a symmetric elliptic Dirichlet operator).

[F2]

Weak solutions: $u \in H_{0}^{1} (Ω)$ is a weak solution of the Dirichlet problem with datum $f \in L^{2} (Ω)$ exactly when $u \in D (L)$ and $L u = f$ (The $L^{2}$ operator associated with a symmetric elliptic form, Weak Dirichlet solutions for a divergence-form operator).

[F3]

Parseval: $∥ f ∥_{L^{2}}^{2} = \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2}$ for every $f \in L^{2} (Ω)$ , and the eigenbasis is orthonormal (Eigenbasis expansion in the form norm, Discrete spectrum of a symmetric elliptic Dirichlet operator).

[F4]

Rayleigh: if $λ_{1} > 0$ then all $λ_{j} \geq λ_{1}$ , and coercivity of $a$ with constant $α^{'} > 0$ implies $λ_{1} \geq α^{'} > 0$ and hence that $0$ is not an eigenvalue (The Rayleigh principle for the first Dirichlet eigenvalue, The $L^{2}$ operator associated with a symmetric elliptic form).

Proof

technique · direct

1.1F1given

Bijectivity and the series. The hypothesis says $0$ is not a weak eigenvalue, so $L$ is injective; since the eigenvalues of $L$ are exactly the list ${λ_{j}}$ , the resolvent corollary [F1] applies with $λ = 0$ and gives that $L$ is bijective with bounded inverse and that the inverse is the displayed series, converging in $L^{2} (Ω)$ and in $H_{0}^{1} (Ω)$ .

2.1F1F2step 1.1given

The weak solution. For $f \in L^{2} (Ω)$ put $u : = L^{- 1} f$ , which lies in $D (L)$ with $L u = f$ ; by [F2] $u$ is the unique weak solution of $a (u, v) = (f, v)_{L^{2}}$ for all $v$ , and by step 1.1 it is the series of the statement. Since $L$ is injective with range $L^{2} (Ω)$ , the weak solution is unique, so this identifies the solution set with the single class $u$ .

2.2F3step 1.1givenalgebra

The norm bound. Put $δ : = \inf_{j} ∣ λ_{j} ∣ > 0$ (positive because $λ_{j} \to + \infty$ and no $λ_{j} = 0$ ). The series of step 1.1 and Parseval [F3] give $∥ u ∥_{L^{2}}^{2} = \sum_{j} \frac{∣ (f, e_{j})_{L^{2}} ∣^{2}}{λ_{j}^{2}} \leq δ^{- 2} \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2} = δ^{- 2} ∥ f ∥_{L^{2}}^{2},$ that is $∥ u ∥_{L^{2}} \leq (\min_{j} ∣ λ_{j} ∣)^{- 1} ∥ f ∥_{L^{2}}$ . If $λ_{1} > 0$ all $λ_{j} \geq λ_{1} > 0$ , so $\min_{j} ∣ λ_{j} ∣ = λ_{1}$ and the sharper bound $∥ u ∥_{L^{2}} \leq λ_{1}^{- 1} ∥ f ∥_{L^{2}}$ holds.

3.1F4step 1.1step 2.1step 2.2given∎

Coercivity gives the hypothesis. If $a$ is coercive with constant $α^{'} > 0$ then $a (u, u) \geq α^{'} ∥ u ∥_{L^{2}}^{2} > 0$ for every nonzero $u$ , so $0$ cannot be a weak eigenvalue and the previous conclusions apply; by [F4] one also has $λ_{1} \geq α^{'} > 0$ , so the sharper bound of step 2.2 is available.

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The first Dirichlet eigenvalue is monotone under domain inclusion

Statement

Assume the Axiom of Choice and Countable Choice. Let $Ω_{1} \subseteq Ω_{2} \subseteq R^{n}$ be nonempty bounded open sets, let $a^{i j}$ be measurable, essentially bounded and uniformly elliptic on $Ω_{2}$ with $a^{i j} = \overline{a^{j i}}$ , and for $i = 1, 2$ let $a^{(i)}$ be the principal Dirichlet form $a^{(i)} (u, v) = \int_{Ω_{i}} a^{i j} D_{j} u \overline{D_{i} v} d x$ on $H_{0}^{1} (Ω_{i})$ ; the model case $a^{i j} = δ^{i j}$ is the Dirichlet Laplacian. Let $λ_{k} (Ω_{i})$ be the eigenvalues of the corresponding symmetric elliptic Dirichlet operator (Discrete spectrum of a symmetric elliptic Dirichlet operator) . Then $λ_{k} (Ω_{2}) \leq λ_{k} (Ω_{1}) for every k \geq 1,$ in particular $λ_{1} (Ω_{2}) \leq λ_{1} (Ω_{1})$ : making the domain smaller raises the Dirichlet frequencies. The mechanism is that extension by zero maps $H_{0}^{1} (Ω_{1})$ isometrically into $H_{0}^{1} (Ω_{2})$ for the energy form and preserves the $L^{2}$ norm (Zero extension of W_0^{1,p} has no boundary derivative), so every $k$ -dimensional competitor in $H_{0}^{1} (Ω_{1})$ is a competitor in $H_{0}^{1} (Ω_{2})$ with the same Rayleigh quotient.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; nonempty bounded open sets $Ω_{1} \subseteq Ω_{2} \subseteq R^{n}$ ; Hermitian uniformly elliptic coefficients $a^{i j}$ on $Ω_{2}$ ; the principal forms $a^{(1)}$ on $H_{0}^{1} (Ω_{1})$ and $a^{(2)}$ on $H_{0}^{1} (Ω_{2})$ ; and $k \geq 1$ .

[F1]

Zero extension: the extension operator $E$ sending a class $u \in W_{0}^{1, 2} (Ω_{1})$ to its zero extension is linear and isometric for the Sobolev norm, with weak derivatives the zero extensions of the weak derivatives (Zero extension of W_0^{1,p} has no boundary derivative, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms).

[F2]

Because $Ω_{1} \subseteq Ω_{2}$ , the inclusion $C_{c}^{\infty} (Ω_{1}) \subseteq C_{c}^{\infty} (Ω_{2})$ induces $E (H_{0}^{1} (Ω_{1})) \subseteq H_{0}^{1} (Ω_{2})$ , and for $u \in H_{0}^{1} (Ω_{1})$ the integrals of the coefficient form over $Ω_{2}$ see only $Ω_{1}$ : $a^{(2)} (E u, E v) = a^{(1)} (u, v)$ and $∥ E u ∥_{L^{2} (Ω_{2})} = ∥ u ∥_{L^{2} (Ω_{1})}$ (Uniformly elliptic divergence-form operators and their sesquilinear forms, The $L^{2}$ operator associated with a symmetric elliptic form, Zero-boundary Sobolev space as a norm closure).

[F3]

Courant--Fischer: for $i = 1, 2$ and every $k \geq 1$ , $λ_{k} (Ω_{i}) = \min {\max_{u \in S ∖ {0}} a^{(i)} (u, u) / ∥ u ∥_{L^{2} (Ω_{i})}^{2} : S \subseteq H_{0}^{1} (Ω_{i}), \dim S = k}$ , the extrema being attained (The Courant-Fischer min-max principle for elliptic eigenvalues, Discrete spectrum of a symmetric elliptic Dirichlet operator, The Rayleigh principle for the first Dirichlet eigenvalue).

Proof

technique · direct

1.1F1F2givenalgebra

Isometry of the extension. By [F1] the map $E$ is linear and isometric for the Sobolev norm, and by [F2] its image lies in $H_{0}^{1} (Ω_{2})$ and the coefficient form and $L^{2}$ norm are preserved: for every $u \in H_{0}^{1} (Ω_{1})$ , $a^{(2)} (E u, E u) = a^{(1)} (u, u), ∥ E u ∥_{L^{2} (Ω_{2})}^{2} = ∥ u ∥_{L^{2} (Ω_{1})}^{2} .$ In particular the Rayleigh quotients agree, $a^{(2)} (E u, E u) / ∥ E u ∥_{L^{2} (Ω_{2})}^{2} = a^{(1)} (u, u) / ∥ u ∥_{L^{2} (Ω_{1})}^{2}$ for $u \neq 0$ .

2.1F3step 1.1givenalgebra∎

Min-max comparison. Fix $k \geq 1$ and let $S_{i}$ be the family of $k$ -dimensional subspaces of $H_{0}^{1} (Ω_{i})$ ; by [F3], $λ_{k} (Ω_{i}) = \min_{S \in S_{i}} \max_{u \in S ∖ {0}} Q_{i} (u)$ with $Q_{i}$ the corresponding Rayleigh quotient. The extension $E$ maps $S_{1}$ into $S_{2}$ (linear isometry preserves dimension), and the quotients agree on corresponding vectors by step 1.1, so $λ_{k} (Ω_{2}) \leq \min_{S \in S_{1}} \max_{u \in S ∖ {0}} Q_{2} (E u) = \min_{S \in S_{1}} \max_{u \in S ∖ {0}} Q_{1} (u) = λ_{k} (Ω_{1}),$ the inequality holding because the minimum over the larger family $S_{2}$ is at most the minimum over the restricted family $E (S_{1})$ . This proves the monotonicity $λ_{k} (Ω_{2}) \leq λ_{k} (Ω_{1})$ for every $k$ ; the case $k = 1$ is the statement about the first Dirichlet eigenvalue.

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The first positive Neumann eigenvalue has the mean-zero Rayleigh characterisation

Statement

Assume the Axiom of Choice and Countable Choice. Let $Ω \subseteq R^{n}$ be a nonempty bounded connected extension domain (Sobolev extension domains and extension operators), put $V : = {u \in H^{1} (Ω) : \int_{Ω} u = 0}$ and $L_{0}^{2} (Ω) : = {f \in L^{2} (Ω) : \int_{Ω} f = 0}$ , and let $a (u, v) = \int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x$ be the principal form with Hermitian uniformly elliptic coefficients $a^{i j} = \overline{a^{j i}}$ (Uniformly elliptic divergence-form operators and their sesquilinear forms); for $a^{i j} = δ^{i j}$ this is the Neumann form of the Laplacian. Then $μ_{1} : = \inf_{u \in V ∖ {0}} \frac{a (u, u)}{∥ u ∥_{L^{2}}^{2}}$ satisfies $μ_{1} > 0$ , the infimum is attained, and the minimisers are exactly the nonzero elements of the eigenspace ${u \in V : a (u, v) = μ_{1} (u, v)_{L^{2}} \forall v \in V}$ . Since both sides of that identity vanish on constants, it actually holds for every $v \in H^{1} (Ω)$ , so $μ_{1}$ is the smallest positive weak Neumann eigenvalue on the mean-zero space and no Neumann eigenvalue of a mean-zero eigenfunction lies in $(0, μ_{1})$ . Moreover $μ_{1} = 1 / ∥ S ∥$ , where the norm is that of the $L_{0}^{2} \to L_{0}^{2}$ realization of the solution map $S : L_{0}^{2} (Ω) \to V$ , which is bounded, compact, self-adjoint and positive in that realization and is defined by $a (S f, v) = (f, v)_{L^{2}}$ for all $v \in V$ . Connectedness supplies Poincare--Wirtinger, and the extension-domain hypothesis supplies that inequality and Rellich compactness; the conclusions are not asserted for arbitrary bounded connected open sets.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded connected extension domain $Ω \subseteq R^{n}$ ; Hermitian uniformly elliptic coefficients $a^{i j} = \overline{a^{j i}}$ with ellipticity constant $θ > 0$ ; the principal form $a (u, v) = \int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x$ ; the mean-zero spaces $V \subseteq H^{1} (Ω)$ and $L_{0}^{2} (Ω) \subseteq L^{2} (Ω)$ .

[F1]

Finiteness and closedness: boundedness of $Ω$ gives $∣ Ω ∣ < \infty$ (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure), so $u \mapsto \int_{Ω} u$ is a bounded linear functional on $H^{1} (Ω)$ and on $L^{2} (Ω)$ , because $∣ \int_{Ω} u ∣ \leq ∣ Ω ∣^{1 / 2} ∥ u ∥_{L^{2}} \leq ∣ Ω ∣^{1 / 2} ∥ u ∥_{H^{1}}$ by Cauchy--Schwarz (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs, Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions). The Hilbert structures are supplied by The Sobolev space $H^{1}$ is a Hilbert space and $L^{2}$ with the integral pairing is a Hilbert space. Hence $V$ and $L_{0}^{2}$ are closed Hilbert subspaces. The open nonempty set contains two disjoint positive-measure boxes by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included; subtracting appropriately weighted indicators gives a nonzero element of $L_{0}^{2}$ .

[F2]

Poincare--Wirtinger: there is $C_{W} = C_{W} (Ω, 2)$ with $∥ u - u_{Ω} ∥_{L^{2}} \leq C_{W} ∥ D u ∥_{L^{2}}$ for all $u \in H^{1} (Ω)$ , so every $u \in V$ satisfies $∥ u ∥_{L^{2}} \leq C_{W} ∥ D u ∥_{L^{2}}$ (Poincare-Wirtinger on bounded connected extension domains by Rellich compactness).

[F3]

The principal form: $a$ is sesquilinear on $H^{1} (Ω)$ and bounded, $a (u, u)$ is real for every $u$ , and $Re a (u, u) = a (u, u) \geq θ ∥ D u ∥_{L^{2}}^{2}$ by uniform ellipticity; consequently, for $u \in V$ , $a (u, u) \geq θ ∥ D u ∥_{L^{2}}^{2} \geq \frac{θ}{1 + C_{W}^{2}} ∥ u ∥_{H^{1}}^{2},$ and on the other hand $a (u, u) \leq (n M_{a}) ∥ D u ∥_{L^{2}}^{2} \leq n M_{a} ∥ u ∥_{H^{1}}^{2}$ (Uniformly elliptic divergence-form operators and their sesquilinear forms, The elliptic form is well defined and bounded on $H^{1}$ , Bounded, coercive and symmetric sesquilinear forms).

[F4]

Lax--Milgram and the solution operator: for $f \in L_{0}^{2} (Ω)$ the functional $v \mapsto (f, v)_{L^{2}}$ is bounded and conjugate-linear on $V$ , so by [F3] and The Lax--Milgram theorem there is a unique $S f \in V$ with $a (S f, v) = (f, v)_{L^{2}}$ for all $v \in V$ ; the map $S$ is linear and $∥ S f ∥_{H^{1}} \leq (1 + C_{W}^{2}) θ^{- 1} ∥ f ∥_{L^{2}}$ , using the coercivity constant $α = θ / (1 + C_{W}^{2})$ and $∥ (f, \cdot)_{L^{2}} ∥_{V^{*}} \leq ∥ f ∥_{L^{2}}$ (Bounded, coercive and symmetric sesquilinear forms).

[F5]

Self-adjointness, positivity, and injectivity: for $f, g \in L_{0}^{2} (Ω)$ , Hermitian symmetry and the defining identity give $a (S f, S g) = (f, S g)_{L^{2}}$ , while conjugating the identity for $a (S g, S f)$ gives $a (S f, S g) = (S f, g)_{L^{2}}$ ; hence $(S f, g)_{L^{2}} = (f, S g)_{L^{2}}$ and $S$ is self-adjoint. Also $(S f, f)_{L^{2}} = a (S f, S f) \geq 0$ . If $S f = 0$ , then $(f, v)_{L^{2}} = 0$ for every $v \in V$ . The density of $C_{c}^{\infty} (Ω)$ in $L^{2} (Ω)$ (Smooth compactly supported functions of an open set are dense in $L^{2}$ ) and boundedness of the mean imply that mean-zero $H^{1}$ functions are dense in $L_{0}^{2}$ : approximate $f$ by smooth compactly supported $φ_{j}$ and replace each by $φ_{j} - (φ_{j})_{Ω} 1$ . Thus $f = 0$ , so $S$ is injective and positive definite.

[F6]

Compactness: the inclusion $H^{1} (Ω) ↪ L^{2} (Ω)$ is compact on the bounded extension domain $Ω$ (Compactness of $W^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded extension domains), and $S : L_{0}^{2} (Ω) \to H^{1} (Ω)$ is bounded by [F4], so the composition $S : L_{0}^{2} (Ω) \to L_{0}^{2} (Ω)$ with the inclusion is compact (Compositions with a compact operator are compact, Compact linear operator).

[F7]

Spectral data: for the compact self-adjoint operator $S$ the nonzero eigenvalues form a finite or countably infinite set of real numbers with finite multiplicities and no accumulation point other than $0$ , eigenspaces for distinct eigenvalues are orthogonal, and with $P_{λ}$ the orthogonal projection onto $E_{λ}$ one has $\overset{}{span} ⋃_{λ} E_{λ} = (\ker S)^{⊥} = L_{0}^{2} (Ω)$ and $S x = \sum_{λ} λ P_{λ} x$ in norm (Spectral theorem for compact self adjoint operators, Eigenspaces of a self adjoint operator are orthogonal); hence $f = \sum_{λ} P_{λ} f$ and $∥ f ∥_{L^{2}}^{2} = \sum_{λ} ∥ P_{λ} f ∥_{L^{2}}^{2}$ for every $f \in L_{0}^{2} (Ω)$ (Orthogonal decomposition by a closed subspace, Fourier expansion in a Hilbert space).

[F8]

Norm and eigenvalues: all eigenvalues of $S$ are positive, and $∥ S ∥$ is the largest eigenvalue of $S$ (Norm point of a compact self adjoint operator is an eigenvalue up to sign, Compact linear operator).

[F9]

Constants: the constant function $1$ lies in $H^{1} (Ω)$ with zero weak gradient, its classical derivative being the weak derivative, so $a (u, 1) = 0$ and the mean-zero condition reads $(u, 1)_{L^{2}} = 0$ for $u \in V$ (Classical derivatives agree with weak derivatives, Integer-order Sobolev spaces and their norms, Zero weak gradient gives componentwise constants).

Proof

technique · direct

1.1F1F2F3givenalgebra

The space $V$ is closed in $H^{1} (Ω)$ and $L_{0}^{2} (Ω)$ is closed in $L^{2} (Ω)$ by [F1]; on $V$ the form $a$ is bounded and satisfies $a (u, u) \geq 0$ for every $u \in V$ by [F3]. The estimate of [F3] is exactly the coercivity statement $a (u, u) \geq α ∥ u ∥_{H^{1}}^{2}, α : = \frac{θ}{1 + C_{W}^{2}} > 0,$ for all $u \in V$ , obtained from Poincare--Wirtinger and ellipticity.

2.1F1F4F5F6step 1.1

Solution operator. For each $f \in L_{0}^{2} (Ω)$ the functional $v \mapsto (f, v)_{L^{2}}$ is bounded and conjugate-linear on $V$ by [F1], so by [F4] there is a unique $S f \in V$ with $a (S f, v) = (f, v)_{L^{2}}$ for every $v \in V$ ; the assignment $f \mapsto S f$ is linear and bounded with the explicit estimate $∥ S f ∥_{H^{1}} \leq α^{- 1} ∥ f ∥_{L^{2}} = (1 + C_{W}^{2}) θ^{- 1} ∥ f ∥_{L^{2}}$ , obtained by testing the defining identity at $v = S f$ and using [F1]. By [F5] the operator $S : L_{0}^{2} (Ω) \to L_{0}^{2} (Ω)$ is self-adjoint, positive definite and injective, and by [F6] it is compact.

3.1F7F8step 2.1algebra

Spectral decomposition. By [F7] and [F8] the nonzero eigenvalues of $S$ are positive real numbers of finite multiplicity with no accumulation point except $0$ ; enumerate the distinct eigenvalues in decreasing order as $ν_{1} > ν_{2} > \dots > 0$ when the set is infinite (with $ν_{k} ↓ 0$ ), and let $P_{k}$ be the orthogonal projection onto the eigenspace $E_{ν_{k}}$ . Since $S$ is injective, [F7] gives $L_{0}^{2} (Ω) = \overset{}{span} ⋃_{k} E_{ν_{k}}$ with orthogonal summands, so for every $f \in L_{0}^{2} (Ω)$ the net of partial sums $f_{n} : = \sum_{k \leq n} P_{k} f$ converges to $f$ in $L^{2}$ and $∥ f ∥_{L^{2}}^{2} = \sum_{k} ∥ P_{k} f ∥_{L^{2}}^{2}$ . Every eigenspace lies in $V$ , because $S g \in V$ and $g = ν^{- 1} S g$ for an eigenvector. Finally, for $g \in E_{ν_{k}}$ and every $v \in V$ one has $a (S g, v) = (g, v)_{L^{2}}$ by definition of $S$ , that is $a (g, v) = ν_{k}^{- 1} (g, v)_{L^{2}}$ .

4.1F7step 1.1step 3.1algebra

Partial sums in the form. Fix $f \in V$ and put $f_{n} = \sum_{k \leq n} P_{k} f$ as in step 3.1. Then step 3.1 gives, for every $v \in V$ , $a (f_{n}, v) = \sum_{k \leq n} ν_{k}^{- 1} (P_{k} f, v)_{L^{2}}$ ; taking $v = f_{n}$ and $v = f$ and using orthogonality of the projections, $a (f_{n}, f_{n}) = \sum_{k \leq n} ν_{k}^{- 1} ∥ P_{k} f ∥_{L^{2}}^{2} = a (f_{n}, f),$ where the last identity is real. Hence $a (f - f_{n}, f - f_{n}) = a (f, f) - a (f_{n}, f_{n}) \geq 0$ by positivity of $a$ on $V$ , and therefore $\sum_{k \leq n} ν_{k}^{- 1} ∥ P_{k} f ∥_{L^{2}}^{2} \leq a (f, f)$ for every $n$ .

5.1F7step 1.1step 4.1algebra

Form-norm expansion. The increasing partial sums of $\sum_{k} ν_{k}^{- 1} ∥ P_{k} f ∥_{L^{2}}^{2}$ are bounded by $a (f, f)$ , so the series converges; consequently, for $m < n$ , $a (f_{n} - f_{m}, f_{n} - f_{m}) = \sum_{m < k \leq n} ν_{k}^{- 1} ∥ P_{k} f ∥_{L^{2}}^{2} ⟶ 0,$ so $(f_{n})$ is Cauchy for the inner product $a$ on $V$ . By the coercivity of step 1.1 it is Cauchy in $H^{1} (Ω)$ , hence converges in $H^{1}$ to some $u \in V$ (closedness of $V$ ). Since $H^{1}$ convergence implies $L^{2}$ convergence and $f_{n} \to f$ in $L^{2}$ , we get $u = f$ ; continuity of $a$ in the $H^{1}$ norm then gives $a (f, f) = \lim_{n} a (f_{n}, f_{n}) = \sum_{k} ν_{k}^{- 1} ∥ P_{k} f ∥_{L^{2}}^{2} .$

6.1F8step 3.1step 5.1algebra

Rayleigh characterisation. Put $μ_{k} : = ν_{k}^{- 1}$ , so that by [F8] $μ_{1} = ν_{1}^{- 1} = 1 / ∥ S ∥$ is the smallest of the $μ_{k}$ and $μ_{k} > 0$ . For $f \in V ∖ {0}$ step 5.1 and $∥ f ∥_{L^{2}}^{2} = \sum_{k} ∥ P_{k} f ∥_{L^{2}}^{2}$ give $\frac{a (f, f)}{∥ f ∥_{L^{2}}^{2}} = \frac{\sum_{k} μ_{k} ∥ P_{k} f ∥_{L^{2}}^{2}}{\sum_{k} ∥ P_{k} f ∥_{L^{2}}^{2}} \geq μ_{1},$ with equality precisely when $P_{k} f = 0$ for every $k$ with $μ_{k} > μ_{1}$ , that is $f \in E_{ν_{1}}$ . Hence $μ_{1} > 0$ is attained and the minimisers are exactly the nonzero elements of $E_{ν_{1}}$ . Moreover $f \in E_{ν_{1}} ∖ {0}$ satisfies $a (f, v) = μ_{1} (f, v)_{L^{2}}$ for all $v \in V$ by step 3.1; conversely, if $0 \neq f \in V$ satisfies $a (f, v) = μ_{1} (f, v)_{L^{2}}$ for all $v \in V$ , then the same expansion gives $\sum_{k} (μ_{k} - μ_{1}) ∥ P_{k} f ∥_{L^{2}}^{2} = 0$ with all coefficients nonnegative, so $P_{k} f = 0$ whenever $μ_{k} > μ_{1}$ and $f \in E_{ν_{1}}$ . Thus the eigenspace ${u \in V : a (u, v) = μ_{1} (u, v)_{L^{2}} \forall v \in V}$ equals $E_{ν_{1}}$ , and no mean-zero weak Neumann eigenvalue $λ \in (0, μ_{1})$ exists, since it would give the same identity with a nonnegative combination vanishing.

7.1F2F6F9step 6.1givenalgebra∎

Extension to $H^{1} (Ω)$ and conclusions. Let $e_{1} \in E_{ν_{1}} ∖ {0}$ . By [F9] the constant $1$ has $a (e_{1}, 1) = 0$ and $(e_{1}, 1)_{L^{2}} = 0$ because $e_{1} \in V$ ; writing an arbitrary $v \in H^{1} (Ω)$ as $v = (v - v_{Ω}) + v_{Ω}$ with $v - v_{Ω} \in V$ , the identity $a (e_{1}, w) = μ_{1} (e_{1}, w)_{L^{2}}$ for $w = v - v_{Ω}$ therefore extends to all $v \in H^{1} (Ω)$ , which is the weak Neumann eigenequation; the same argument extends the eigenspace description of step 6.1, showing that $μ_{1}$ is the smallest positive weak Neumann eigenvalue on the mean-zero space. Together with $μ_{1} = 1 / ∥ S ∥$ from step 6.1 this proves all the assertions; connectedness is used only through Poincare--Wirtinger [F2] (a disconnected domain admits the componentwise constants in $V$ with $a = 0$ , so the infimum would be $0$ ), and the extension-domain hypothesis is used only through [F2] and [F6].

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The Neumann spectrum and the constant zero mode

Statement

Assume the Axiom of Choice (AC) and Countable Choice (CC). The explicit CC premise is used by Uniformly elliptic divergence-form operators and their sesquilinear forms, while AC matches the cited componentwise-constancy and mean-zero Neumann results as currently stated. Let $Ω \subseteq R^{n}$ be a nonempty bounded open set and let $a (u, v) = \int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x$ be the principal form with Hermitian uniformly elliptic coefficients (Uniformly elliptic divergence-form operators and their sesquilinear forms). Then $Re a (u, u) = \int_{Ω} Re (a^{i j} D_{j} u \overline{D_{i} u}) d x \geq θ ∥ D u ∥_{L^{2}}^{2} \geq 0,$ with equality if and only if $D u = 0$ a.e., i.e. if and only if $u$ is constant on each connected component of $Ω$ (Zero weak gradient gives componentwise constants). Hence the constant functions are weak Neumann eigenfunctions with eigenvalue $0$ , and on a domain with exactly $m$ connected components the zero eigenspace of the principal Neumann problem is exactly the $m$ -dimensional space of componentwise constants. In particular the lowest weak Neumann eigenvalue on $H^{1} (Ω)$ is $0$ ; a first positive eigenvalue, when it exists, lies above this zero mode. If, in addition, $Ω$ is connected and is a Sobolev extension domain, The first positive Neumann eigenvalue has the mean-zero Rayleigh characterisation gives a positive first Neumann level after restricting to the global mean-zero subspace. This positivity conclusion is not asserted for a general bounded open $Ω$ ; when $Ω$ has multiple components, nonzero componentwise constants can also have global mean zero.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; Hermitian uniformly elliptic coefficients $a^{i j}$ with ellipticity constant $θ > 0$ ; the principal form $a (u, v) = \int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x$ ; and $u \in H^{1} (Ω)$ .

[F1]

Pointwise ellipticity: $Re (a^{i j} (x) D_{j} u (x) \overline{D_{i} u (x)}) \geq θ ∣ D u (x) ∣^{2} \geq 0$ for almost every $x \in Ω$ , and $a (u, u)$ is real because the coefficients are Hermitian (Uniformly elliptic divergence-form operators and their sesquilinear forms, Integer-order Sobolev spaces and their norms).

[F2]

A nonnegative measurable function has zero integral if and only if it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[F3]

A class with zero weak gradient is constant on each connected component of $Ω$ (Zero weak gradient gives componentwise constants, Connected components, quasicomponents, and totally disconnected spaces).

[F4]

On a connected bounded Sobolev extension domain the mean-zero restriction of the principal form has a positive first level (The first positive Neumann eigenvalue has the mean-zero Rayleigh characterisation).

Proof

technique · direct

1.1F1F2F3givenalgebra

Since $Re a (u, u) = \int_{Ω} Re (a^{i j} D_{j} u \overline{D_{i} u}) d x$ and the integrand is at least $θ ∣ D u ∣^{2} \geq 0$ almost everywhere by [F1], the integral is nonnegative. If $a (u, u) = 0$ , then $0 \leq θ \int_{Ω} ∣ D u ∣^{2} \leq Re a (u, u) = 0$ , so $\int_{Ω} ∣ D u ∣^{2} = 0$ and [F2] gives $∣ D u ∣ = 0$ almost everywhere; conversely $D u = 0$ a.e. makes the integrand vanish a.e. and hence $a (u, u) = 0$ . By [F3], $D u = 0$ a.e. holds exactly when $u$ is constant on each connected component of $Ω$ .

2.1F1F3step 1.1givenalgebra

If $Ω$ has finitely many connected components, write them as $Ω_{1}, \dots, Ω_{m}$ . The componentwise constants $\sum_{j = 1}^{m} c_{j} 1_{Ω_{j}}$ with $c_{j} \in K$ form a linear subspace of $H^{1} (Ω)$ of dimension $m$ , because the components are nonempty disjoint open sets of positive finite measure and each indicator has zero weak gradient: every compactly supported test function meets only finitely many components, and its integral derivative on each component is zero; by step 1.1 this subspace is exactly the zero set of the quadratic form $Re a$ , that is, the kernel of the symmetric form $a$ on $H^{1} (Ω)$ .

3.1F1F3step 1.1step 2.1givenalgebra

Weak Neumann eigenfunctions of eigenvalue $0$ are exactly the nonzero elements of that kernel: $a (u, v) = 0$ for all $v \in H^{1} (Ω)$ implies $a (u, u) = 0$ , hence $u$ is componentwise constant by step 1.1; conversely a componentwise constant $u$ has $D u = 0$ a.e., so $a (u, v) = \int_{Ω} a^{i j} D_{j} u \overline{D_{i} v} d x = 0$ for every $v \in H^{1} (Ω)$ , and every nonzero constant function supplies such an eigenfunction. Thus $0$ is the lowest weak Neumann eigenvalue, since testing any weak eigenpair at its eigenfunction gives a nonnegative eigenvalue. Its eigenspace consists of the componentwise constants in $H^{1} (Ω)$ and has dimension $m$ when there are exactly $m$ components.

4.1F4step 2.1givenalgebra∎

The mean-zero refinement requires the extra hypotheses: if $Ω$ is connected and a Sobolev extension domain, [F4] supplies a positive first level on the global mean-zero subspace. Without connectedness this can fail: on a domain with several components, a nonzero componentwise constant such as $1_{Ω_{1}} - \frac{∣ Ω_{1} ∣}{∣ Ω_{2} ∣} 1_{Ω_{2}}$ has global mean zero and zero form value, so no positive lower bound on the mean-zero space follows from the present hypotheses.

5 · Examples, counterexamples and false statements

None yet.

Fredholm Elliptic Problems and the Elliptic Spectrum

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Garding's inequality for a divergence-form elliptic operator

Statement

Facts & Assumptions

Proof

A sufficiently large shift is coercive

Statement

Facts & Assumptions

Proof

The shifted elliptic solution operator

Definition

The shifted solution operator is compact on L2L2

Statement

Facts & Assumptions

Proof

On bounded domains, the unshifted equation is an identity-minus-compact equation

Statement

Facts & Assumptions

Proof

The formal adjoint and the adjoint weak Dirichlet problem

Definition

The adjoint solution operator solves the adjoint form problem

Statement

Facts & Assumptions

Proof

The elliptic Fredholm range condition is orthogonality to the adjoint kernel

Statement

Facts & Assumptions

Proof

The Fredholm alternative for weak elliptic Dirichlet problems

Statement

Facts & Assumptions

Proof

The elliptic kernel and cokernel are finite dimensional

Statement

Facts & Assumptions

Proof

Uniqueness implies existence for the elliptic Dirichlet problem

Statement

Facts & Assumptions

Proof

Smooth compactly supported functions of an open set are dense in L2L2

Statement

Facts & Assumptions

Proof

The L2L2 operator associated with a symmetric elliptic form

Definition

The associated elliptic operator is densely defined, symmetric and lower bounded

Statement

Facts & Assumptions

Proof

The symmetric elliptic form operator is self-adjoint with compact resolvent

Statement

Facts & Assumptions

Proof

Symmetric elliptic weak eigenpairs

Definition

The symmetric shifted solution operator is positive and self-adjoint

Statement

Facts & Assumptions

Proof

Discrete spectrum of a symmetric elliptic Dirichlet operator

Statement

Facts & Assumptions

Proof

Eigenbasis expansion in the form norm

Statement

Facts & Assumptions

Proof

The Rayleigh principle for the first Dirichlet eigenvalue

Statement

Facts & Assumptions

Proof

The Courant-Fischer min-max principle for elliptic eigenvalues

Statement

Facts & Assumptions

The shifted solution operator is compact on $L^{2}$

Smooth compactly supported functions of an open set are dense in $L^{2}$

The $L^{2}$ operator associated with a symmetric elliptic form

Eigenfunctions for distinct symmetric elliptic eigenvalues are $L^{2}$ -orthogonal