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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.

Depends on

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Sources

John K. Hunter, Notes on Partial Differential Equations (UC Davis, revised 18 June 2014, complete 242-page notes) (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)
Leon Simon, Lectures on Partial Differential Equations (Stanford, complete 223-page author scan) (standard reference, not scraped)