Original text: CC0 1.0. Prerequisite proofs and component terms.

Closing energy domains and comparing resolvents

CC0 1.0.

The operator applications use the complete closed-form representation and form-sum proofs, the self-adjoint spectral calculus and the bounded factorization lemma.

A quadratic expression on a convenient set of vectors does not yet determine a closed energy form. One must check whether limits in the energy norm still identify vectors uniquely in the original Hilbert space. This unit makes that test explicit, constructs the smallest closed extension, and explains which dense sets actually determine a form. It then recovers domain inclusion from the order of one pair of resolvents.

The mathematical antecedents include Takesaki, Theory of Operator Algebras II and III, Appendix A.8(ii), the core definition in A.8, A.10, and the remark following A.9. The course's proofs use an energy-space inclusion and a bounded factorization, organized around the domains required by later weight constructions.

OA-MOD-FC-01 — Objects and dependencies

Let $H$ be any complex Hilbert space, with inner product linear in its first variable. Let $q$ be a nonnegative sesquilinear form on a linear subspace $D\subseteq H$. Density of $D$ is not assumed. Write

A form $r$ extends $q$ if $D\subseteq D(r)$ and $r(x,y)=q(x,y)$ for $x,y\in D$. We call $q$ closable if it has a closed nonnegative extension on $H$. Here closedness means completeness of the form domain in its form norm. Extension is a statement about domains and values; it does not mean the ordering of extended diagonals defined in OA-MOD-QF-02.

The first four results use only completion, bounded Hilbert-space adjoints, Riesz representation, and orthogonal complements. These facts are proved in Hilbert spaces and compact operators. They use neither a spectral theorem nor a von Neumann algebra.

The operator applications additionally use closed-form representation OA-MOD-QF-03, finite form sums OA-MOD-QF-10, and the exact self-adjoint Borel-calculus contract OA-MOD-DEP-SPECTRAL. The self-adjoint spectral-calculus proof is supplied in the linked programme provider. The order proof uses the bounded factorization proved in OA-MOD-DW-02. The last form-sum assertion uses the elementary reversal of graph inclusion under adjoints; its proof is recalled there.

OA-MOD-FC-02 — The completion test and canonical closure

Theorem. Complete the inner-product space $(D,\langle\cdot,\cdot\rangle_q)$ to a Hilbert space $V$, and denote its isometric dense inclusion by $\iota:D\to V$. There is a unique contraction $j:V\to H$ satisfying $j\iota x=x$. The following conditions are equivalent:

1. $q$ has a closed nonnegative extension on $H$.
2. If $x_n\in D$, $x_n\to0$ in $H$, and $q[x_n-x_m]\to0$ as $n,m\to\infty$, then $q[x_n]\to0$.
3. $j$ is injective.

When they hold, the closure $\overline q$ has the exact domain and values

It is closed, extends $q$, and its original domain $D$ is dense in $D(\overline q)$ for the form norm. Every closed extension of $q$ extends $\overline q$.

Proof. The inclusion $D\to H$ has norm at most one for the form norm, so it extends uniquely to the contraction $j$.

Suppose $r$ is a closed extension. Under the hypotheses of condition 2, the sequence $x_n$ is Cauchy for the $r$-norm: its Hilbert-norm differences and its form differences both tend to zero. Completeness gives a limit in $D(r)$. The continuous inclusion into $H$ identifies that limit as zero. Hence $q[x_n]=r[x_n]\to0$. This proves $1\Rightarrow2$.

If $u\in\ker j$, choose $x_n\in D$ with $\iota x_n\to u$ in $V$. Then $x_n\to0$ in $H$ and $q[x_n-x_m]\to0$. Condition 2 gives

Thus $j$ is injective. Conversely, a sequence as in condition 2 is Cauchy in the form norm, so $\iota x_n$ converges to some $u\in V$. Its Hilbert limit gives $ju=0$. Injectivity forces $u=0$, proving condition 2.

Now assume injectivity. The displayed formula is well-defined because a vector of $j(V)$ has a unique preimage. It is sesquilinear and nonnegative since $j$ is a contraction. Moreover,

Thus $j$, viewed as a map from $V$ onto the new form domain with its form norm, is an isometry onto a complete space. This proves closedness. Substitution of $u=\iota x$ and $v=\iota y$ shows extension. Density of $\iota(D)$ proves the asserted density in form norm.

Finally, let $r$ be any closed extension. For $u\in V$, choose $\iota x_n\to u$. The sequence is Cauchy for the $r$-norm and hence converges there to a vector $y\in D(r)$. In $H$ it converges to $ju$, so $y=ju$. Passing to the limit in the $r$-inner product, for two such sequences, proves $r(ju,jv)=\overline q(ju,jv)$. Hence $r$ extends $\overline q$. $\square$

The range $j(V)$ need not be closed in $H$. It is complete for a stronger norm. No bounded inverse $j^{-1}:j(V)\to V$ for the Hilbert norm is asserted.

An equivalent concrete description is useful:

The completion theorem proves both existence and independence of the limit; Hilbert-norm convergence by itself does not.

OA-MOD-FC-03 — Closedness and lower semicontinuity on all vectors

For any form $q$, define $\widetilde q:H\to[0,\infty]$ by $q[x]$ on $D(q)$ and $+\infty$ elsewhere.

Theorem. A nonnegative form is closed if and only if $\widetilde q$ is lower semicontinuous in the norm topology of $H$. No density hypothesis is required.

Proof of necessity without spectral calculus. If $q$ is closed, put $V=D(q)$ with its form norm and let $j:V\to H$ be its inclusion. For $x\in H$, set

We claim that

First, $\operatorname{ran}j^*$ is dense in $V$: a vector orthogonal to that range lies in $\ker j=\{0\}$. If $x=ju$, the expression in the supremum equals

Density gives $G(ju)=\|u\|_V^2=\|ju\|^2+q[ju]$.

If $G(x)=C<\infty$, the term $f=0$ gives $C\geq0$. Varying a complex scalar multiplying $f$, and minimizing the corresponding quadratic, gives

When $j^*f=0$, the same conclusion follows by scaling the linear term: it must vanish. Therefore $j^*f\mapsto\langle x,f\rangle$ is a well-defined bounded conjugate-linear functional on the dense subspace $\operatorname{ran}j^*$. By Riesz representation it has the form $\langle u,j^*f\rangle_V$ for some $u\in V$. The adjoint identity then gives $\langle x,f\rangle=\langle ju,f\rangle$ for every $f$, so $x=ju$. Consequently $G(x)=+\infty$ outside $j(V)$, proving the claim.

Each function in the supremum defining $G$ is norm continuous and real-valued. Hence $G$ is lower semicontinuous. Subtraction of the continuous function $x\mapsto\|x\|^2$ proves that $\widetilde q$ is lower semicontinuous.

Proof of sufficiency. Let $x_n$ be Cauchy in the form norm. It has a Hilbert-norm limit $x$. The numbers $q[x_n]$ are bounded, so lower semicontinuity implies $\widetilde q[x]<\infty$, or $x\in D(q)$. For any $\varepsilon>0$, choose $N$ with $q[x_n-x_m]\leq\varepsilon^2$ for $n,m\geq N$. For each fixed $n\geq N$, lower semicontinuity applied as $m\to\infty$ gives

Thus $x_n\to x$ in the form norm. The domain is complete. $\square$

The proof uses sequences only for completeness in a metric norm; it places no countability condition on a basis or on later families of forms. The supremum formula proves topological lower semicontinuity directly.

OA-MOD-FC-04 — Closability and lower semicontinuity on the given domain

Theorem. A nonnegative form $q$ on $D\subseteq H$ is closable if and only if the function $x\mapsto q[x]$ is lower semicontinuous on $D$ for the topology inherited from the Hilbert norm.

Equivalently, for every sequence $x_n\in D$ with Hilbert-norm limit $x\in D$,

Proof. If $q$ is closable, its closure is closed by OA-MOD-FC-02. The extended diagonal of that closure is lower semicontinuous by OA-MOD-FC-03. Restriction to $D$ gives the assertion.

Conversely, suppose the stated relative lower semicontinuity holds. Let $x_n$ satisfy condition 2 of OA-MOD-FC-02. For fixed $n$, the sequence $x_n-x_m$ lies in $D$ and converges in $H$ to $x_n\in D$. Hence

The right side is arbitrarily small for all sufficiently large $n$, by the form-Cauchy hypothesis. Thus $q[x_n]\to0$, and OA-MOD-FC-02 supplies a closed extension. $\square$

The distinction between the two topologies of the assertion is substantive. A zero form on a proper dense subspace is closable. Its extension by infinity is not lower semicontinuous on $H$, so the original form is not closed. Both claims are checked explicitly in OA-MOD-FC-08.

OA-MOD-FC-05 — Cores determine the domain, not just a dense set of vectors

For a closed form $q$, a linear subspace $E\subseteq D(q)$ is a form core if it is dense in $D(q)$ for $\|\cdot\|_q$.

Proposition. The restriction $q|_E$ is closable, and its closure is exactly $q$ if and only if $E$ is a form core. More generally, its closed domain is the closure of $E$ inside the form Hilbert space $D(q)$, with the restricted form.

Proof. The original $q$ is a closed extension. Complete $E$ in the restricted form norm. Its completion is the closed subspace $\overline E^{\|\cdot\|_q}$ of the Hilbert space $D(q)$. The inclusion of this subspace into $H$ is injective. OA-MOD-FC-02 therefore identifies the closure with exactly this subspace and this form. It equals $q$ precisely when the subspace is all of $D(q)$. $\square$

Spectral core proposition. Let $A\geq0$ be self-adjoint and $q_A[x]=\|A^{1/2}x\|^2$. Then $D(A)$ is a form core. The subspace

is both a form core for $q_A$ and an operator core for $A$. An operator core means density in $D(A)$ for $(\|x\|^2+\|Ax\|^2)^{1/2}$.

Proof. For $x\in D(A^{1/2})$, write $x_n=E_A([0,n])x$. It lies in $D(A)$, and

The integral is finite because $x\in D(A^{1/2})$. The same truncations applied to $x\in D(A)$ give convergence in the operator graph norm by replacing $1+t$ with $1+t^2$. This proves both assertions. $\square$

Any operator core for $A$ is a form core: graph-norm convergence implies form-norm convergence because $t\leq1+t^2$, and $D(A)$ is already a form core. Mere density in $H$ is insufficient; a dense subspace can be closed and proper in the form norm, as OA-MOD-FC-08 demonstrates.

OA-MOD-FC-06 — One resolvent inequality recovers the whole form order

Theorem. For nonnegative self-adjoint operators $A_1,A_2$ on $H$, the following are equivalent:

1. $D(A_2^{1/2})\subseteq D(A_1^{1/2})$, and $\|A_1^{1/2}x\|^2\leq\|A_2^{1/2}x\|^2$ on $D(A_2^{1/2})$.
2. For some $\lambda>0$, $$(A_2+\lambda I)^{-1}\leq(A_1+\lambda I)^{-1}.$$
3. The inequality in condition 2 holds for every $\lambda>0$.

The operators need not commute. The inequality in condition 1 is between forms, not an assertion that the unbounded operators have the same domain.

Proof. OA-MOD-QF-05 gives $1\Rightarrow3$, and $3\Rightarrow2$ is immediate. Fix a $\lambda$ as in condition 2 and put

Both operators are bounded, positive, and injective. Apply the bounded factorization of OA-MOD-DW-02 to $B_2\leq B_1$. In that lemma's left-factor notation, $B_2^{1/2}=V B_1^{1/2}$ for a contraction $V$. Taking adjoints, using that the square roots are self-adjoint, gives

The spectral calculus gives actual range equalities and norms

For $x\in D(A_2^{1/2})$, let $\xi=B_2^{-1/2}x$. Then

Thus $x\in D(A_1^{1/2})$. Injectivity of $B_1^{1/2}$ gives $B_1^{-1/2}x=C\xi$, and therefore

Canceling the Hilbert-norm term proves condition 1. $\square$

This argument reads the form domain directly from the range of a bounded square root. It does not infer domain inclusion from a pointwise inequality on a previously chosen common subspace.

OA-MOD-FC-07 — When a form sum equals an operator sum

Let $A_1,\ldots,A_m\geq0$ be self-adjoint, and assume

is dense in $H$. The sum $q[x]=\sum_k\|A_k^{1/2}x\|^2$ is closed on $D$, by the finite-sum proof in OA-MOD-QF-10. Its representing nonnegative self-adjoint operator is the form sum, denoted $A_{\mathrm{form}}$.

Proposition. On $D_{\mathrm{alg}}=\bigcap_k D(A_k)$, the algebraic sum $Sx=\sum_k A_kx$ is a restriction of $A_{\mathrm{form}}$. If $S$ is densely defined and essentially self-adjoint on that domain, then

Proof. If $x\in D_{\mathrm{alg}}$ and $v\in D$, the spectral pairing identity gives

The graph characterization in OA-MOD-QF-03 yields $x\in D(A_{\mathrm{form}})$ and $A_{\mathrm{form}}x=Sx$. Since $A_{\mathrm{form}}$ is closed, $\overline S\subseteq A_{\mathrm{form}}$.

For densely defined operators, $T\subseteq U$ implies $U^*\subseteq T^*$: the adjoint identity holding for every vector in the larger domain also holds in the smaller domain, with the same representing vector. If $S$ is essentially self-adjoint, both $\overline S$ and $A_{\mathrm{form}}$ are self-adjoint. Taking adjoints of their inclusion gives the reverse inclusion. Hence equality holds. $\square$

Density of $D$ alone does not assert essential self-adjointness of $S$, or even density of $D_{\mathrm{alg}}$. Those are separate hypotheses in the stated implication.

Editable source · Proof dependencies and component terms