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Spectral form domain and core of a semibounded operator

Statement

Assume the Axiom of Choice. Let A be self-adjoint on a complex Hilbert space H, with A≥cI meaning ⟨Ax,x⟩≥c∥x∥2 for every x∈D(A), for some real c. Put Q(A):=D((A−cI)1/2),qA[x]:=c∥x∥2+∥(A−cI)1/2x∥2(x∈Q(A)). Then Q(A) and qA do not depend on the choice of the constant c≤inf⁡σ(A) (the form qA[x]=∫λ dEx is itself unchanged, while the summand ∥(A−cI)1/2x∥2 changes by the constant (c−c′)∥x∥2 when c is replaced by c′≤c), qA[x]=∫λ dEx(λ)=⟨Ax,x⟩ for x∈D(A), the domain D(A) is dense in Q(A) for the norm ∥x∥Q=(∥x∥2+∥(A−cI)1/2x∥2)1/2, and qA is a closed quadratic form with qA[x]≥c∥x∥2.

Here the square root is the Borel calculus of max⁡(λ−c,0); the proof shows that E is carried on [c,∞). A closed semibounded quadratic form means the diagonal of a Hermitian sesquilinear form on a dense linear domain, complete in the displayed shifted form norm. If H={0}, use the unique PVM and operator and the convention inf(empty spectrum)=+infinity.

Facts & Assumptions

[A1]

The spectral theorem gives D(A)={x:∫λ2dEx<∞} and A as the coordinate integral on nonzero H. The unbounded integral is closed, has linear domain, squared norm integral, and real-function pairing ⟨f(E)x,x⟩=∫f dEx. Real f gives a self-adjoint operator. The zero-space integral is defined directly. Spectral theorem for unbounded self-adjoint operators (PVM form) The unbounded PVM integral is densely defined, closed and normal Integral of a measurable function against a projection-valued measure

[A2]

Projections multiply by intersection, and Ex(B)=∥E(B)x∥2 is a finite measure of mass ∥x∥2. Consequently EE(J)x(B)=Ex(B∩J), using E(B)E(J)=E(B∩J) and the norm formula; complementary projections give the analogous complementary restriction. The spectrum of A is the essential range of the coordinate function. Projection valued measure Unbounded Borel functional calculus: domains, products, spectral mapping

[A3]

Scalar dominated convergence applies to the finite measures E_x. Dominated convergence

[A4]

H is complete and its inner product is first-linear. Self-adjoint operators have dense linear domains. The assumed AC directly supplies every choice function required by the PVM, closed-integral and spectral-theorem interfaces. Hilbert space Symmetric, self-adjoint and essentially self-adjoint operators The Axiom of Choice

Proof

technique · direct

Given: AC, self-adjoint A and its lower bound c.

1.1A1A2A4given

If H={0}, every domain and form consists of zero, all norms vanish, and every assertion follows directly from the zero-space convention in [A1]. Otherwise obtain E from [A1], under the choice assumption in [A4]. For Jm=[−m,c−1/m] (empty intervals allowed), a vector v=E(Jm)x belongs to D(A) by [A2] and boundedness of lambda on J_m. If v were nonzero then ⟨Av,v⟩=∫Jmλ dEv≤(c−1/m)∥v∥2, contradicting the lower bound. Thus E(J_m)=0 for all positive integers m. Their union is (−∞,c), whose scalar measures therefore vanish by countable subadditivity. The projection norm formula gives E((-infinity,c))=0. The essential-range description in [A2] implies σ(A)⊆[c,∞).

2.1A1A2step 1.1

Put B=max⁡(λ−c,0)(E). It is closed and self-adjoint by [A1], and step 1.1 gives D(B)={x:∫(λ−c)dEx<∞} and ∥Bx∥2=∫(λ−c)dEx. Integrals here and below can be restricted to [c,infinity). Since ∣λ∣≤(λ−c)+∣c∣ there, lambda is absolutely integrable for x in Q(A). Hence qA[x]=c∥x∥2+∥Bx∥2=∫λdEx and qA[x]≥c∥x∥2.

3.1A2step 2.1

For any other lower spectral bound c'<=c, λ−c′=(λ−c)+(c−c′) on the carrier. Since E_x has finite mass, the two domain integrals are finite simultaneously. Adding the appropriate constant times the mass gives the same q_A, while the square-root squared norm increases by (c−c′)∥x∥2. Two arbitrary admissible lower bounds can be compared in their numerical order, so this proves full independence. Their squared form norms differ by that same multiple of ∥x∥2, hence are equivalent since each dominates ∥x∥2.

3.2A1A2step 2.1

If x belongs to D(A), then λ−c≤1+λ2+∣c∣ on the carrier, so x belongs to Q(A). By [A1] and step 2.1, qA[x]=∫λdEx=⟨Ax,x⟩.

3.3A1A2A3step 2.1

For x in Q(A), set xn=E([−n,n])x, n>=1. By [A2], ∫λ2dExn≤n2∥x∥2, so x_n belongs to D(A). The same restriction identity gives ∥x−xn∥Q2=∫∣λ∣>n(1+λ−c)dEx→0 by dominated convergence, with nonnegative integrable majorant 1+λ−c on the carrier. This is the asserted form-norm density, with the exact identity ∥z∥Q2=qA[z]+(1−c)∥z∥2.

4.1A1A4step 2.1step 3.2

The form is the diagonal of a(x,y)=c⟨x,y⟩+⟨Bx,By⟩ on the linear domain D(B); this is Hermitian and sesquilinear by [A4]. Its domain is dense in H because it contains D(A) by step 3.2. For a Cauchy sequence in the form norm, both x_n and Bx_n are Cauchy in H. Completeness gives limits x and y. Closedness of B implies x in D(B) and Bx=y. Therefore ∥xn−x∥Q2=∥xn−x∥2+∥Bxn−Bx∥2→0, proving completeness and closedness in the stated sense.

5.1A4step 1.1step 2.1step 3.1step 3.2step 3.3step 4.1∎

Steps 2.1 and 3.1 establish the domain, integral identity, lower bound and independence; steps 3.2 and 3.3 give the operator-domain identity and core, and step 4.1 gives the closed quadratic form. Positive integer cutoffs are specified without choices. AC is inherited through [A4]; negative and zero lower bounds are allowed without taking a square root of q_A itself. The zero Hilbert space was handled in step 1.1.

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