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Unitary groups converge under strong resolvent convergence

Statement

Assume the Axiom of Choice. Let An,A be self-adjoint operators on a complex Hilbert space H with AnA in the strong resolvent sense. Then eitAnxeitAxfor every tR and every xH. If, in addition, A and all An are bounded below by a common real constant γ (meaning Sx,xγx2 for xD(S) and S=A,An), then etAnetA strongly for every t0.

Facts & Assumptions

[A1]

For a self-adjoint S with spectral PVM, eitS is the Borel calculus of λeitλ and is a strongly continuous unitary group, with infinitesimal generator iS (A self-adjoint operator generates a strongly continuous unitary group, Infinitesimal generator of a unitary group, Strongly continuous one-parameter unitary group).

[A2]

Under AC, strong resolvent convergence gives g(An)xg(A)x for every bounded continuous g:RC and every x (Continuous functional calculus under resolvent convergence, The Axiom of Choice).

[A3]

On nonzero H, each self-adjoint S has a spectral PVM ES with D(S)={x:λ2dExS<} and S=λdES. For domain vectors, Sx,x=λdExS, using the quadratic pairing identity and the reality of λ in the first-variable-linear convention. Functions agreeing off a measurable ES-null set have the same integral operator and domain; the zero-space calculus 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).

[A4]

PVM projections satisfy E(B)E(C)=E(BC) and E(R)=I; the scalar measures are positive of mass x2, and scalar monotone convergence holds (Projection valued measure, Scalar and complex measures from a pvm, Monotone convergence for the integral).

Proof

technique · direct

Given: AC and the self-adjoint operators with strong resolvent convergence in the statement.

1.1

If H={0}, all operators in either conclusion are its unique operator, so both conclusions hold. Otherwise the spectral PVMs exist by [A3]. Fix any tR, including negative times. The function λeitλ is continuous and has absolute value one everywhere. Thus [A2] gives eitAnxeitAx for every x; [A1] identifies these as the stated unitary groups. At t=0 each operator is I.

A1A2A3given
1.2

For the second claim suppose S is one of A,An and satisfies the common lower bound. Let Bm=[m,γ1/m] for integers m1, with an empty interval interpreted as empty. If ES(Bm)0, choose x=ES(Bm)y0 for some y. The projection identities give ES(RBm)x=0, so the scalar measure of x is supported on Bm. As Bm is bounded, [A3] gives xD(S) and Sx,x=λdExS(γ1/m)x2<γx2, a contradiction. Hence each ES(Bm)=0. The sets Bm increase to (,γ); monotone convergence gives ExS((,γ))=0 for every x. Since ES(B)x2=ES(B)x,x, the projection ES((,γ)) itself is zero.

A3A4given
2.1

Now fix t0 and define gt(λ)=etmax(λ,γ). It is continuous and bounded by etγ, and it agrees with etλ on [γ,). Step 1.2 and null-set invariance in [A3] give equality of the integral operators gt(S)=etS, including domains, for S=A,An. In particular each exponential here has full domain and is bounded: its defining squared integral is at most e2tγx2, and its quadratic norm identity gives the same operator bound. Applying [A2] to gt proves etAnxetAx for every x.

A2A3A4step 1.2
3.1

The first conclusion holds for every real time by step 1.1, and the second for every nonnegative time by step 2.1. At time zero both reduce to the identity. The lower bound is assumed for the limit and every approximant; no preservation-of-lower-bound theorem is assumed. AC supplies the spectral and calculus hypotheses, including their Countable Choice assumptions.

A1A2A3step 1.1step 2.1

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