Section 5 9 October 2026

Ordered resolvents for an unbounded closed-form interaction

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23259571

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5 Ordered resolvents for an unbounded closed-form interaction

Theorem 5.1 (Closed-form factor transfer and physical weights)

Let $A_0$ and $A=A_0+W^*JW$ be self-adjoint operators defined by closed semibounded forms on the same $\mathcal V$ , with equivalent positive shifted form norms. Here $W:\mathcal V\to\mathcal H_{\rm aux}$ is bounded and $J=J^*$ is bounded on $\mathcal H_{\rm aux}$ . The adjoint $W^*$ maps $\mathcal H_{\rm aux}$ to $\mathcal V^*$ . Actual form closure is an assumption; a formal factorization alone does not supply it.

For $\zeta=E+i\eta$ , $\eta>0$ , set

R_0=(A_0-\zeta)^{-1},\quad R=(A-\zeta)^{-1},\quad G_0=WR_0W^*,\quad G=WRW^* .

All products are bounded between the indicated form spaces, and

T=(I+JG_0)^{-1}=I-JG,\qquad RW^*=R_0W^*T,\qquad G=G_0T .

(5.1)

In particular

\begin{align}\operatorname{Im}G &=T^*(\operatorname{Im}G_0)T =\eta(RW^*)^*(RW^*), \tag{5.2}\\ (RW^*)^*T_{\rm ph}(RW^*) &=T^*\big[(R_0W^*)^*T_{\rm ph}(R_0W^*)\big]T . \tag{5.3}\end{align}

The second formula is an equality of bounded auxiliary-space forms for any positive closed physical form $T_{\rm ph}$ with $\mathcal V\subset\operatorname{Dom}T_{\rm ph}^{1/2}$ continuously. No commutation with $A_0,A,W$ or $J$ is required.

Proof

Choose $K_0=A_0+c_0\ge I$ . The spectral theorem gives the exact identity

\|K_0^{1/2}R_0K_0^{1/2}\| =\sup_{\lambda\in\sigma(A_0)} \frac{\lambda+c_0}{|\lambda-\zeta|}<\infty .

(5.4)

Thus $R_0$ extends boundedly $\mathcal V^*\to\mathcal V$ . The same argument with a positive shift of $A$ and equivalence of form norms gives this extension for $R$ . Their form inverse identities therefore give both ordered resolvent identities

R-R_0=-R_0W^*JWR=-RW^*JWR_0.

Sandwiching by $W,W^*$ yields $G-G_0=-G_0JG=-GJG_0$ . It follows by multiplication in the displayed order that both products of $I+JG_0$ and $I-JG$ equal $I$ . This proves invertibility and the formula for $T$ . The first resolvent identity now gives $RW^*=R_0W^*(I-JG)$ , proving (5.1).

For a self-adjoint resolvent, $\operatorname{Im}R=\eta R^*R$ . The identity extends to dual inputs by continuity: both sides define bounded sesquilinear forms on $\mathcal V^*$ , and Hilbert inputs are dense there. Sandwiching with $W^*$ proves the second equality in (5.2). Substitution of $RW^*=R_0W^*T$ proves its first equality. The same substitution into the quadratic form of $T_{\rm ph}$ gives (5.3). Continuity of that form on $\mathcal V$ justifies each product even when $T_{\rm ph}$ is unbounded on $\mathcal H$ .

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The resolvent inverse in this theorem exists off the real axis because the closed self-adjoint parents exist. A Neumann criterion is sufficient for a quantitative bound, not necessary for the identity:

\|JG_0\|\le\theta<1\quad\Longrightarrow\quad \|T\|\le(1-\theta)^{-1}.

(5.5)

If $LF^{-1}=W^*b$ with bounded $b$ , an earned $\operatorname{Im}G_0/\pi\le s_{0,\eta}I$ implies

\frac1\pi\operatorname{Im} [(LF^{-1})^*R(LF^{-1})] \le \frac{\|b\|^2s_{0,\eta}}{(1-\theta)^2}I

when its full form-dual sandwich is used. Likewise (5.3) transfers an earned physical-energy resolvent bound. Absorptive response alone does not replace that physical-energy bound. For a form-dual coupling this is first a resolvent sandwich: one uses the bounded near factors of Theorem 4.3, or the ordered temporal estimate below, rather than treating $LF^{-1}$ as a bounded Hilbert-space coupling without proof.

Equality in (5.4) is over the actual spectrum. Replacing it by a whole half-line above a spectral floor gives an upper bound, not in general an equality. Spectral gaps matter for that distinction, whereas the shift that makes $K_0$ positive does not create such gaps. At a real eigenvalue no inverse has been asserted. As $\eta$ decreases, genuine poles and thresholds may make both the baseline response and the quantitative inverse ceiling large.