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# Closing a domain after spectral localization

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## OA-FLOW.GRAPH.CONTRACT — Functional calculus used here

Let $P$ be self-adjoint on a Hilbert space $H$, of any Hilbert dimension, with spectral measure $E$. The spectral theorem gives the strongly continuous unitary group $U_t=e^{itP}$ and the closed normal operator $B=f(P)$ for any Borel $f:\mathbb R\to\mathbb C$ finite $E$-almost everywhere. Its domain is

$$\operatorname{Dom}(B)=\left\{\xi\in H:\int_{\mathbb R}|f(p)|^2\,d\langle E(p)\xi,\xi\rangle<\infty\right\}.$$

The spectral projections commute with $B$ on its domain and with $U_t$; $BU_t\xi=U_tB\xi$ there. If a closed subspace reduces every $U_t$, it reduces every $E(S)$ for Borel $S$. This last assertion is the spectral-theorem consequence that the spectral projections of a self-adjoint generator belong to the von Neumann algebra generated by its unitary group. These are the exact functional-calculus inputs to the proof.

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## OA-FLOW.GRAPH.MAIN — The graph-localization theorem

**Theorem.** In the preceding setting, let $D$ be a linear subspace of $\operatorname{Dom}(f(P))$. Assume that $D$ is dense in $H$ and $U_tD\subset D$ for every real $t$. Then $D$ is a core for $f(P)$: the closure of the graph of $f(P)|_D$ is the full graph of $f(P)$.

There is no requirement that $D\subset\operatorname{Dom}(P)$, nor that $f$ be continuous, polynomially bounded or real-valued. Finiteness almost everywhere refers to the spectral measure, not Lebesgue measure.

**Proof.** Write $B=f(P)$ and let

$$K=\overline{\{(d,Bd):d\in D\}}^{\,H\oplus H}.$$

Closedness of $B$ gives $K\subset\operatorname{Graph}(B)$. Since $BU_t=U_tB$ on $\operatorname{Dom}(B)$, the hypothesis on $D$ implies

$$ (U_t\oplus U_t)(d,Bd)=(U_td,BU_td)\in\operatorname{Graph}(B|_D).$$

The same holds with $-t$, so $K$ reduces every $U_t\oplus U_t$. The generator on $H\oplus H$ is $P\oplus P$. Its spectral projection for a Borel set $S$ is $E(S)\oplus E(S)$. The functional-calculus contract therefore gives

$$ (E(S)\oplus E(S))K\subset K. \tag{G1}$$

For $n\geq1$, set

$$S_n=\{p:|p|\leq n,\ |f(p)|\leq n\},\qquad Q_n=E(S_n).$$

The projections increase strongly to $1$, since $f$ is finite spectral-almost everywhere. Moreover $BQ_n$ is bounded with norm at most $n$. Fix $\xi\in H$ and choose $d_j\in D$ with $d_j\to\xi$ in $H$. Formula (G1) shows that each

$$ (Q_nd_j,BQ_nd_j)=(Q_nd_j,Q_nBd_j)$$

belongs to $K$. For fixed $n$, boundedness of $Q_n$ and $BQ_n$ implies convergence in $H\oplus H$ to $(Q_n\xi,BQ_n\xi)$. Therefore this pair belongs to $K$.

Now take $\xi\in\operatorname{Dom}(B)$. Strong convergence gives $Q_n\xi\to\xi$ and, because $BQ_n\xi=Q_nB\xi$, also $BQ_n\xi\to B\xi$. Hence $(\xi,B\xi)\in K$. Thus $\operatorname{Graph}(B)\subset K$, giving equality. $\square$

The proof deliberately places $Q_nd$ in the **closed graph** $K$. The hypothesis does not say that $D$ itself is invariant under discontinuous spectral cutoffs, and the proof never needs that stronger assertion.

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## OA-FLOW.GRAPH.POWERS — Imaginary-power invariance controls real powers

**Corollary.** Let $A$ be a positive, nonsingular self-adjoint operator on $H$, and let $r\in\mathbb R$. If $D\subset\operatorname{Dom}(A^r)$ is dense in $H$ and $A^{it}D\subset D$ for all $t\in\mathbb R$, then $D$ is a core for $A^r$.

**Proof.** Nonsingularity means $\ker A=0$; it does not mean $A^{-1}$ is bounded. The spectral calculus defines the densely defined self-adjoint operator $P=\log A$ and gives $A^{it}=e^{itP}$ and $A^r=f(P)$ with $f(p)=e^{rp}$. Apply the theorem. When $r=0$, $A^r=1$ and the result is the original norm density; when $r<0$, the same proof handles the possible singular behavior of the inverse near zero. No change of domain convention is made. $\square$

**Application to a dual Hilbert algebra.** Suppose a proposed left Hilbert algebra has dense underlying vector domain $D$, an antilinear operator $S_0$ on $D$, and operators $J,\Delta$ with $J$ antiunitary, $\Delta$ positive nonsingular self-adjoint, and

$$ S_0d=J\Delta^{1/2}d\quad(d\in D),\qquad
D\subset\operatorname{Dom}(\Delta^{1/2}),\quad\Delta^{it}D\subset D.$$

The corollary with $r=1/2$ implies that $D$ is a graph core for $\Delta^{1/2}$. Applying the isometry $1\oplus J$ to the graphs then gives

$$\overline{S_0}=J\Delta^{1/2}.$$

The isometry is real-linear on $H\oplus H$ if $J$ is antilinear; this suffices to preserve closure and convergence of graphs. It also proves closability directly. This closes the analytic bridge once all the displayed hypotheses have been proved for the dual Hilbert algebra. It does **not** establish its density, invariance or the displayed formula; those remain concrete obligations in the general dual-weight construction.

