Spectral cutoff cores and unitary identifications of completed scales

Interpret $A\ge I$ as $⟨Au,u⟩\ge \mathrm{\parallel }u{\mathrm{\parallel }}_H^2$ for $u\in D(A)$. Apply [F4] with $c=1$: $E((-\mathrm{\infty },1))=0$. Thus all scalar spectral measures are carried on $[1,\mathrm{\infty })$. For every real $t$, define $A^t$ by the Borel multiplier ${\lambda }^t$ there, extended by $1$ below $1$; values below $1$ do not affect the operator. For $u\in \mathcal{C}$ every such power is defined, since an integer $m\ge 1$ with $2m\ge 2t$ gives ${\lambda }^{2t}\le {\lambda }^{2m}$ on this carrier.

For $u\in H$ and every positive integer $k$, the norm and product formulas give $\mathrm{\parallel }{A}^k{E}_{N}u{\mathrm{\parallel }}_H^2={\int }_{[1,N]}{\lambda }^{2k}d{E}_u\le N^{2k}\mathrm{\parallel }u{\mathrm{\parallel }}_H^2$, so $E_{N}H\subset \mathcal{C}$. Also $\mathrm{\parallel }(I-E_N)u{\mathrm{\parallel }}_H^2=E_u((N,\mathrm{\infty }))\to 0$, proving density of $\mathcal{C}$ in $H$. For $u\in \mathcal{C}$ and real $s$, the integrable measure ${\lambda }^{s}d{E}_u$ has tails tending to zero, so $\mathrm{\parallel }(I-E_N)u{\mathrm{\parallel }}_s^2={\int }_{(N,\mathrm{\infty })}{\lambda }^{s}d{E}_u\to 0$. The same formulas give $\mathrm{\parallel }{E}_{N}u{\mathrm{\parallel }}_s\le \mathrm{\parallel }u{\mathrm{\parallel }}_s$.

The linear space $\mathcal{C}$ has the first-linear Hermitian pairing $⟨u,v{⟩}_s=⟨A^{s\mathrm{/}2}u,A^{s\mathrm{/}2}v{⟩}_H$. It is positive definite: if its diagonal is zero, ${\int }_{[1,\mathrm{\infty })}{\lambda }^{s}d{E}_u=0$; positivity of ${\lambda }^s$ implies $E_u([1,\mathrm{\infty }))=0$, hence $u=0$. Its norm is exactly $\mathrm{\parallel }u{\mathrm{\parallel }}_s$. Therefore [F5] and [F6], with countable choice from [F3], construct the Hilbert completion $H_A^s$ with dense embedded core, which we identify with $\mathcal{C}$. Apply [F7] to each core contraction $E_N:\mathcal{C}\to H_A^s$. For $x\in H_A^s$ and $u\in \mathcal{C}$, the extension satisfies $\mathrm{\parallel }(I-E_N)x{\mathrm{\parallel }}_s\le 2\mathrm{\parallel }x-u{\mathrm{\parallel }}_s+\mathrm{\parallel }(I-E_N)u{\mathrm{\parallel }}_s$. First approximate $x$ by $u$, then use step 2.1; this proves strong convergence on every completed scale.

The map $J_s:\mathcal{C}\to H$, $J_{s}u=A^{s\mathrm{/}2}u$, is linear and isometric. By [F7] it extends to $H_A^s$, and passage to limits preserves the norm and pairing. Its range is closed: if $J_s{x}_n$ converges in $H$, isometry makes $x_n$ Cauchy; completeness gives $x_n\to x$ and its image converges to $J_{s}x$. Its range contains $E_{N}H$: if $v=E_{N}v$, then $u=A^{-s\mathrm{/}2}v$ has spectral support in $[1,N]$, lies in $\mathcal{C}$, and $J_{s}u=v$ by the product formula. Step 2.1 makes the union of these cutoff ranges dense in $H$, so $J_s$ is onto and unitary. The core identity $J_s{E}_N=E_N{J}_s$ extends by continuity; since $A^{-s\mathrm{/}2}{E}_{N}H\subset \mathcal{C}$, it also identifies every completed cutoff range with core vectors.

For any real $r$ and $u\in \mathcal{C}$, $A^{-r\mathrm{/}2}u$ belongs to $\mathcal{C}$: for each integer $k\ge 0$, choose an integer $m\ge 1$ with $2m\ge 2k-r$; then the relevant moment ${\int }_{[1,\mathrm{\infty })}{\lambda }^{2k-r}d{E}_u$ is bounded by ${\int }_{[1,\mathrm{\infty })}{\lambda }^{2m}d{E}_u<\mathrm{\infty }$. The same argument with $-r$ shows that $A^{r\mathrm{/}2}$ preserves $\mathcal{C}$, and the product formula makes these core maps inverse. Moreover, $\mathrm{\parallel }{A}^{-r\mathrm{/}2}u{\mathrm{\parallel }}_{s+r}^2={\int }_{[1,\mathrm{\infty })}{\lambda }^{s}d{E}_u=\mathrm{\parallel }u{\mathrm{\parallel }}_s^2$. Extend both core maps using [F7] between the Hilbert completions. Norm equality passes to limits, and their compositions are identities by density and uniqueness of extension. They are thus inverse onto isometries at all real indices.

For $s\ge 0$, $A^{-s\mathrm{/}2}$ is a bounded contraction on $H$ and is injective, since its multiplier is strictly positive on the spectral carrier. The map ${\iota }_s=A^{-s\mathrm{/}2}{J}_s:H_A^s\to H$ is therefore a contractive injection and equals the identity on $\mathcal{C}$, so it is the canonical extension of that identity. The exact product-domain formula gives $\mathrm{ran}{A}^{-s\mathrm{/}2}=D(A^{s\mathrm{/}2})$: every $v\in H$ has $A^{-s\mathrm{/}2}v\in D(A^{s\mathrm{/}2})$ and $A^{s\mathrm{/}2}{A}^{-s\mathrm{/}2}v=v$, while every $u\in D(A^{s\mathrm{/}2})$ satisfies $u=A^{-s\mathrm{/}2}{A}^{s\mathrm{/}2}u$. Surjectivity of $J_s$ gives the asserted range of ${\iota }_s$, and $\mathrm{\parallel }{A}^{s\mathrm{/}2}{\iota }_{s}x{\mathrm{\parallel }}_H=\mathrm{\parallel }{J}_{s}x{\mathrm{\parallel }}_H=\mathrm{\parallel }x{\mathrm{\parallel }}_s$. Spectral truncations converge in this exact domain norm by its integrable spectral tails, so no graph-norm factor is introduced.

If $b\ge a$, then ${\lambda }^a\le {\lambda }^b$ on $[1,\mathrm{\infty })$, so the identity on $\mathcal{C}$ extends by [F7] to a contraction $i_{b,a}:H_A^b\to H_A^a$. On the core, and hence on the completions, $J_a{i}_{b,a}=A^{(a-b)\mathrm{/}2}{J}_b$. The multiplier $A^{(a-b)\mathrm{/}2}$ is injective and has range $D(A^{(b-a)\mathrm{/}2})$ by the inverse-product argument of step 5.1. Therefore $i_{b,a}$ is injective, and for $v=J_a{i}_{b,a}x$ one has $\mathrm{\parallel }{A}^{(b-a)\mathrm{/}2}v{\mathrm{\parallel }}_H=\mathrm{\parallel }{J}_{b}x{\mathrm{\parallel }}_H=\mathrm{\parallel }x{\mathrm{\parallel }}_b$. This proves precisely the claimed couple and its second norm. For $c\ge b\ge a$, $i_{b,a}{i}_{c,b}=i_{c,a}$ because both maps agree on the dense core; consequently a finite collection embeds compatibly in the Hilbert space at its minimum index, which is Hausdorff.