CC realization of the bounded spatial operator on a finite-component H² domain

[Concrete multiplier, including the adjoint domain] For any real finite-valued measurable function $b$ on ${\mathbb{R}}^{3N}$, multiplication on $D(M_b)=\{f:bf\in \mathcal{H}\}$ is densely defined: $f_L={\mathbf{1}}_{\mathrm{\mid }b\mathrm{\mid }\le L}f$ belongs to this domain and tends to $f$ by dominated convergence. It is symmetric because $b$ is real and the pairings are integrable by Cauchy–Schwarz. If $g\in D(M_b^{\ast })$ with representing vector $h$, every $f\in \mathcal{H}$ supported on $E_L=\{\mathrm{\mid }b\mathrm{\mid }\le L\}$ belongs to the multiplier domain and satisfies $⟨f,bg-h⟩=0$ on that set. The vector ${\mathbf{1}}_{E_L}(bg-h)$ lies in $\mathcal{H}$, so testing with that vector proves $h=bg$ a.e. on $E_L$. Increasing the integer $L$ covers the space up to null sets, proving $bg=h\in \mathcal{H}$ and $g\in D(M_b)$. Conversely the symmetry identity gives every vector of $D(M_b)$ in the adjoint domain with adjoint value $bg$. Thus $M_b$ is self-adjoint on the exact maximal domain, including for the weight $w$. This proof is valid for the finite vector-valued space by its component-sum pairing; it makes no arbitrary family of choices.

[Fourier transfer, exact H² and normalization] Apply the componentwise unitary ${\mathcal{F}}_2$ from [F1] to step 1.1 with $b=w$. For a vector $y$ in the adjoint domain of $K$, changing variables $f={\mathcal{F}}_{2}u$ in its representing functional gives ${\mathcal{F}}_{2}y\in D(M_w^{\ast })=D(M_w)$ with the corresponding adjoint value; the converse follows from symmetry. Hence $K$ is self-adjoint on ${\mathcal{F}}_2^{-1}D(M_w)$. Since $w=4{\pi }^{2}a\mathrm{\mid }\xi {\mathrm{\mid }}^2$ with $a>0$, $1+w$ is bounded above and below by positive multiples of $1+\mathrm{\mid }\xi {\mathrm{\mid }}^2$. For $u\in L^2$, $w{\mathcal{F}}_{2}u\in L^2$ therefore holds exactly when $(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2){\mathcal{F}}_{2}u\in L^2$, which is $u\in Y$ by [F1]. Moreover $\mathrm{\parallel }{\mathcal{F}}_{2}u{\mathrm{\parallel }}_2+\mathrm{\parallel }w{\mathcal{F}}_{2}u{\mathrm{\parallel }}_2$ is equivalent to $\mathrm{\parallel }(1+w){\mathcal{F}}_{2}u{\mathrm{\parallel }}_2$ by the triangle inequality and nonnegativity of $w$, and hence to $\mathrm{\parallel }u{\mathrm{\parallel }}_Y$. This proves graph-norm equivalence. The weak-derivative multiplier in [F1] gives ${\mathcal{F}}_2(-a\mathrm{\Delta }u)=4{\pi }^{2}a\mathrm{\mid }\xi {\mathrm{\mid }}^2{\mathcal{F}}_{2}u$, proving the exact differential action for all $u\in Y$. Finally $⟨Ku,u⟩=\int w\mathrm{\parallel }{\mathcal{F}}_{2}u{\mathrm{\parallel }}^2\ge 0$, with integrability by Cauchy–Schwarz.

[Explicit perturbed inverses on the fixed domain] Multiplication by $V$ is bounded and self-adjoint on $\mathcal{H}$, with norm at most $v$, since $V$ is real and essentially bounded. Thus $H=K+M_V$ is symmetric on the dense $Y$ and is closed there: if $u_l\to u$ and $H{u}_l\to h$, boundedness gives $K{u}_l\to h-M_{V}u$, and closedness of $K$ gives $u\in Y$ with $Hu=h$. For $z=\pm i\mu$, the concrete multiplier $(w-z{)}^{-1}$ is bounded by ${\mu }^{-1}$ and $\mathrm{\mid }w\mathrm{/}(w-z)\mathrm{\mid }\le 1$. Therefore $R_0(z)$ maps $\mathcal{H}$ into $Y$ and satisfies both inverse equations of $K-z$, with range exactly $Y$. On $Y$ the correctly ordered factorization is $H-z=(I+M_V{R}_0(z))(K-z)$. Since $\mathrm{\parallel }{M}_V{R}_0(z)\mathrm{\parallel }\le v\mathrm{/}\mu <1$, the geometric series in the statement is the inverse of the first factor: finite-sum multiplication leaves a remainder tending to zero in norm. The bounded operator space is complete, because an operator-norm Cauchy sequence has a bounded pointwise limit on the complete target and the same uniform Cauchy estimate gives operator-norm convergence. Thus $R_H(z)$ exists, maps into $Y$, and its two ordered inverse equations follow from the factorization. Surjectivity of this range onto $Y$ follows from the second inverse equation. Its norm is at most ${\mu }^{-1}\mathrm{/}(1-v\mathrm{/}\mu )=(\mu -v{)}^{-1}$.

[Self-adjointness without an AC perturbation theorem] Let $y\in D(H^{\ast })$. By step 3.1, $H-i\mu :Y\to \mathcal{H}$ is onto; solve $(H-i\mu )x=(H^{\ast }-i\mu )y$ with its explicit inverse, so $x\in Y$. Symmetry implies $H^{\ast }x=Hx$, whence $y-x\in \ker (H^{\ast }-i\mu )$. For a densely defined operator with first-linear pairing, this kernel is $\mathrm{ran}(H+i\mu {)}^{\perp }$, directly from the adjoint identity. The other inverse from step 3.1 makes that range all of $\mathcal{H}$, so $y=x\in Y$. Since $H\subseteq H^{\ast }$ by symmetry, the adjoint domains and actions are equal, proving self-adjointness on exactly $Y$. Also $⟨Hu,u⟩=⟨Ku,u⟩+\int V\mathrm{\parallel }u{\mathrm{\parallel }}^2\ge -v\mathrm{\parallel }u{\mathrm{\parallel }}^2$ for every $u\in Y$, proving the claimed lower bound. Only the two explicit imaginary resolvents were used.

[Graph norms, cores and scope] For $u\in Y$, boundedness of $V$ gives $\mathrm{\parallel }Hu\mathrm{\parallel }\le \mathrm{\parallel }Ku\mathrm{\parallel }+v\mathrm{\parallel }u\mathrm{\parallel }$ and $\mathrm{\parallel }Ku\mathrm{\parallel }\le \mathrm{\parallel }Hu\mathrm{\parallel }+v\mathrm{\parallel }u\mathrm{\parallel }$. Thus the graph norms of $H$ and $K$ are equivalent, and step 2.1 identifies both with the Sobolev norm. The finite-component smooth density from [F3] therefore makes $C_c^{\mathrm{\infty }}({\mathbb{R}}^{3N};{\mathbb{C}}^M)$ graph dense in the exact domain of each operator, which is precisely the core claim. Steps 1.1–4.1 supply the self-adjointness, exact differential action, lower bound and explicit inverses. All primitive suppliers used here assume CC; neither an AC multiplier theorem nor Kato–Rellich nor any assertion concerning a singular potential or an arbitrary spin basis was invoked.