Onto radial continuum transform

[Projected norm identity on compact sources] For compactly supported $g$, [F1] with $a\downarrow 0$, $b\uparrow \mathrm{\infty }$ (or the monotone convergence of the positive measures on the increasing intervals $(1\mathrm{/}k,k)$) gives $\mathrm{\parallel }{T}_{l}g{\mathrm{\parallel }}_{L^2((0,\mathrm{\infty }),dE)}^2={\int }_0^{\mathrm{\infty }}{\rho }_l(E)\mid \int {\phi }_l(\sqrt{E},r)g(r)dr{\mid }^{2}dE=⟨E_{A_l}((0,\mathrm{\infty }))g,g⟩$; the right side is finite because it is the mass of the finite measure $E_{A_l,g}$ on $(0,\mathrm{\infty })$. This is the norm identity of (a) on the dense compact-source domain $C_c$.

[Extension through the positive projection] Step 1.1 gives $\mathrm{\parallel }{T}_{l}g\mathrm{\parallel }\le \mathrm{\parallel }g\mathrm{\parallel }$ on compact sources, so extend $T_l$ first to a contraction $T$ on all $L^2$. Passing to compact-source approximations yields $\mathrm{\parallel }Tg{\mathrm{\parallel }}^2=⟨P_{+}g,g⟩$ for every $g$, where $P_{+}=E_{A_l}((0,\mathrm{\infty }))$. Thus $T(I-P_{+})=0$ and $\widehat{T}=T{\mathrm{\mid }}_{P_{+}H}$ is an isometry, with $T=\widehat{T}{P}_{+}$. If $P_{+}g=P_{+}h$ for compact sources, the norm identity gives $T_l(g-h)=0$, so the induced map $P_{+}g\mapsto T_{l}g$ is well-defined and isometric. The projected core $P_{+}{C}_c$ is dense in $P_{+}H$: project compact-source approximations to any vector in $P_{+}H$ and use $\mathrm{\parallel }{P}_{+}\mathrm{\parallel }\le 1$. This proves the unique-extension and factorization assertions of (a). Its range is closed because Cauchy sequences in that range pull back to Cauchy sequences in the complete positive subspace. In particular compact-source images are dense in its range, without incorrectly calling $T_l$ isometric on all compact sources.

[Compressed resolvents and the missing norm equality] Write $W=\widehat{T}:H_{+}\to K$, $M=\mathrm{ran}W$ and $Q$ for the orthogonal projection onto this closed range. Define $W^{\ast }=W^{-1}Q$: uniqueness of orthogonal decomposition gives directly $⟨Wu,v⟩=⟨u,W^{\ast }v⟩$, $W^{\ast }W=I_{H_{+}}$ and $W{W}^{\ast }=Q$, without invoking another representation theorem. Polarize [F1] on compact sources; the mixed measures are finite, and equality on positive intervals extends to all positive Borel sets by the interval-generated finite-measure uniqueness argument. Integration of bounded simple functions and bounded approximation then gives $⟨M_{b}Tg,Th⟩=⟨b_{+}(A_l)g,h⟩$, where $b_{+}$ is $b$ on the positive half-line and zero elsewhere. Boundedness extends this identity from compact sources to all $g,h\in H$. Thus $W^{\ast }{M}_{b}W=b(A_{+})$, initially only a compression. For $z=x+iy$, $y\mathrm{\ne }0$, put $m_z(E)=(E-z{)}^{-1}$ and $R_z=(A_{+}-z{)}^{-1}$. The identity gives $Q{M}_{m_z}Wu=W{R}_{z}u$. Pointwise $\mathrm{Im}{m}_z=y\mathrm{\mid }{m}_z{\mathrm{\mid }}^2$, so $\mathrm{Im}⟨M_{m_z}Wu,Wu⟩=y\mathrm{\parallel }{M}_{m_z}Wu{\mathrm{\parallel }}^2$. Setting $v=R_{z}u$ and $u=(A_{+}-z)v$, self-adjointness and the linear-first convention give $\mathrm{Im}⟨R_{z}u,u⟩=\mathrm{Im}⟨v,(A_{+}-z)v⟩=y\mathrm{\parallel }{R}_{z}u{\mathrm{\parallel }}^2$. The compressed pairing identifies these imaginary parts. Dividing by the nonzero real $y$ proves $\mathrm{\parallel }{M}_{m_z}Wu{\mathrm{\parallel }}^2=\mathrm{\parallel }{R}_{z}u{\mathrm{\parallel }}^2=\mathrm{\parallel }W{R}_{z}u{\mathrm{\parallel }}^2=\mathrm{\parallel }Q{M}_{m_z}Wu{\mathrm{\parallel }}^2$. Pythagoras forces $(I-Q)M_{m_z}Wu=0$. Apply this to both $z$ and $\bar{z}$, whose multipliers are adjoints: $M$ and $M^{\perp }$ reduce these multipliers, $Q$ commutes with them, and $M_{m_z}W=W{R}_z$. This is the required compression-to-reduction proof before any onto claim.

[Resolvent reduction implies indicator reduction] For $0<a<b$ and $\epsilon >0$, define $p_{\epsilon ,a,b}(E)={\pi }^{-1}{\int }_a^b\epsilon \mathrm{/}[(E-t{)}^2+{\epsilon }^2]dt$. Its multiplier commutes with $Q$: it is the operator-norm Riemann-integral limit of the imaginary parts of $M_{m_{t+i\epsilon }}$, since $\mathrm{\parallel }{M}_{m_{t+i\epsilon }}-M_{m_{s+i\epsilon }}\mathrm{\parallel }\le \mathrm{\mid }t-s\mathrm{\mid }\mathrm{/}{\epsilon }^2$. One has $0\le p_{\epsilon ,a,b}\le 1$, and these functions tend as $\epsilon =1\mathrm{/}n\downarrow 0$ to ${\mathbf{1}}_{(a,b)}$ except at the two Lebesgue-null endpoints. Dominated convergence against $\mathrm{\mid }h{\mathrm{\mid }}^2$ for each $h\in K$ gives strong convergence of the multipliers, so $Q$ commutes with $M_{{\mathbf{1}}_{(a,b)}}$. The Borel sets whose multipliers commute with $Q$ form a sigma-algebra: complements use $I-M_{{\mathbf{1}}_B}$, intersections use products, and disjoint countable unions use strongly convergent finite sums; disjointification gives arbitrary countable unions. Positive intervals generate the Borel sigma-algebra, so all Borel indicators commute with $Q$. Bounded simple approximation gives all bounded Borel multipliers; completed-measurable sets have the same multiplier as a Borel representative modulo null sets.

[The reducing projection is a scalar indicator] Set $q(E)=e^{-E}>0$ in $K$, $\psi =Qq$ and $a=\psi \mathrm{/}q$ using measurable representatives. Commutation gives $Q(bq)=b\psi$ for bounded measurable $b$. For every measurable $B$, contractivity gives ${\int }_B\mathrm{\mid }\psi {\mathrm{\mid }}^2\le {\int }_B{q}^2$, hence $\mathrm{\mid }a\mathrm{\mid }\le 1$ a.e.; otherwise intersecting $\{\mathrm{\mid }a\mathrm{\mid }>1+1\mathrm{/}n\}$ with a bounded positive interval gives a contradiction to this inequality. Vectors $bq$ are dense: for $h\in K$, $h_n={\mathbf{1}}_{\{1\mathrm{/}n<E<n,\mathrm{\mid }h\mathrm{/}q\mathrm{\mid }\le n\}}h$ have that form and converge in $K$ by dominated convergence. Therefore $Q=M_a$ on $K$. Self-adjointness and idempotence force $a=\bar{a}=a^2$ a.e., testing the multiplier identities on bounded-interval indicators. Thus $Q=M_{{\mathbf{1}}_S}$ for a measurable $S$, and $M=L^2(S)$.

[Countable compact tests eliminate the complement] For the explicit family $g_j$ of [F5], each $T_l{g}_j$ lies in $M=L^2(S)$ and vanishes a.e. on $S^c$. Remove the union of the countably many exceptional null sets. For every remaining $E\in S^c$, positivity of ${\rho }_l(E)$ gives $\int {\phi }_l(\sqrt{E},r)g_j(r)dr=0$ for every $j$. On each $(1\mathrm{/}k,k)$, $k\ge 2$, this is a continuous linear functional by Cauchy-Schwarz and local square integrability of ${\phi }_l$. Its vanishing on the fixed dense family implies its vanishing on all of $L^2(1\mathrm{/}k,k)$, including $\overline{{\phi }_l}{\mathbf{1}}_{(1\mathrm{/}k,k)}$. Hence ${\int }_{1\mathrm{/}k}^k\mathrm{\mid }{\phi }_l{\mathrm{\mid }}^2=0$. Exhausting $(0,\mathrm{\infty })$ contradicts the nonzero continuous regular solution of [F1]. The remaining part of $S^c$ is therefore empty, so $S^c$ is null, $Q=I_K$, and $W$ is unitary with isometric inverse $W^{\ast }$. Only countably many conull sets were intersected. This proves (b).

[Exact bounded and unbounded diagonalization] Steps 3.1, 4.1 and 6.1 give $\widehat{T}{E}_{A_l}(B){\mathrm{\mid }}_{P_{+}H}=M_{{\mathbf{1}}_B}\widehat{T}$ for every positive Borel set. Linear combinations and bounded pointwise simple approximation therefore give $\widehat{T}h(A_l){\mathrm{\mid }}_{P_{+}H}{\widehat{T}}^{\ast }=M_{h{\mathrm{\mid }}_{(0,\mathrm{\infty })}}$ for every bounded Borel $h$. This uses the onto-ness of step 6.1 and does not incorrectly pair negative components through $T$. For unbounded energy, spectral truncation and the norm formula give $\int E^2\mathrm{\mid }\widehat{T}g(E){\mathrm{\mid }}^{2}dE={\int }_{(0,\mathrm{\infty })}{\lambda }^{2}d{E}_g(\lambda )$, so the exact generator domain is $\{h:Eh\in L^2(dE)\}$ and $\widehat{T}{A}_l{\mathrm{\mid }}_{P_{+}H}{\widehat{T}}^{\ast }=M_E$ there.

[Inverse integral and exact scope of convergence] For $h\in K$ put $h_n={\mathbf{1}}_{(1\mathrm{/}n,n)}h$ and $v_n(r)={\int }_{1\mathrm{/}n}^n\sqrt{{\rho }_l(E)}{\phi }_l(\sqrt{E},r)h(E)dE$. The compact parameter bounds of [F5] and Cauchy-Schwarz give $v_n\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2(0,\mathrm{\infty })$. Fubini for compact-source $g$ gives $⟨Tg,h_n⟩=⟨g,v_n⟩$. Since $T=W{P}_{+}$ and $W^{\ast }{h}_n\in H_{+}$, the left side also equals $⟨g,W^{\ast }{h}_n⟩$. All compact-source tests therefore identify $v_n$ locally a.e. with $W^{\ast }{h}_n$, so $v_n$ represents a global $L^2$ element of $H_{+}$. Finally $h_n\to h$ in $K$ and $W^{\ast }$ is an isometry, giving $W^{\ast }h={\lim }_{n\to \mathrm{\infty }}{v}_n$ in $L^2(dr)$. This explicitly realizes the inverse without asserting pointwise convergence at infinity. If compact source means bounded support that may approach zero, cutting it off below $1\mathrm{/}n$ and using the regular $r^{l+1}$ bound of [F5] extends all compact-energy integral identities by Cauchy-Schwarz and $L^2$ continuity.

[Complete resolution of $C$] By the block identity of Transported radial endpoint realization and the angular unitary of Angular radial unitary reduction, $C$ is unitarily equivalent to the direct sum over $(l,m)$ of the operators $A_l$, and by the previous items $A_l$ is unitarily equivalent to $M_E$ on $L^2((0,\mathrm{\infty }),dE)$ together with the discrete eigenvalues $-{\gamma }^2\mathrm{/}(4n^2)$, $n\ge l+1$, each simple, whose eigenspaces are spanned by the transported regular solutions (Negative-subspace completeness and degeneracy, Laguerre bound energy exhaustion); the essential spectrum is $[0,\mathrm{\infty })$ and the positive spectrum is purely absolutely continuous by Coulomb continuum spectral density. The absence of a zero eigenvector follows from Coulomb continuum spectral density. For $\gamma >0$, PVM transport from [F3] through the angular unitary and the exact block identity shows that the normalized global negative eigenfamily restricts to a complete radial negative family in each channel; otherwise placing a missing radial vector in one channel contradicts global completeness. Thus each radial unitary sends $u$ to $((⟨u,u_{nl}\mathrm{/}\mathrm{\parallel }{u}_{nl}\mathrm{\parallel }⟩{)}_{n\ge l+1},W{P}_{+}u)$, with inverse the $L^2$ bound-family sum plus step 7.2. Assembling these unitaries in the fixed countable channel order gives the stated onto unitary $V$, since orthogonal norm sums give convergence and surjectivity. The exact block domain and step 7.1 identify its diagonal domain with square summability of all energy-multiplied discrete and continuum coordinates. For $\gamma =0$, use the transported free operator: its explicit Fourier multiplier $\mathrm{\mid }p{\mathrm{\mid }}^2$ has zero projection on $(-\mathrm{\infty },0]$, since $\{p=0\}$ is null. PVM uniqueness and unitary transport in [F3] give ${⨁}_{l,m}{E}_{A_l}((-\mathrm{\infty },0])=0$, so every individual nonpositive block projection is zero. Each free radial block is consequently wholly positive, and the same positive transforms assemble a continuum-only unitary. No zero-energy bound copies are introduced.

[Conclusion] Clauses (a), (b), (c), (d) are steps 1.1 and 2.1; 3.1 through 6.1; 7.1; and 8.1. Step 7.2 explicitly realizes the isometric adjoint inverse.