Transported radial endpoint realization

Put $D_{0}u=u^{\mathrm{\prime }}-u\mathrm{/}r$ on $C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$. This operator is closable: if $u_n\to 0$ and $D_0{u}_n\to g$ in $L^2$, then on every compact subinterval $u_n\mathrm{/}r\to 0$ in $L^2$ and $u_n^{\mathrm{\prime }}\to 0$ distributionally, so $g=0$ distributionally and almost everywhere. Integration of $(\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}r{)}^{\mathrm{\prime }}$ on the core gives $\mathrm{\parallel }{D}_{0}u{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }u\mathrm{/}r{\mathrm{\parallel }}_2^2=-2\mathrm{Re}\int (u\mathrm{/}r)\overline{D_{0}u}$; consequently $\mathrm{\parallel }u\mathrm{/}r{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }{D}_{0}u{\mathrm{\parallel }}_2$. Applied to differences, these identities make the defining $V_l$ norm equivalent to the $H^1$ norm for each fixed $l$, and identify its completion in $L^2$ with $H_0^1(0,\mathrm{\infty })$. In particular the graph completion has injective first projection, its derivative is distributional, and it is a Hilbert space densely and continuously embedded in $L^2$. No fixed-$l$ equivalence constant will be used in an infinite channel sum.

Let $\mathcal{W}$ be the Hilbert direct sum of $V_l$ over $(l,m)$, with squared norm $\sum \mathrm{\parallel }{u}_{lm}{\mathrm{\parallel }}_{V_l}^2$. On finite compact smooth radial sums, the corrected angular-radial block identity and Euclidean integration by parts give $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{H^1}^2=\sum \mathrm{\parallel }{u}_{lm}{\mathrm{\parallel }}_{V_l}^2$ and $q(\psi ,\phi )=\sum a_l(u_{lm},v_{lm})$, where $q=\int \mathrm{\nabla }\psi \cdot \overline{\mathrm{\nabla }\phi }-\gamma \int \psi \bar{\phi }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$. Polar integration gives the actual $r^{-1}$ potential weight. To establish the necessary full $H^1$ density, first use smooth density and the origin-removing cutoffs of [F7], followed by compact support and mollification, to reduce to $\psi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\setminus \{0\})$.

The trace at zero of an $H^1(0,1)$ function exists by $\mathrm{\mid }u(t)-u(s)\mathrm{\mid }\le \mathrm{\mid }t-s{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_2$, and is continuous in its $H^1$ norm: choose a point where $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(0,1)}$ and use this estimate to bound $\mathrm{\mid }u(0)\mathrm{\mid }$ by $\mathrm{\parallel }u{\mathrm{\parallel }}_2+\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_2$. Thus the completion in step 1.1 has zero trace. Conversely $u,u^{\mathrm{\prime }}-u\mathrm{/}r,u\mathrm{/}r\in L^2$ gives $u^{\mathrm{\prime }}\in L^2$ and a trace; its trace must be zero since otherwise $\int \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{r}^2$ diverges. Take a smooth cutoff ${\chi }_{\epsilon }$ vanishing below $\epsilon$, one beyond $2\epsilon$, with $\mathrm{\mid }{\chi }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\le c\mathrm{/}\epsilon$; then $\mathrm{\parallel }{\chi }_{\epsilon }^{\mathrm{\prime }}u{\mathrm{\parallel }}_2^2\le 4c^2{\int }_0^{2\epsilon }\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{r}^2\to 0$, and all omitted function, derivative and weighted tails tend to zero. A cutoff between $R$ and $2R$ adds a derivative bounded by $(c\mathrm{/}R)\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(R,2R)}$. Mollify the resulting compactly supported functions on intervals away from zero; $H^1$ convergence and boundedness of $1\mathrm{/}r$ there give core convergence in $V_l$. This proves exactly the stated distributional realization, including $l=0$, and the first part of (a).

For such an annular smooth $\psi$, let ${\psi }_N=P_N\psi$ be the angular truncation $l\le N$. All its radial coefficients are smooth with the same annular support. The polar formula gives $\mathrm{\Delta }{\psi }_N=P_N\mathrm{\Delta }\psi$: radial differentiation commutes with the finite projection, and angular integration by parts gives $⟨-{\mathrm{\Delta }}_{S^2}\psi (r\cdot ),Y_{lm}⟩={\lambda }_l⟨\psi (r\cdot ),Y_{lm}⟩$. This latter identity uses only the supplied finite-span eigenidentity, with a local Green argument for the other smooth function: excise caps of angular radius $\delta$, integrate the chart expression, cancel periodic azimuth boundaries, and let $\delta \downarrow 0$; smooth intrinsic gradients are bounded and cap boundaries have length $O(\delta )$, so their terms vanish. Angular completeness and polar integration give ${\psi }_N\to \psi$ and $\mathrm{\Delta }{\psi }_N\to \mathrm{\Delta }\psi$ in $L^2$. For the smooth annular residual $R_N$, integration by parts gives $\mathrm{\parallel }\mathrm{\nabla }{R}_N{\mathrm{\parallel }}^2=⟨-\mathrm{\Delta }{R}_N,R_N⟩\le \mathrm{\parallel }\mathrm{\Delta }{R}_N\mathrm{\parallel }\mathrm{\parallel }{R}_N\mathrm{\parallel }\to 0$. Thus finite compact smooth radial sums are $H^1$-dense with the gradient tail actually controlled.

Write $Du=u^{\mathrm{\prime }}-u\mathrm{/}r$ and ${\lambda }_l=l(l+1)$. Hardy and Cauchy–Schwarz give $\mathrm{\mid }\int u\bar{v}\mathrm{/}r\mathrm{\mid }\le 2\mathrm{\parallel }Du{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2$; the derivative and angular terms are bounded by their corresponding $V_l$ norms. Hence $a_l$ is bounded and symmetric. The inequality $2\gamma xy\le x^2\mathrm{/}2+2{\gamma }^2{y}^2$ gives the uniform bound $a_l[u]\ge \frac{1}{2}\mathrm{\parallel }Du{\mathrm{\parallel }}_2^2+{\lambda }_l\mathrm{\parallel }u\mathrm{/}r{\mathrm{\parallel }}_2^2-2{\gamma }^2\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. With $\mu =2{\gamma }^2+1$, the shifted form $b_l=a_l+\mu ⟨\cdot ,\cdot ⟩$ is a complete inner product with $\frac{1}{2}\mathrm{\parallel }u{\mathrm{\parallel }}_{V_l}^2\le b_l[u]\le c_{\gamma }\mathrm{\parallel }u{\mathrm{\parallel }}_{V_l}^2$, uniformly in $l$. Riesz representation on $(V_l,b_l)$ solves $b_l(u,v)=⟨f,v⟩$ for every $f\in L^2$, so the actual domain definition gives $u\in D(A_l)$ and $(A_l+\mu )u=f$. Testing $v=u$ gives uniqueness. This proves closedness and semiboundedness of the form and surjectivity of its associated shifted operator.

For $u\in C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$, integration by parts gives $a_l(u,v)=⟨L_{l}u,v⟩$ first for core $v$, where $L_{l}u=-u^{\mathrm{\prime }\mathrm{\prime }}+({\lambda }_l\mathrm{/}{r}^2-\gamma \mathrm{/}r)u\in L^2$. Both sides are continuous in $V_l$, so the identity extends to every $v\in V_l$. Thus the core is contained in $D(A_l)$ and this operator domain is $L^2$-dense before any adjoint is taken. The representing $f$ is unique by $V_l$ density. Symmetry of the form gives symmetry of $A_l$. For $y\in D(A_l^{\ast })$, solve $(A_l+\mu )x=(A_l^{\ast }+\mu )y$ by step 3.1; then $y-x\in \ker (A_l^{\ast }+\mu )=\mathrm{ran}(A_l+\mu {)}^{\perp }=0$. Hence $A_l=A_l^{\ast }$ on exactly its defined domain, and $⟨A_{l}u,u⟩=a_l[u]\ge -2{\gamma }^2\mathrm{\parallel }u{\mathrm{\parallel }}^2$.

Finite-support families with core coordinates are dense in $\mathcal{W}$, since each $V_l$ is the core completion. Step 2.2 therefore extends the finite-core isometry onto $\mathcal{W}$, with its $L^2$ extension agreeing with $U$: $U(H^1)=\mathcal{W}$. The full polarized form identity extends by continuity. In particular the infinite Coulomb sum is uniformly bounded by $\sum \mathrm{\mid }\int u_{lm}\overline{v_{lm}}\mathrm{/}r\mathrm{\mid }\le 2(\sum \mathrm{\parallel }D{u}_{lm}{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}(\sum \mathrm{\parallel }{v}_{lm}{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}$; kinetic and angular sums have their corresponding Cauchy–Schwarz bounds. This uses no equivalence constant growing with $l$.

If $w\in V_l$ is $b_l$-orthogonal to $D(A_l)$, the solution $u_f=(A_l+\mu {)}^{-1}f$ from step 3.1 gives $0=b_l(u_f,w)=⟨f,w⟩$ for every $f\in L^2$; therefore $w=0$. Orthogonal decomposition in the Hilbert space $(V_l,b_l)$ implies that the closure of $D(A_l)$ is all $V_l$. Norm equivalence makes this the required density in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{V_l}$ and completes (a),(b). This proves form density directly, without a spectral-form assertion.

For $u\in D(A_l)$, test its form identity against core $v$. Distributional integration by parts gives $-u^{\mathrm{\prime }\mathrm{\prime }}+({\lambda }_l\mathrm{/}{r}^2-\gamma \mathrm{/}r)u=A_{l}u$; the cross terms in $(u^{\mathrm{\prime }}-u\mathrm{/}r)\overline{(v^{\mathrm{\prime }}-v\mathrm{/}r)}$ cancel by integrating $(u\bar{v}\mathrm{/}r{)}^{\mathrm{\prime }}$. On a compact interval away from zero the potential coefficient is bounded, so $u^{\mathrm{\prime }\mathrm{\prime }}\in L^2$ there and step 2.1 gives $u,u^{\mathrm{\prime }}\in L^2$. Thus $u\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$, and [F9] gives its $C^1$ representative and locally absolutely continuous derivative. If $A_{l}u=Eu$, the right side for $u^{\mathrm{\prime }\mathrm{\prime }}$ is continuous, so integration makes $u^{\mathrm{\prime }}$ of class $C^1$ and the equation holds everywhere with $u\in C^2$. This proves (c) without regularity at the origin.

Define $\widehat{A}=⨁A_l$ on exactly the square-summable graph domain in (d). Finite-support domain vectors are dense and coordinate symmetry gives symmetry. Self-adjointness of each $A_l$ gives $\mathrm{\parallel }(A_l\pm i{)}^{-1}\mathrm{\parallel }\le 1$, by $\mathrm{\parallel }(A_l\pm i)u{\mathrm{\parallel }}^2=\mathrm{\parallel }{A}_{l}u{\mathrm{\parallel }}^2+\mathrm{\parallel }u{\mathrm{\parallel }}^2$. For $f\in K$, take $u_{lm}=(A_l\pm i{)}^{-1}{f}_{lm}$; the coordinate norm sum is bounded by $\mathrm{\parallel }f{\mathrm{\parallel }}^2$, and $A_l{u}_{lm}=f_{lm}\mp i{u}_{lm}$ has square-summable norms. Thus both ranges of $\widehat{A}\pm i$ are all $K$, proving self-adjointness by the symmetric range criterion. Coordinate closedness also directly proves the graph is closed.

For $\psi \in H^2=D(C)$ and $\phi \in H^1$, weak Fourier integration by parts and three-dimensional Hardy give $q(\psi ,\phi )=⟨C\psi ,\phi ⟩$: prove it first for core $\phi$ and extend by $H^1$ continuity, using $\psi \mathrm{/}\mathrm{\mid }x\mathrm{\mid }\in L^2$. By step 4.2, $u=U\psi \in \mathcal{W}$. Fix one coordinate $j=(l,m)$ and $v\in V_l$; then $\phi =U^{-1}(v{\delta }_j)\in H^1$ and the full form identity gives $a_l(u_j,v)=⟨(UC\psi {)}_j,v⟩$. Hence $u_j\in D(A_l)$ and $A_l{u}_j=(UC\psi {)}_j$. The latter coordinates are square summable, so $C\subseteq U^{\ast }\widehat{A}U$. Both operators are self-adjoint; taking adjoints reverses the inclusion, giving equality. This proves the full iff graph-domain statement and action formula (d), also for $\gamma =0$ using its explicit free operator.

If $A_{l}u=Eu$, the one-block family $u{\delta }_{(l,0)}$ belongs to $D(\widehat{A})$; step 6.1 gives $\psi =U^{-1}(u{\delta }_{(l,0)})\in H^2$, $C\psi =E\psi$, and $\psi =u(r)Y_{l0}(\mathrm{\Omega })\mathrm{/}r$ almost everywhere. Its $H^2$ Fourier transform lies in $L^1$ by Cauchy–Schwarz with $(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-1}\in L^2$, so the absolutely convergent inverse defines a bounded continuous representative by dominated convergence, agreeing with the $L^2$ inverse by truncation. Step 5.2 and continuity of $Y_{l0}$ make the polar expression continuous away from zero; two continuous functions agreeing almost everywhere on that open set agree everywhere, since any nonzero difference persists on a positive-measure neighborhood. Choose ${\mathrm{\Omega }}_0$ with $Y_{l0}({\mathrm{\Omega }}_0)\mathrm{\ne }0$; then $\mathrm{\mid }u(r)\mathrm{/}r\mathrm{\mid }\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}\mathrm{\mid }{Y}_{l0}({\mathrm{\Omega }}_0)\mathrm{\mid }$ for $0<r\le 1$. This proves (e) with genuine operator-domain transport and justifies evaluation on a fixed ray. All clauses (a)–(e) follow.