Angular radial unitary reduction

[Polar measurability and the angular expansion] For continuous compactly supported $\psi$, $(r,\mathrm{\Omega })\mapsto \psi (r\mathrm{\Omega })$ is continuous on $(0,\mathrm{\infty })\times S^2$, and the polar formula [F3] gives ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }\psi {\mathrm{\mid }}^{2}dx={\int }_0^{\mathrm{\infty }}{r}^2{\int }_{S^2}\mathrm{\mid }\psi (r\mathrm{\Omega }){\mathrm{\mid }}^{2}d\sigma dr$. For a general $L^2$ class, take a finite Lebesgue-measurable representative and use [F8] on its real and imaginary parts to obtain Borel representatives, setting them to zero on the Borel null sets where they are not finite; their combination is a finite Borel representative ${\psi }_0$. If ${\psi }_0$ and ${\tilde{\psi }}_0$ are two such representatives, their Borel disagreement set is Lebesgue-null; applying [F3] to its indicator shows that their polar pullbacks agree for $r^{2}drd\sigma$-almost every $(r,\mathrm{\Omega })$, so the angular coefficient functions agree for almost every $r$. Thus the coefficients depend only on the $L^2$ class. Since $(r,\mathrm{\Omega })\mapsto {\psi }_0(r\mathrm{\Omega })$ is Borel, completed-product Fubini [F3] gives measurable sections and Parseval [F2] gives ${\int }_{S^2}\mathrm{\mid }{\psi }_0(r\mathrm{\Omega }){\mathrm{\mid }}^{2}d\sigma ={\sum }_{l,m}\mathrm{\mid }{c}_{lm}(r){\mathrm{\mid }}^2$ for almost every $r$. The polar formula [F3] then yields ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }\psi {\mathrm{\mid }}^{2}dx={\int }_0^{\mathrm{\infty }}{r}^2{\sum }_{l,m}\mathrm{\mid }{c}_{lm}(r){\mathrm{\mid }}^{2}dr$.

[Polar form of the Laplacian] Let $f\in C^2({\mathbb{R}}^3\setminus \{0\})$, put $F(r,\mathrm{\Omega }):=f(r\mathrm{\Omega })$, and use the chart $\mathrm{\Omega }(\theta ,\phi )=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$ for $0<\theta <\pi$. Write $D^{2}f$ for the Hessian matrix. The chain rule gives ${\mathrm{\partial }}_{r}F=\mathrm{\nabla }f(r\mathrm{\Omega })\cdot \mathrm{\Omega }$ and ${\mathrm{\partial }}_r^{2}F=D^{2}f(r\mathrm{\Omega })[\mathrm{\Omega },\mathrm{\Omega }]$. For $a\in \{\theta ,\phi \}$ one has ${\mathrm{\partial }}_{a}F=r\mathrm{\nabla }f(r\mathrm{\Omega })\cdot {\mathrm{\partial }}_a\mathrm{\Omega }$ and ${\mathrm{\partial }}_a^{2}F=r^2{D}^{2}f(r\mathrm{\Omega })[{\mathrm{\partial }}_a\mathrm{\Omega },{\mathrm{\partial }}_a\mathrm{\Omega }]+r\mathrm{\nabla }f(r\mathrm{\Omega })\cdot {\mathrm{\partial }}_a^2\mathrm{\Omega }$. Using ${\mathrm{\Delta }}_{S^2}={\mathrm{\partial }}_{\theta }^2+\cot \theta {\mathrm{\partial }}_{\theta }+{\sin }^{-2}\theta {\mathrm{\partial }}_{\phi }^2$, the identities ${\mathrm{\partial }}_{\theta }\mathrm{\Omega }\otimes {\mathrm{\partial }}_{\theta }\mathrm{\Omega }+{\sin }^{-2}\theta {\mathrm{\partial }}_{\phi }\mathrm{\Omega }\otimes {\mathrm{\partial }}_{\phi }\mathrm{\Omega }=I-\mathrm{\Omega }\otimes \mathrm{\Omega }$ and ${\mathrm{\partial }}_{\theta }^2\mathrm{\Omega }+\cot \theta {\mathrm{\partial }}_{\theta }\mathrm{\Omega }+{\sin }^{-2}\theta {\mathrm{\partial }}_{\phi }^2\mathrm{\Omega }=-2\mathrm{\Omega }$ give ${\mathrm{\Delta }}_{S^2}F=r^2(\mathrm{\Delta }f-D^{2}f[\mathrm{\Omega },\mathrm{\Omega }])-2r{\mathrm{\partial }}_{r}F$. Substitution of ${\mathrm{\partial }}_r^{2}F=D^{2}f[\mathrm{\Omega },\mathrm{\Omega }]$ yields ${\mathrm{\partial }}_r^{2}F+2r^{-1}{\mathrm{\partial }}_{r}F+r^{-2}{\mathrm{\Delta }}_{S^2}F=\mathrm{\Delta }f(r\mathrm{\Omega })$, proving (c) on this chart. Both sides are continuous on ${\mathbb{R}}^3\setminus \{0\}$, so the identity also holds on the polar axis by continuity.

[The isometry] By step 1.1 and Tonelli [F3] applied to the nonnegative function $r^2{\sum }_{l,m}\mathrm{\mid }{c}_{lm}(r){\mathrm{\mid }}^2$ one has ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }\psi {\mathrm{\mid }}^{2}dx={\int }_0^{\mathrm{\infty }}{r}^2{\sum }_{l,m}\mathrm{\mid }{c}_{lm}(r){\mathrm{\mid }}^{2}dr={\sum }_{l,m}{\int }_0^{\mathrm{\infty }}\mathrm{\mid }{u}_{lm}(r){\mathrm{\mid }}^{2}dr$ for continuous compactly supported $\psi$, and both sides are finite; the linear map $U$ is thus isometric on the dense subspace $C_c({\mathbb{R}}^3)$ (density [F5] applies to $C_c^{\mathrm{\infty }}\subseteq C_c$), and for $\psi \in L^2$ the a.e. definition of $u_{lm}$ gives the same isometry by continuity and [F6].

[Reduction of the free Laplacian on one block] Let $u\in C_c^{\mathrm{\infty }}((0,\mathrm{\infty }))$ and $\psi (x):=f(r)Y_{lm}(\mathrm{\Omega })$ with $f(r):=u(r)\mathrm{/}r$ and $x=r\mathrm{\Omega }$. The polynomial restriction formula in the statement follows from $(\sin \theta {)}^m{e}^{im\phi }=({\mathrm{\Omega }}_1+i{\mathrm{\Omega }}_2{)}^m$ and [F4], including the cut and poles by continuity. Then $\psi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\setminus \{0\})$, because $\mathrm{supp}u$ is a compact subset of $(0,\mathrm{\infty })$ and $Y_{lm}$ is smooth on $S^2$. Since $Y_{lm}$ lies in the finite span, [F4] gives ${\mathrm{\Delta }}_{S^2}\psi (r\cdot )=-l(l+1)\psi (r\cdot )$ on the chart; by step 1.2, $-\mathrm{\Delta }\psi =-{\mathrm{\partial }}_r^2\psi -2r^{-1}{\mathrm{\partial }}_r\psi -r^{-2}{\mathrm{\Delta }}_{S^2}\psi =(-f^{\mathrm{\prime }\mathrm{\prime }}(r)-2f^{\mathrm{\prime }}(r)\mathrm{/}r+l(l+1)f(r)\mathrm{/}{r}^2)Y_{lm}$, and substituting $f=u\mathrm{/}r$ gives $-f^{\mathrm{\prime }\mathrm{\prime }}-2f^{\mathrm{\prime }}\mathrm{/}r+l(l+1)f\mathrm{/}{r}^2=(-u^{\mathrm{\prime }\mathrm{\prime }}+l(l+1)u\mathrm{/}{r}^2)\mathrm{/}r$, which is the first identity of (d).

[Surjectivity and inverse] For a finite-support vector $(u_{lm})\in K$, extend each radial entry by zero to $\mathbb{R}$, apply [F8] with $n=1$ to its real and imaginary parts, set any Borel null set of nonfinite values to zero, and restrict the resulting finite-valued Borel representatives to $(0,\mathrm{\infty })$. Define $V(u_{lm}):={\sum }_{l,m}{u}_{lm}(r)Y_{lm}(\mathrm{\Omega })\mathrm{/}r$ for $r>0$ (assign any value at the origin). This is Borel; using [F3] and [F2], $\mathrm{\parallel }V(u_{lm}){\mathrm{\parallel }}_{L^2}^2={\int }_0^{\mathrm{\infty }}{r}^2{\int }_{S^2}\mathrm{\mid }\sum u_{lm}(r)Y_{lm}(\mathrm{\Omega })\mathrm{/}r{\mathrm{\mid }}^{2}d\sigma dr={\sum }_{l,m}\int \mathrm{\mid }{u}_{lm}{\mathrm{\mid }}^{2}dr$, the cross terms vanishing because $\int Y_{lm}\stackrel{‾}{Y_{l^{\mathrm{\prime }}{m}^{\mathrm{\prime }}}}d\sigma ={\delta }_{l{l}^{\mathrm{\prime }}}{\delta }_{m{m}^{\mathrm{\prime }}}$ by [F2]; hence $V$ is isometric on the dense finite-support subspace of $K$ and extends to a linear isometry $K\to L^2$. On finite sums the definitions give $U(V(u_{lm}))=(u_{lm})$, and on $C_c({\mathbb{R}}^3)$ the angular expansion shows $V(U\psi )=\psi$; by density and continuity these identities extend, so $V=U^{-1}$ and $U$ is unitary.

[The free form on one block] With $u,\psi$ as in step 2.2, ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }\mathrm{\nabla }\psi {\mathrm{\mid }}^{2}dx=⟨-\mathrm{\Delta }\psi ,\psi ⟩$ since [F9] applied componentwise on a ball containing the support to the compactly supported $C^1$ field $\bar{\psi }\mathrm{\nabla }\psi$ gives $\int \mathrm{\mid }\mathrm{\nabla }\psi {\mathrm{\mid }}^2=\int (-\mathrm{\Delta }\psi )\bar{\psi }$ with zero boundary flux; substituting step 2.2 and using $\mathrm{\parallel }{Y}_{lm}{\mathrm{\parallel }}_{L^2(S^2)}=1$ yields ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }\mathrm{\nabla }\psi {\mathrm{\mid }}^{2}dx={\int }_0^{\mathrm{\infty }}(-u^{\mathrm{\prime }\mathrm{\prime }}(r)\overline{u(r)}+l(l+1)\mathrm{\mid }u(r){\mathrm{\mid }}^2\mathrm{/}{r}^2)dr$. Integration by parts from [F9] in the compact support of $u$ gives ${\int }_0^{\mathrm{\infty }}-u^{\mathrm{\prime }\mathrm{\prime }}\bar{u}dr={\int }_0^{\mathrm{\infty }}\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^{2}dr$, and the elementary identity $-2\mathrm{Re}(u^{\mathrm{\prime }}\bar{u})\mathrm{/}r+\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{r}^2=-(\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}r{)}^{\mathrm{\prime }}$ integrates to zero over the compact support, so $\int \mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2=\int \mathrm{\mid }{u}^{\mathrm{\prime }}-u\mathrm{/}r{\mathrm{\mid }}^2$; combining, the right-hand side equals ${\int }_0^{\mathrm{\infty }}(\mathrm{\mid }{u}^{\mathrm{\prime }}-u\mathrm{/}r{\mathrm{\mid }}^2+l(l+1)\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}{r}^2)dr$, which is the second identity of (d).

[Conclusion] On the dense subspace of finite sums of one-block functions the map $U$ is given by the displayed coefficient formula, so clause (a) follows from step 2.1 for all $\psi \in L^2$; clause (b) is step 3.1; clauses (c) and (d) are steps 1.2, 2.2 and 3.2.