Laguerre bound energy exhaustion

[The nonterminating tail] Suppose $j\mathrm{\notin }{\mathbb{N}}_0$. Every $v_i$ is real and nonzero. Choose $\epsilon \in (0,1\mathrm{/}2)$ and an integer $I>\max (j,0)$ large enough that $(i-j)\mathrm{/}(i+2l+2)\ge 1-\epsilon$ for all $i\ge I$. All coefficients from $I$ onward have one sign and $\mathrm{\mid }{v}_{i+1}\mathrm{\mid }\ge (1-\epsilon )\mathrm{\mid }{v}_i\mathrm{\mid }\mathrm{/}(i+1)$; induction gives $\mathrm{\mid }{v}_i\mathrm{\mid }\ge c(1-\epsilon {)}^i\mathrm{/}i!$ for $i\ge I$, where $c=\mathrm{\mid }{v}_I\mathrm{\mid }I!(1-\epsilon {)}^{-I}>0$. Since the series is absolutely convergent, on $x\ge 0$ the absolute value of its same-sign tail is its sum of absolute values. The reverse triangle inequality gives $\mathrm{\mid }v(x)\mathrm{\mid }\ge c{e}^{(1-\epsilon )x}-P(x)$, where one may take the nonnegative polynomial $P(x)={\sum }_{i<I}\mathrm{\mid }{v}_i\mathrm{\mid }{x}^i+c{\sum }_{i<I}(1-\epsilon {)}^i{x}^i\mathrm{/}i!$. Thus for sufficiently large $r$, $\mathrm{\mid }{u}_{\mathrm{r}\mathrm{e}\mathrm{g}}(r)\mathrm{\mid }\ge (c\mathrm{/}2)r^{l+1}{e}^{k(1-2\epsilon )r}$. Its squared integral diverges, since the exponent is positive; hence $u_{\mathrm{r}\mathrm{e}\mathrm{g}}\mathrm{\notin }{L}^2(0,\mathrm{\infty })$.

[Terminating solutions belong to the form domain] Suppose $j\in {\mathbb{N}}_0$ and write $u=r^{l+1}{e}^{-kr}v(2kr)$. Near zero, $u=O(r^{l+1})$, $u^{\mathrm{\prime }}=O(r^l)$ and $u\mathrm{/}r=O(r^l)$; at infinity each decays exponentially times a polynomial. Therefore $u,u^{\mathrm{\prime }},u\mathrm{/}r\in L^2$, and the trace at zero is zero. To verify $u\in V_l$ directly, choose a smooth cutoff ${\chi }_{\delta }$ that vanishes below $\delta$, equals one above $2\delta$ and has $\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}\mathrm{\mid }\le C\mathrm{/}\delta$. The additional derivative term obeys $\mathrm{\parallel }{\chi }_{\delta }^{\mathrm{\prime }}u{\mathrm{\parallel }}_2^2\le 4C^2{\int }_{\delta }^{2\delta }\mathrm{\mid }u\mathrm{/}r{\mathrm{\mid }}^{2}dr\to 0$. The other omitted terms tend to zero by integrability; cutting off the exponential tail and then smoothing on compact subsets of $(0,\mathrm{\infty })$ gives compact smooth approximants in the $V_l$ norm. This is membership in the specified endpoint realization, rather than membership inferred from $L^2$ alone.

[The actual origin condition excludes the second solution] If $u\in D(A_l)$ and $A_{l}u=-k^{2}u$, then $u$ solves the classical radial equation by [F2]. By [F1] it is $c_1{u}_{\mathrm{r}\mathrm{e}\mathrm{g}}+c_2{u}_2$. The regular quotient $u_{\mathrm{r}\mathrm{e}\mathrm{g}}\mathrm{/}r=O(r^l)$ is bounded, whereas $u_2\mathrm{/}r\sim -r^{-l-1}\mathrm{/}(2l+1)$. If $c_2\mathrm{\ne }0$, the reverse triangle inequality therefore makes $\mathrm{\mid }u(r)\mathrm{\mid }\mathrm{/}r\to \mathrm{\infty }$, contrary to [F2]. Consequently $u=c_1{u}_{\mathrm{r}\mathrm{e}\mathrm{g}}$, and every negative eigenspace has dimension at most one. Together with step 1.1 this proves (a).

[The form identity gives the operator domain and $H^2$ transport] The substitution of [F1] shows that $u$ solves $-u^{\mathrm{\prime }\mathrm{\prime }}+(l(l+1)\mathrm{/}{r}^2-\gamma \mathrm{/}r)u=-k^{2}u$ on $(0,\mathrm{\infty })$. Against $w\in C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$, integration by parts gives $a_l(u,w)=\int (u^{\mathrm{\prime }}\overline{w^{\mathrm{\prime }}}+l(l+1)u\bar{w}\mathrm{/}{r}^2-\gamma u\bar{w}\mathrm{/}r)dr=-k^2⟨u,w⟩$. The equality between the kinetic forms follows by integrating $(u\bar{w}\mathrm{/}r{)}^{\mathrm{\prime }}$, with no boundary term because $w$ is compactly supported away from zero. Boundedness of $a_l$ and density in [F2] extend the identity to every $w\in V_l$. The actual definition of $D(A_l)$ now gives $u\in D(A_l)$ and $A_{l}u=-k^{2}u$. The vector with this one $(l,m)$ component belongs to the direct-sum graph domain, so the exact block identity gives $\psi =u(r)Y_{lm}(\mathrm{\Omega })\mathrm{/}r\in D(C)=H^2({\mathbb{R}}^3)$ and $C\psi =-k^2\psi$ for every $m$. This argument uses no assertion that $e^{-k\mathrm{\mid }x\mathrm{\mid }}$ is smooth at $x=0$.

[All radial negative eigenvalues] Termination means $j=n-l-1\in {\mathbb{N}}_0$, so $n\ge l+1$ and $k=\gamma \mathrm{/}(2n)$. Step 2.2 supplies a nonzero eigenvector at each $E_n=-{\gamma }^2\mathrm{/}(4n^2)$, and step 2.1 shows its eigenspace is exactly $\mathbb{C}{u}_{\mathrm{r}\mathrm{e}\mathrm{g}}$. All other negative energies are excluded by steps 1.1 and 2.1. This proves (b), including every integer $n\ge l+1$, with no finite truncation of the bound family.

[All negative eigenvalues and eigenvectors of $C$] By the exact block/domain identity, $C\psi =E\psi$ is equivalent to $A_l{u}_{lm}=E{u}_{lm}$ in every component, where $U\psi =(u_{lm})$. If $\psi \mathrm{\ne }0$, at least one component is nonzero. Step 3.1 therefore forces $E=E_n$ for some $n\ge 1$, and all components with $l\ge n$ vanish; the remaining finitely many components are scalar multiples of the regular solutions at $k=\gamma \mathrm{/}(2n)$. Conversely each $E_n$ is a $C$ eigenvalue by taking the $l=m=0$ vector from step 2.2. Thus (c) and (d) hold. In fact every such eigenfunction is a finite linear combination of the transported regular solutions, which implies the asserted closed-span membership.

[Entire negative spectrum, including radial blocks] By [F3], every negative spectral point of $C$ is an eigenvalue, so step 4.1 proves $\sigma (C)\cap (-\mathrm{\infty },0)=\{E_n:n\ge 1\}$. Fix $l$ and realize $A_l$ as the reducing $(l,0)$ block of $UC{U}^{\ast }$. If $\lambda <0$ is outside $\sigma (C)$, the restricted inverse of $C-\lambda$ supplies a bounded inverse for $A_l-\lambda$. If $\lambda =E_n$ with $n<l+1$, choose an open interval $J=(\lambda -\delta ,\lambda +\delta )$ whose intersection with $\sigma (C)$ is just $\{\lambda \}$, as allowed by [F3]. The restricted projection on $J$ equals the restricted projection on $\{\lambda \}$ by [F4]; this is zero, since step 3.1 excludes a radial eigenvector at this energy. The restricted PVM is consequently supported outside $J$. Define $b(t)=(t-\lambda {)}^{-1}$ outside $J$ and $b(t)=0$ inside $J$. Both $b$ and $tb$ are bounded. By the exact-domain calculus [F4], its block multiplier maps into $D(A_l)$ and gives both $(A_l-\lambda )b(A_l)=I$ and $b(A_l)(A_l-\lambda )=I$ on the respective exact domains. Thus $\lambda \mathrm{\notin }\sigma (A_l)$. The remaining negative points are eigenvalues from step 3.1, proving $\sigma (A_l)\cap (-\mathrm{\infty },0)=\{E_n:n\ge l+1\}$. This establishes the final exhaustion sentence for both operators, including absence of additional continuous negative radial spectrum.

[Conclusion] Step 2.1 proves (a), step 3.1 proves (b), step 4.1 proves (c) and (d), and step 5.1 proves the stated exhaustion of the entire negative spectra.