No positive compact-potential eigenvalues

(Bounded-potential operator on its stated domain.) Let $H_0={\mathcal{F}}_2^{-1}{M}_q{\mathcal{F}}_2$, $q(\xi )=4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$, on the domain in [F4]. This domain is dense because truncations of an $L^2$ Fourier function to $\{q\le n\}$ belong to it and converge in $L^2$. Since $q$ is real, $H_0$ is symmetric, and for $z=\pm i\lambda$, $\lambda >0$, the explicit multiplier inverse $R_0(z)={\mathcal{F}}_2^{-1}{M}_{(q-z{)}^{-1}}{\mathcal{F}}_2$ has norm at most ${\lambda }^{-1}$ and maps $L^2$ into $D(H_0)$, because $q\mathrm{/}(q-z)$ is bounded; hence $H_0-z$ is onto. The operator $H=H_0+M_V$ on the same dense domain is symmetric because $V$ is real and $M_V$ is bounded. Choose $\lambda >\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}$. Then $\mathrm{\parallel }{M}_V{R}_0(\pm i\lambda )\mathrm{\parallel }<1$, so the norm-convergent geometric series gives $(I+M_V{R}_0(z){)}^{-1}$ and $(H-z{)}^{-1}=R_0(z)(I+M_V{R}_0(z){)}^{-1}$. In particular $H\pm i\lambda$ are onto. To verify self-adjointness directly, for $y\in D(H^{\ast })$ solve $(H+i\lambda )x=(H^{\ast }+i\lambda )y$ with $x\in D(H)$; then $y-x\in \ker (H^{\ast }+i\lambda )=\mathrm{ran}(H-i\lambda {)}^{\perp }=\{0\}$. Thus $D(H^{\ast })\subseteq D(H)$, and symmetry gives equality and $H=H^{\ast }$. The free multiplier identity in [F4] identifies this $H$ with $-\mathrm{\Delta }+V$ distributionally on $H^2$. Only explicit norm-convergent series and the CC Fourier interfaces are used, with no Kato–Rellich or arbitrary-choice multiplier supplier.

(Angular test functions and the exterior radial equation.) The harmonics of [F1] are smooth on the whole sphere: for $m\ge 0$, their formula is the restriction of a constant multiple of $(x_1+i{x}_2{)}^m{P}_l^{(m)}(x_3)$ to $S^2$, and the negative indices are conjugates. In angle coordinates, with $t=\cos \theta$, direct differentiation gives $-{\sin }^{-1}\theta {\mathrm{\partial }}_{\theta }(\sin \theta {\mathrm{\partial }}_{\theta }w(t))=-((1-t^2)w^{\mathrm{\prime }}(t){)}^{\mathrm{\prime }}$ and $-{\sin }^{-2}\theta {\mathrm{\partial }}_{\phi }^2{e}^{im\phi }=m^2(1-t^2{)}^{-1}{e}^{im\phi }$. Thus the associated Legendre equation gives $-{\mathrm{\Delta }}_{S^2}{Y}_{lm}=l(l+1)Y_{lm}$ away from the poles and hence everywhere by smoothness. Define $f_{lm}(r)={\int }_{S^2}\psi (r\omega )\overline{Y_{lm}(\omega )}d\sigma (\omega )$ for almost every $r>R$ and ${\alpha }_{lm}(r)=r{f}_{lm}(r)$. On every compact radial interval $[a,b]\subset (R,\mathrm{\infty })$, the smooth polar changes of coordinates and angular integration send the locally $H^2$ function $\psi$ to $f_{lm}\in H^2(a,b)$: this follows by mollifying $\psi$ on a slightly larger annulus, applying the ordinary chain rule to the smooth approximants, and passing in $H^2$; the coordinate Jacobians and coefficients and their first two derivatives are bounded on a finite sphere-chart cover of that annulus. The polar Laplacian formula is valid here without a domain assumption on an infinite angular diagonal operator: for a smooth Cartesian $h$, writing $G(r,\omega )=h(r\omega )$, tangential differentiation gives ${\mathrm{\Delta }}_{S^2}G=r^2{\sum }_{i,j}({\delta }_{ij}-{\omega }_i{\omega }_j){\mathrm{\partial }}_i{\mathrm{\partial }}_{j}h-2r{\mathrm{\partial }}_{r}G$, while ${\mathrm{\partial }}_r^{2}G={\sum }_{i,j}{\omega }_i{\omega }_j{\mathrm{\partial }}_i{\mathrm{\partial }}_{j}h$; addition yields $\mathrm{\Delta }h={\mathrm{\partial }}_r^{2}G+2r^{-1}{\mathrm{\partial }}_{r}G+r^{-2}{\mathrm{\Delta }}_{S^2}G$. The same identity passes to distributions by the smooth approximants just described. Testing the exterior equation $-\mathrm{\Delta }\psi =k^2\psi$ against $\eta (r)Y_{lm}(\omega )$, $\eta \in C_c^{\mathrm{\infty }}((R,\mathrm{\infty }))$, and using the polar Laplacian and angular integration by parts gives $(r^2{f}_{lm}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}+(k^2{r}^2-l(l+1))f_{lm}=0$ distributionally. Consequently ${\alpha }_{lm}^{\mathrm{\prime }\mathrm{\prime }}+(k^2-l(l+1)r^{-2}){\alpha }_{lm}=0$ on $(R,\mathrm{\infty })$. The smooth approximants just constructed converge in radial $H^2$; [F5] therefore supplies a $C^1$ representative of ${\alpha }_{lm}$ on each compact radial interval. Its distributional second derivative is the continuous function $-[k^2-l(l+1)r^{-2}]{\alpha }_{lm}$, so twice taking its continuous primitive gives a $C^2$ function with the same second derivative; the difference is affine, since local mollifications have second derivative zero. Thus ${\alpha }_{lm}$ is $C^2$, and repeated differentiation of this equation gives smoothness. This requires no extra regularity of $V$.

(The two reduced exterior solutions.) Define spherical Bessel functions for $z>0$ by $j_0(z)=\sin z\mathrm{/}z$, $y_0(z)=-\cos z\mathrm{/}z$ and the recurrences $j_{l+1}(z)=l{j}_l(z)\mathrm{/}z-j_l^{\mathrm{\prime }}(z)$, $y_{l+1}(z)=l{y}_l(z)\mathrm{/}z-y_l^{\mathrm{\prime }}(z)$. Put $J_l(z)=z{j}_l(z)$ and $Q_l(z)=z{y}_l(z)$, so either sequence obeys $u_{l+1}=(l+1)u_l\mathrm{/}z-u_l^{\mathrm{\prime }}$. If $u_l^{\mathrm{\prime }\mathrm{\prime }}+(1-l(l+1)\mathrm{/}{z}^2)u_l=0$, differentiating this recurrence twice shows $u_{l+1}^{\mathrm{\prime }\mathrm{\prime }}+(1-(l+1)(l+2)\mathrm{/}{z}^2)u_{l+1}=0$; the initial functions $\sin z$ and $-\cos z$ solve the $l=0$ equation. Induction also shows each $J_l,Q_l$ is a finite sum of sine and cosine times polynomials in $1\mathrm{/}z$. The recurrence therefore gives $J_l(z)=\sin (z-l\pi \mathrm{/}2)+O(z^{-1})$, $Q_l(z)=-\cos (z-l\pi \mathrm{/}2)+O(z^{-1})$, and the differentiated expansions $J_l^{\mathrm{\prime }}(z)=\cos (z-l\pi \mathrm{/}2)+O(z^{-1})$, $Q_l^{\mathrm{\prime }}(z)=\sin (z-l\pi \mathrm{/}2)+O(z^{-1})$. Their Wronskian $J_l{Q}_l^{\mathrm{\prime }}-J_l^{\mathrm{\prime }}{Q}_l$ has derivative zero by the reduced equation and limit $1$ by these expansions, so it is identically $1$. Hence $r{j}_l(kr)=J_l(kr)\mathrm{/}k$ and $r{y}_l(kr)=Q_l(kr)\mathrm{/}k$ are independent solutions of the radial reduced equation with $r$-Wronskian $1\mathrm{/}k$. Every solution in step 1.2 is therefore ${\alpha }_{lm}(r)=A_{lm}r{j}_l(kr)+B_{lm}r{y}_l(kr)$ for fixed complex constants $A_{lm},B_{lm}$. Indeed, choose the constants matching the value and derivative at one $r_0>R$ using this nonzero Wronskian; the difference has zero Cauchy data. Writing the difference equation as a first-order system with continuous bounded coefficients on any compact radial interval, its integral equation forces zero on sufficiently short successive subintervals (the supremum norm is at most the interval length times the coefficient bound times itself), and finitely many such intervals cover every compact interval. This proves uniqueness and the asserted representation on all of $(R,\mathrm{\infty })$.

(Complex coefficients cannot be square-integrable at infinity.) For fixed $(l,m)$, write $A=A_{lm}$, $B=B_{lm}$ and $\vartheta (r)=kr-l\pi \mathrm{/}2$. Step 2.1 gives ${\alpha }_{lm}(r)=k^{-1}(A\sin \vartheta -B\cos \vartheta )+O(r^{-1})$, with constants allowed to depend on this fixed channel. For complex $A,B$ the squared leading term is $k^{-2}(\mathrm{\mid }A{\mathrm{\mid }}^2{\sin }^2\vartheta +\mathrm{\mid }B{\mathrm{\mid }}^2{\cos }^2\vartheta -2\mathrm{Re}(A\bar{B})\sin \vartheta \cos \vartheta )$, not the square of a real expression. Since the leading term is bounded and the error is $O(r^{-1})$, the error in its squared modulus is also $O(r^{-1})$. The integral of that leading term from any fixed $R_1>R$ to $T$ is $\frac{\mathrm{\mid }A{\mathrm{\mid }}^2+\mathrm{\mid }B{\mathrm{\mid }}^2}{2k^2}(T-R_1)+O(1)$, because the remaining trigonometric terms have bounded primitives for $k>0$; the integrated error is $O(\log T)$. Thus if $(A,B)\mathrm{\ne }(0,0)$, ${\int }_{R_1}^T\mathrm{\mid }{\alpha }_{lm}(r){\mathrm{\mid }}^{2}dr\ge \frac{\mathrm{\mid }A{\mathrm{\mid }}^2+\mathrm{\mid }B{\mathrm{\mid }}^2}{4k^2}(T-R_1)$ for all sufficiently large $T$, contradicting square-integrability. But square-integrability follows directly from Cauchy–Schwarz and the polar formula: ${\int }_R^{\mathrm{\infty }}\mathrm{\mid }{\alpha }_{lm}(r){\mathrm{\mid }}^{2}dr\le {\int }_R^{\mathrm{\infty }}{r}^2{\int }_{S^2}\mathrm{\mid }\psi (r\omega ){\mathrm{\mid }}^{2}d\sigma dr\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_2^2$. Therefore $A_{lm}=B_{lm}=0$ for every channel. No estimate uniform in $l,m$ and no interchange of an asymptotic expansion with an infinite angular sum is used.

(Exterior vanishing and unique continuation.) Step 3.1 makes every angular coefficient zero for almost every $r>R$. There are countably many channels, so intersecting their conull radial sets and the set on which $\psi (r\cdot )\in L^2(S^2)$ gives a conull set of radii where all coefficients vanish simultaneously; completeness in [F1] then gives $\psi (r\cdot )=0$ in $L^2(S^2)$. Equivalently, angular Parseval and Tonelli give ${\int }_{\mathrm{\mid }x\mathrm{\mid }>R}\mathrm{\mid }\psi (x){\mathrm{\mid }}^{2}dx={\sum }_{l,m}{\int }_R^{\mathrm{\infty }}\mathrm{\mid }{\alpha }_{lm}(r){\mathrm{\mid }}^{2}dr=0$. Thus $\psi$ vanishes almost everywhere on the nonempty open exterior. On the connected open set ${\mathbb{R}}^3$, the original equation is $\mathrm{\Delta }\psi =(V-k^2)\psi$ with $V-k^2\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^{\mathrm{\infty }}({\mathbb{R}}^3;\mathbb{C})$ and $\psi \in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$. The exact weak-UCP statement [F2] applies and yields $\psi =0$ almost everywhere on ${\mathbb{R}}^3$, so $k^2$ is not an eigenvalue.