Outgoing Fredholm invertibility

Concrete kernel approximation: put $h_k(x,y)=e^{ik\mathrm{\mid }x-y\mathrm{\mid }}V(y)\mathrm{/}(4\pi \mathrm{\mid }x-y\mathrm{\mid })$ on $B_R\times B_R$, zero elsewhere. Its squared norm is at most $\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}^2\mathrm{\mid }{B}_R\mathrm{\mid }(16{\pi }^2{)}^{-1}{\int }_{\mathrm{\mid }z\mathrm{\mid }<2R}\mathrm{\mid }z{\mathrm{\mid }}^{-2}dz<\mathrm{\infty }$. Thus [F1] gives boundedness, compactness and finite-rank approximation on $\mathcal{H}$. More explicitly, enumerate the rational complex box-step functions restricted to $B_R$, and apply Gram-Schmidt in that order, discarding zero remainders; this uniquely specifies an orthonormal basis $(e_j)$ because the original enumeration is dense. The product basis $e_i(x)\overline{e_j(y)}$ is complete in product $L^2$: rectangular simple functions are dense by the same kernel lemma's box approximation, and approximation of each factor approximates each finite product. Its finite square partial sums $h_{k,n}={\sum }_{i,j\le n}{a}_{ij}{e}_i(x)\overline{e_j(y)}$, with $a_{ij}=\int h_k(x,y)\overline{e_i(x)}{e}_j(y)dxdy$, converge in kernel norm. The associated $F_n$ has range in $\mathrm{span}\{e_1,\dots ,e_n\}$ and $\mathrm{\parallel }{K}_k-F_n\mathrm{\parallel }\le \mathrm{\parallel }{h}_k-h_{k,n}{\mathrm{\parallel }}_2\to 0$. Select the least $n$ making this bound smaller than $1\mathrm{/}2$. No maximal-family or arbitrary-basis selection occurs.

Local Green regularity: for $g=Vf$ zero-extended from $B_R$, one has $g\in L^2\cap L^1$ with compact support. Set $u=-G_k\ast g$, $G_k(z)=e^{ik\mathrm{\mid }z\mathrm{\mid }}\mathrm{/}(4\pi \mathrm{\mid }z\mathrm{\mid })$. On any fixed bounded $x$-region the convolution and its first weak derivatives use bounded-radius restrictions of $G_k$ and $\mathrm{\nabla }{G}_k$; both are in $L^1$, because their singularities are $\mathrm{\mid }z{\mathrm{\mid }}^{-1}$ and $\mathrm{\mid }z{\mathrm{\mid }}^{-2}$. The bound $\mathrm{\parallel }q\ast g{\mathrm{\parallel }}_2\le \mathrm{\parallel }q{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_2$ follows by weighted Cauchy-Schwarz $\mathrm{\mid }\int q(z)g(x-z)dz{\mathrm{\mid }}^2\le \mathrm{\parallel }q{\mathrm{\parallel }}_1\int \mathrm{\mid }q(z)\mathrm{\mid }\mathrm{\mid }g(x-z){\mathrm{\mid }}^{2}dz$ and integration in $x$. Hence $u\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. Direct radial differentiation away from zero gives $(-\mathrm{\Delta }-k^2)G_k=0$ there, while integration over a shrinking sphere gives the unit delta coefficient from $-{\mathrm{\partial }}_r(1\mathrm{/}(4\pi r))$; the locally integrable remainder tends to zero. Thus $(-\mathrm{\Delta }-k^2)G_k={\delta }_0$ distributionally and $(-\mathrm{\Delta }-k^2)u=-g$. The right side and $u$ are locally square integrable, so [F2] gives $u\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$. This argument requires the actual $L^2$ source, and makes no assertion for arbitrary compact $L^1$ sources.

Kernel-to-eigenfunction and flux: suppose $(I+K_k)f=0$ and use $u$ from step 1.2. On $B_R$ one has $u=-K_{k}f=f$, hence $g=Vu$ and $(-\mathrm{\Delta }+V-k^2)u=0$ globally. For $r>2R$, Taylor expansion uniformly for $y\in B_R$ gives $G_k(r\omega -y)=e^{ikr}{e}^{-ik\omega \cdot y}\mathrm{/}(4\pi r)+O(r^{-2})$ and ${\mathrm{\partial }}_r{G}_k(r\omega -y)=ik{e}^{ikr}{e}^{-ik\omega \cdot y}\mathrm{/}(4\pi r)+O(r^{-2})$. Integrating against $g\in L^1$ yields $u(r\omega )=e^{ikr}A(\omega )\mathrm{/}r+O(r^{-2})$ and ${\mathrm{\partial }}_{r}u(r\omega )=ik{e}^{ikr}A(\omega )\mathrm{/}r+O(r^{-2})$, where $A(\omega )=-(4\pi {)}^{-1}\int e^{-ik\omega \cdot y}g(y)dy$ is smooth because all $y$-moments are integrable. The weak Green identity, justified by $H_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$ and radial cutoffs (the solution is smooth outside $B_R$), gives $\mathrm{Im}{\int }_{\mathrm{\mid }x\mathrm{\mid }=r}\bar{u}{\mathrm{\partial }}_{r}udS=0$: its interior integral is ${\int }_{B_r}(\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2+(V-k^2)\mathrm{\mid }u{\mathrm{\mid }}^2)$, which is real. Substituting the expansions gives $0=k{\int }_{S^2}\mathrm{\mid }A{\mathrm{\mid }}^{2}d\sigma +O(r^{-1})$, so $A=0$. Consequently $u=O(r^{-2})$ at infinity and $u\in L^2({\mathbb{R}}^3)$. Its equation gives $\mathrm{\Delta }u=(V-k^2)u\in L^2$, so the global Fourier criterion [F2] gives $u\in H^2$. The exact repaired positive-eigenvalue conclusion [F3] then yields $u=0$ and $f=0$. Thus $I+K_k$ is injective, by that supplier.

Explicit finite-rank inversion: take $F=F_n$ from step 1.1 and write $E=K_k-F$, $L=I+E$. Since $\mathrm{\parallel }E\mathrm{\parallel }<1\mathrm{/}2$, the series $L^{-1}={\sum }_{m\ge 0}(-E{)}^m$ converges in operator norm, is a two-sided inverse, and has norm at most $2$. Put $T=L^{-1}F$ and $B=I+T$, so $I+K_k=LB$. The subspace $Y=\mathrm{ran}T$ is finite dimensional. Obtain an orthonormal basis of $Y$ by Gram-Schmidt on the finite list $L^{-1}{e}_1,\dots ,L^{-1}{e}_n$ after applying the finite coefficient matrix of $F$, discarding zero vectors; thus the orthogonal projection $P_Y$ is a finite sum of coordinate pairings. Because $T(\mathcal{H})\subseteq Y$, $B(Y)\subseteq Y$; injectivity of $I+K_k$ from step 2.1 makes $B{\mathrm{\mid }}_Y$ injective and therefore bijective by finite-dimensional linear algebra. Given $h\in \mathcal{H}$, put $z=(I-P_Y)h$ and $y=(B{\mathrm{\mid }}_Y{)}^{-1}(P_{Y}h-Tz)$. Then $B(z+y)=z+Tz+By=h$. These finite matrix and projection operations are bounded; explicitly $\mathrm{\parallel }z+y\mathrm{\parallel }\le [1+\mathrm{\parallel }(B{\mathrm{\mid }}_Y{)}^{-1}\mathrm{\parallel }(1+\mathrm{\parallel }T\mathrm{\parallel })]\mathrm{\parallel }h\mathrm{\parallel }$. Hence $B^{-1}$ and $(I+K_k{)}^{-1}=B^{-1}{L}^{-1}$ are bounded. If $Y=\{0\}$ the same formula means $B=I$. This proves full surjectivity and the lower bound without a Banach Fredholm alternative, kernel/range-chain selection or DC.

Uniform positive-energy boundary: for $\zeta ,\eta$ in a bounded complex neighborhood of $[a,b]$, with $\mathrm{\mid }\mathrm{Im}\zeta \mathrm{\mid },\mathrm{\mid }\mathrm{Im}\eta \mathrm{\mid }\le 1$, the integral mean-value identity gives $\mathrm{\mid }{e}^{i\zeta d}-e^{i\eta d}\mathrm{\mid }\le d{e}^{2R}\mathrm{\mid }\zeta -\eta \mathrm{\mid }$ for $0\le d\le 2R$. Thus $\mathrm{\parallel }{K}_{\zeta }-K_{\eta }\mathrm{\parallel }\le \mathrm{\parallel }{h}_{\zeta }-h_{\eta }{\mathrm{\parallel }}_2\le C_R\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\zeta -\eta \mathrm{\mid }$. At each real $k>0$, step 3.1 gives $A_k=(I+K_k{)}^{-1}$; whenever $\mathrm{\parallel }{A}_k(K_{\zeta }-K_k)\mathrm{\parallel }<1\mathrm{/}2$, the factorization $I+K_{\zeta }=(I+K_k)[I+A_k(K_{\zeta }-K_k)]$ and its Neumann inverse give $\mathrm{\parallel }(I+K_{\zeta }{)}^{-1}\mathrm{\parallel }\le 2\mathrm{\parallel }{A}_k\mathrm{\parallel }$ and continuity at $k$. In particular $k\mapsto A_k$ is continuous on $[a,b]$, so $M={\max }_{k\in [a,b]}\mathrm{\parallel }{A}_k\mathrm{\parallel }<\mathrm{\infty }$. Choose $\delta <1$ with $C_R\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}M\delta <1\mathrm{/}2$ (if $V=0$, any $\delta <1$ works). For any $\zeta$ at distance less than $\delta$ from the interval, choose its nearest real point $k\in [a,b]$ by the explicit clipping formula; the same factorization gives invertibility, bound $2M$, and convergence to $A_k$ in operator norm. Energy $z={\zeta }^2$ with the outgoing square-root branch near $[a^2,b^2]$ therefore has the same compact-positive-energy boundary behavior. Complex conjugation $Cf=\bar{f}$ satisfies $C{K}_{k}C=K_{-k}$ because $V$ is real, so incoming real-axis inverses and their boundary limits have the same bounds. All original claims and the stated compact-positive-energy uniformity follow.