Birman–Schwinger principle for bounded compactly supported potentials

Write $M=\mathrm{\parallel }W{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$. The multiplication operators $M_W$ and $M_s$ are bounded, nonnegative, and self-adjoint, with $\mathrm{\parallel }{M}_W\mathrm{\parallel }\le M$ and $\mathrm{\parallel }{M}_s{\mathrm{\parallel }}^2\le M$. Since $H_0\ge 0$, Kato–Rellich makes $H=H_0-M_W$ self-adjoint on $D(H_0)=W^{2,2}$, and $H\ge -MI$. On $W^{1,2}$ define $q_H(u,v)=q_0(u,v)-⟨M_{s}u,M_{s}v⟩$. The perturbation has form bound $a=0$, $b=M$, because $⟨M_{s}u,M_{s}u⟩=⟨M_{W}u,u⟩\le M\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$; hence [F1] makes $q_H$ closed and bounded below by $-M$.

Let $\tilde{H}$ be the self-adjoint operator representing $q_H$ from [F1]. For $u\in W^{2,2}$ and $v\in W^{1,2}$, the free-form identity [A1] and $⟨M_{s}u,M_{s}v⟩=⟨M_{W}u,v⟩$ give $q_H(u,v)=⟨(H_0-M_W)u,v⟩$. Thus $H\subseteq \tilde{H}$; self-adjointness gives $\tilde{H}\subseteq H^{\ast }=H$, so $\tilde{H}=H$. The representation theorem's square-root identity on the full form domain, with shift $M+1$, and [F3] then identify $q_H$ with the spectral form of $H$.

The resolvent convolution formula in [F2] gives, for a.e. $x$, $(K_{\kappa }f)(x)=s(x)\int G_{\kappa }(x-y)s(y)f(y)dy$; thus the displayed function is the integral kernel of $K_{\kappa }$. If $S$ is a compact set outside which $W=0$ almost everywhere and $M=\mathrm{\parallel }W{\mathrm{\parallel }}_{\mathrm{\infty }}$, then $\iint \mathrm{\mid }s(x)G_{\kappa }(x-y)s(y){\mathrm{\mid }}^{2}dxdy\le M^2\mathrm{\mid }S\mathrm{\mid }{\int }_{S-S}\mathrm{\mid }{G}_{\kappa }(z){\mathrm{\mid }}^{2}dz<\mathrm{\infty },$ because $S-S$ is bounded and the $\mathrm{\mid }z{\mathrm{\mid }}^{-1}$ singularity of $G_{\kappa }$ is square-integrable locally in three dimensions. This is an $L^2$ kernel, so [F5] makes its integral operator $K_{\kappa }$ Hilbert–Schmidt and compact.

In the Fourier representation of [A1], $R_{\kappa }=(H_0+{\kappa }^{2}I{)}^{-1}={\mathcal{F}}_2^{\ast }{M}_{(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2+{\kappa }^2{)}^{-1}}{\mathcal{F}}_2\mathrm{.}$ The multiplier is real, strictly positive, and bounded by ${\kappa }^{-2}$; its positive square root is the bounded self-adjoint multiplier $R_{\kappa }^{1\mathrm{/}2}={\mathcal{F}}_2^{\ast }{M}_{(4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2+{\kappa }^2{)}^{-1\mathrm{/}2}}{\mathcal{F}}_2\mathrm{.}$ Consequently $K_{\kappa }=(M_s{R}_{\kappa }^{1\mathrm{/}2})(M_s{R}_{\kappa }^{1\mathrm{/}2}{)}^{\ast }$, so it is positive and self-adjoint.

If $Hu=-{\kappa }^{2}u$, then $(H_0+{\kappa }^{2}I)u=M_{W}u$. Put $f=M_{s}u$. If $f=0$, invertibility of $H_0+{\kappa }^{2}I$ forces $u=0$; hence a nonzero eigenvector gives $f\mathrm{\ne }0$ and $K_{\kappa }f=M_s{R}_{\kappa }{M}_s(M_{s}u)=M_s{R}_{\kappa }{M}_{W}u=M_{s}u=f\mathrm{.}$ Conversely, if $K_{\kappa }f=f$ with $f\mathrm{\ne }0$, set $u=R_{\kappa }{M}_{s}f$. Then $M_{s}u=K_{\kappa }f=f$, so $(H_0+{\kappa }^{2}I)u=M_{s}f=M_s^{2}u=M_{W}u$ and $Hu=-{\kappa }^{2}u$. These maps are inverse because $R_{\kappa }{M}_{W}u=u$ for an eigenvector and $M_s{R}_{\kappa }{M}_{s}f=f$ for a $1$-eigenvector; therefore the eigenspaces are isomorphic.

Let $A=(H_0+{\kappa }^{2}I{)}^{1\mathrm{/}2}$, which maps $W^{1,2}$ bijectively onto $L^2$ by [A1] and Fourier calculus, and put $T=M_s{A}^{-1}$. Then $T{T}^{\ast }=K_{\kappa }$. The compactness of $K_{\kappa }$ implies $T$ is compact: for finite-rank spectral projections $P_n$ of $K_{\kappa }$ increasing to its support, $\mathrm{\parallel }(I-P_n)T{\mathrm{\parallel }}^2=\mathrm{\parallel }(I-P_n)K_{\kappa }(I-P_n)\mathrm{\parallel }⟶0,$ so $T$ is an operator-norm limit of the finite-rank maps $P_{n}T$. Hence $C=T^{\ast }T$ is positive compact. For $v=Au$, $u\in W^{1,2}$, the exact form identity is $q_H[u]+{\kappa }^2\mathrm{\parallel }u{\mathrm{\parallel }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2-\mathrm{\parallel }Tv{\mathrm{\parallel }}^2=⟨(I-C)v,v⟩\mathrm{.}$ The maps $u\mapsto v$ and $v\mapsto A^{-1}v$ are inverse bijections.

The maximal dimension of a linear subspace on which the right side of the form identity in step 4.2 is strictly negative equals $\mathrm{rank}{E}_C((1,\mathrm{\infty }))$: that spectral subspace is negative for $I-C$, and any subspace of larger dimension contains a nonzero vector orthogonal to it, where the spectral integral of $1-C$ is nonnegative. By [F4], this rank is the number of eigenvalues of $C$ above $1$, with multiplicity.

The same maximal negative dimension equals $\mathrm{rank}{E}_H((-\mathrm{\infty },-{\kappa }^2))$ by [F3]: the spectral range below the threshold is supported in $[-M,-{\kappa }^2)$ by step 1.1, so it lies in the form domain and is strictly negative for the shifted form. Any larger subspace has a nonzero vector in its orthogonal complement, where the shifted form is nonnegative. Step 5.1 makes this rank finite; its finite-dimensional spectral range therefore consists of eigenvectors, so its dimension is the number of eigenvalues of $H$ below $-{\kappa }^2$, counted with multiplicity. No essential spectral point lies strictly below $-{\kappa }^2$: a sufficiently small interval around any such point has finite-rank projection as a subprojection of $E_H((-\mathrm{\infty },-{\kappa }^2))$, contrary to [F6]. Hence the essential threshold is at least $-{\kappa }^2$ (or $+\mathrm{\infty }$), and [F7] identifies the same count with the form min-max levels strictly below the cutoff. Finally, $T$ maps each positive eigenspace of $C=T^{\ast }T$ bijectively onto the same-eigenvalue eigenspace of $K_{\kappa }=T{T}^{\ast }$, with inverse ${\mu }^{-1}{T}^{\ast }$ at eigenvalue $\mu >0$. Thus the number above $1$ for $C$ equals that for $K_{\kappa }$, proving the asserted count.