Rank-one resolvent formula for the delta interface operator

[Zero-coupling identification and nonvanishing denominator] Let $A_0^{\mathrm{i}\mathrm{n}\mathrm{t}}$ be the interface operator with parameter $\alpha =0$. If $u\in D(A_0^{\mathrm{i}\mathrm{n}\mathrm{t}})$, its piecewise second derivative $q$ is in $L^2(\mathbb{R})$ and the two one-sided derivative traces agree. For every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$, integration by parts on the two half-lines gives ${\int }_{\mathbb{R}}{u}^{\mathrm{\prime }}{\phi }^{\mathrm{\prime }}=(u^{\mathrm{\prime }\ast }(0-)-u^{\mathrm{\prime }\ast }(0+))\phi (0)-{\int }_{\mathbb{R}}q\phi =-{\int }_{\mathbb{R}}q\phi$; hence $D^{2}u=q$ globally and $u\in H^2(\mathbb{R})$. Conversely, if $u\in H^2(\mathbb{R})$, then $u^{\mathrm{\prime }}\in H^1(\mathbb{R})$ has a single continuous representative, so its one-sided traces at zero agree; the restrictions of $u$ lie in $H^2$ on each half-line and satisfy the zero-jump condition. Thus $D(A_0^{\mathrm{i}\mathrm{n}\mathrm{t}})=H^2(\mathbb{R})$ and both $A_0^{\mathrm{i}\mathrm{n}\mathrm{t}}$ and $A_{\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}$ act as $-D^2$ there, proving the identification asserted in the Statement. Now, if $1+\alpha \mathrm{/}(2k)=0$, then $2k=-\alpha \in \mathbb{R}$; but $k^2=-z$ is nonreal because $z$ is nonreal, so $k$ itself is nonreal. Hence $1+\alpha G_z(0)=1+\alpha \mathrm{/}(2k)\mathrm{\ne }0$.

[Construction of the preimage] Fix $f\in L^2(\mathbb{R})$ and put $r:=(A_0-z{)}^{-1}f$, with $A_0=A_{\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}$ by step 1.1. Exactly as in the range construction of The one-dimensional delta interface operator, the vector $u:=r+c{G}_z,c:=-\frac{\alpha r(0)}{1+\alpha G_z(0)},$ lies in $D(A_{\alpha })$ and satisfies $(A_{\alpha }-z)u=f$; here $r(0)={\int }_{\mathbb{R}}{G}_z(y)f(y)dy={\mathrm{\ell }}_z(f)$ because $r(x)=\int G_z(x-y)f(y)dy$ evaluated at $x=0$.

[The resolvent formula] Since $A_{\alpha }-z$ is injective by [F1], the vector $u$ of step 2.1 is the unique preimage of $f$, hence $(A_{\alpha }-z{)}^{-1}f=u=r+c{G}_z=(A_0-z{)}^{-1}f-\frac{\alpha {\mathrm{\ell }}_z(f)}{1+\alpha G_z(0)}{G}_z,$ which is (a).

[Rank one and bounds] The map $K_z$ displayed in (b) equals $f\mapsto {\mathrm{\ell }}_z(f)w$ with $w:=-\alpha G_z\mathrm{/}(1+\alpha G_z(0))$; its range is contained in $\mathbb{C}{G}_z$. If $\alpha \mathrm{\ne }0$, then $w\mathrm{\ne }0$, and $f=\overline{G_z}\in L^2$ satisfies ${\mathrm{\ell }}_z(f)={\int }_{\mathbb{R}}\mathrm{\mid }{G}_z(y){\mathrm{\mid }}^{2}dy=\mathrm{\parallel }{G}_z{\mathrm{\parallel }}_{L^2}^2>0$. Hence the range contains a nonzero multiple of $G_z$, so $K_z$ has rank exactly one; if $\alpha =0$, then $w=0$ and $K_z=0$. By [F2], $\mathrm{\parallel }{\mathrm{\ell }}_z\mathrm{\parallel }\le \mathrm{\parallel }{G}_z{\mathrm{\parallel }}_{L^2}$, and therefore [step 3.1, F2, algebra] $\mathrm{\parallel }{K}_z\mathrm{\parallel }\le \mathrm{\parallel }{\mathrm{\ell }}_z\mathrm{\parallel }\mathrm{\parallel }w\mathrm{\parallel }\le \mathrm{\parallel }{G}_z{\mathrm{\parallel }}_{L^2}\frac{\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }{G}_z{\mathrm{\parallel }}_{L^2}}{\mathrm{\mid }1+\alpha G_z(0)\mathrm{\mid }}=\frac{\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }{G}_z{\mathrm{\parallel }}_{L^2}^2}{\mathrm{\mid }1+\alpha G_z(0)\mathrm{\mid }}\mathrm{.}$ Substituting $\mathrm{\parallel }{G}_z{\mathrm{\parallel }}_{L^2}^2=(4\mathrm{\mid }k{\mathrm{\mid }}^2\mathrm{Re}k{)}^{-1}$ from [F2] gives the displayed value, and the triangle inequality for operator norms gives $\mathrm{\parallel }(A_{\alpha }-z{)}^{-1}\mathrm{\parallel }\le \mathrm{\parallel }(A_0-z{)}^{-1}\mathrm{\parallel }+\mathrm{\parallel }{K}_z\mathrm{\parallel }$. This is (b).