The one-dimensional delta interface operator

[Density] Let $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\setminus \{0\})$. Then ${\phi }^{\ast }=\phi$ vanishes near $0$, so ${\phi }^{\mathrm{\prime }\ast }(0+)-{\phi }^{\mathrm{\prime }\ast }(0-)=0=\alpha {\phi }^{\ast }(0)$ and both half-line restrictions are smooth with compact support: $\phi \in D(A_{\alpha })$. Now let $f\in L^2(\mathbb{R})$ and $\epsilon >0$. By [F4] there is $\psi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\mathrm{\parallel }f-\psi \mathrm{\parallel }<\epsilon$. Choose a function $\chi \in C^{\mathrm{\infty }}(\mathbb{R})$ with $\chi =0$ near $0$ and $\chi =1$ outside $[-1,1]$, and put ${\chi }_{\delta }(x):=\chi (x\mathrm{/}\delta )$. Then ${\chi }_{\delta }\psi \in C_c^{\mathrm{\infty }}(\mathbb{R}\setminus \{0\})$ and $\mathrm{\parallel }{\chi }_{\delta }\psi -\psi {\mathrm{\parallel }}_{L^2}\to 0$ as $\delta \to 0$ by dominated convergence, since $\mathrm{\mid }{\chi }_{\delta }\psi -\psi \mathrm{\mid }\le 2\mathrm{\mid }\psi \mathrm{\mid }\in L^1\cap L^2$ and the left side tends to $0$ pointwise. For small $\delta$, $\mathrm{\parallel }{\chi }_{\delta }\psi -\psi \mathrm{\parallel }<\epsilon$, so $D(A_{\alpha })$ contains a dense subspace.

[Well-definedness] Let $u\in D(A_{\alpha })$. By definition the restrictions $u_{\pm }$ lie in $H^2$ on the two half-lines, so their second weak derivatives $D^2{u}_{\pm }$ are $L^2$ classes on those half-lines; the class $A_{\alpha }u$ obtained by gluing $-D^2{u}_{+}$ and $-D^2{u}_{-}$ is in $L^2(\mathbb{R})$ and depends only on the class $u$. The assignment is linear because weak differentiation and restriction are linear.

[Symmetry] Let $u,v\in D(A_{\alpha })$. Splitting the integrals and integrating by parts twice on each half-line, using [F2] and the vanishing of the boundary products at $\pm \mathrm{\infty }$, $⟨A_{\alpha }u,v⟩=(u^{\mathrm{\prime }\ast }(0+)-u^{\mathrm{\prime }\ast }(0-))\overline{v^{\ast }(0)}+{\int }_{\mathbb{R}}(Du)\overline{Dv}=\alpha u^{\ast }(0)\overline{v^{\ast }(0)}+{\int }_{\mathbb{R}}(Du)\overline{Dv}\mathrm{.}$ (On each half-line the two integrations by parts produce the boundary contribution of the derivative jump, and the intermediate terms $\int Du\overline{Dv}$ combine.) Hence $⟨A_{\alpha }u,v⟩-⟨u,A_{\alpha }v⟩=\alpha u^{\ast }(0)\overline{v^{\ast }(0)}-\overline{\alpha v^{\ast }(0)\stackrel{‾}{u^{\ast }(0)}}=0,$ so $A_{\alpha }$ is symmetric.

[Range at nonreal $z$] Fix $z\in \mathbb{C}\setminus \mathbb{R}$, $f\in L^2(\mathbb{R})$, and put $r:=R_0(z)f\in H^2(\mathbb{R})$, so that $(A_0-z)r=f$ by [F3]. Let $k$ with $k^2=-z$, $\mathrm{Re}k>0$, and put $u:=r+c{G}_z,c:=-\frac{\alpha r(0)}{1+\alpha G_z(0)}=-\frac{\alpha r(0)}{1+\alpha \mathrm{/}(2k)}\mathrm{.}$ The denominator does not vanish: if $1+\alpha \mathrm{/}(2k)=0$, then $2k=-\alpha \in \mathbb{R}$, contradicting $\mathrm{Im}k\mathrm{\ne }0$ (which holds because $k^2=-z$ is nonreal). Since $r,G_z\in H^1(\mathbb{R})$ (steps (a) of The free-line resolvent kernel and its inverse operator), $u\in H^1(\mathbb{R})$; on each half-line $u$ is a linear combination of $H^2$ functions, hence satisfies the half-line regularity in the definition of $D(A_{\alpha })$. The trace of $Du$ at $0$ jumps only through $G_z$: $r\in H^2(\mathbb{R})$ has a continuous derivative representative at $0$, while $G_z^{\mathrm{\prime }}(0+)-G_z^{\mathrm{\prime }}(0-)=-1$ by [F3]. Therefore $u^{\mathrm{\prime }\ast }(0+)-u^{\mathrm{\prime }\ast }(0-)=-c=\alpha (r(0)+\frac{c}{2k})=\alpha u^{\ast }(0),$ where the middle equality uses the definition of $c$ and $G_z(0)=(2k{)}^{-1}$. So $u\in D(A_{\alpha })$, and away from $0$ $(A_{\alpha }-z)u=-D^2(r+c{G}_z)-z(r+c{G}_z)=(f+zr)-zr+0=f,$ since $-D^{2}r-zr=f$ on $\mathbb{R}$ and $-D^2{G}_z-z{G}_z=0$ on each half-line (direct computation from $G_z^{\mathrm{\prime }\mathrm{\prime }}=k^2{G}_z$ and $k^2=-z$). Hence $\mathrm{ran}(A_{\alpha }-z)=L^2(\mathbb{R})$ for every nonreal $z$.

[Self-adjointness] Steps 1.1 and 1.3 give a densely defined symmetric operator, and step 2.1 gives $\mathrm{ran}(A_{\alpha }-z)=L^2(\mathbb{R})$ for $z=\pm i$; the range criterion [F4] makes $A_{\alpha }$ self-adjoint.