Green resolvent for separated Robin boundary conditions

[Density, well-definedness, symmetry] $C_c^{\mathrm{\infty }}(0,L)\subseteq D(A)$ because test functions have vanishing traces at $0$ and $L$, and $C_c^{\mathrm{\infty }}(0,L)$ is dense in $L^2(0,L)$ by [F1], so $D(A)$ is dense. If $u\in D(A)$, then $D^{2}u$ and $Vu$ are $L^2$ classes, so $Au\in L^2$ is well defined and depends only on the class $u$; linearity is clear. For $u,v\in D(A)$, the first-order identity of Integration by parts and the endpoint-vanishing space on intervals clause (a), applied to the pairs $(u^{\mathrm{\prime }},v)$ and $(v^{\mathrm{\prime }},u)$, gives $⟨Au,v⟩-⟨u,Av⟩=[u^{\ast }\overline{v^{\mathrm{\prime }\ast }}-u^{\mathrm{\prime }\ast }\overline{v^{\ast }}{]}_0^L=(u(L)\overline{\beta v(L)}-\beta u(L)\overline{v(L)})-(u(0)\overline{\alpha v(0)}-\alpha u(0)\overline{v(0)})=0,$ because both $u$ and $v$ satisfy the two boundary conditions and $\alpha ,\beta$ are real. Hence $A$ is symmetric.

[The two solutions and the Wronskian] By Regular interval IVP: existence, a priori bound and entire spectral-parameter dependence the solution $u_a$ with data $(1,\alpha )$ at $0$ and the solution $u_b$ with data $(1,\beta )$ at $L$ exist uniquely. By [F2] $W$ is constant. If $W=0$, then $u_a,u_b$ are linearly dependent; since $u_a(0)=1\mathrm{\ne }0$, $u_b=c{u}_a$ for some $c\mathrm{\ne }0$, and the right boundary condition $u_b^{\mathrm{\prime }}(L)=\beta u_b(L)$ gives $u_a^{\mathrm{\prime }}(L)=\beta u_a(L)$, so $u_a\in D(A)$ and $A{u}_a=z{u}_a$ with $u_a\mathrm{\ne }0$. A symmetric operator has only real eigenvalues ($⟨A{u}_a,u_a⟩=z\mathrm{\parallel }{u}_a{\mathrm{\parallel }}^2$ equals its own conjugate by symmetry), so $z\in \mathbb{R}$, contradicting $z\mathrm{\notin }\mathbb{R}$. Hence $W(z)\mathrm{\ne }0$.

[The Green kernel is well defined and inverts $A-z$] For $f\in L^2(0,L)$ put $a(x):={\int }_0^x{u}_a(y)f(y)dy$ and $b(x):={\int }_x^L{u}_b(y)f(y)dy$. These functions are absolutely continuous with $a^{\mathrm{\prime }}=u_{a}f$ and $b^{\mathrm{\prime }}=-u_{b}f$ almost everywhere. The kernel formula gives $u(x)=-(u_b(x)a(x)+u_a(x)b(x))\mathrm{/}W$. Its derivative is $u^{\mathrm{\prime }}(x)=-(u_b^{\mathrm{\prime }}(x)a(x)+u_a^{\mathrm{\prime }}(x)b(x))\mathrm{/}W$, since the two terms containing $f$ cancel. The latter expression is absolutely continuous and, using $u_a^{\mathrm{\prime }\mathrm{\prime }}=(V-z)u_a$, $u_b^{\mathrm{\prime }\mathrm{\prime }}=(V-z)u_b$ and $u_b^{\mathrm{\prime }}{u}_a-u_a^{\mathrm{\prime }}{u}_b=W$, its derivative is $u^{\mathrm{\prime }\mathrm{\prime }}=(V-z)u-f$ almost everywhere. Thus $u\in H^2(0,L)$ and $-u^{\mathrm{\prime }\mathrm{\prime }}+Vu-zu=f$. At zero, $u(0)=-W^{-1}{\int }_0^L{u}_{b}f$ and $u^{\mathrm{\prime }}(0)=-\alpha W^{-1}{\int }_0^L{u}_{b}f=\alpha u(0)$; at $L$, $u(L)=-W^{-1}{\int }_0^L{u}_{a}f$ and $u^{\mathrm{\prime }}(L)=-\beta W^{-1}{\int }_0^L{u}_{a}f=\beta u(L)$. Therefore $u\in D(A)$ and $(A-z)u=f$.

[Self-adjointness] Step 3.1 shows $\mathrm{ran}(A-z)=L^2(0,L)$ for every nonreal $z$ (in particular for $z=\pm i$); with the density and symmetry of step 1.1, the range criterion [F3] makes $A$ self-adjoint.

[Compactness] The kernel $G_z$ is continuous on the compact square $[0,L{]}^2$ (it is built from continuous functions, and the apparent singularity at $x=y$ is removable because $\min$ and $\max$ are continuous), hence bounded. By [F4] the resolvent $(A-z{)}^{-1}$ is compact for every nonreal $z$: the uniform-continuity approximation of $G_z$ by piecewise-constant kernels exhibits it as a norm limit of finite-rank integral operators. For an arbitrary $z$ in the resolvent set, the second resolvent identity $(A-z{)}^{-1}=(A-i{)}^{-1}+(z-i)(A-i{)}^{-1}(A-z{)}^{-1}$ (an identity of bounded operators on $L^2(0,L)$, obtained from $(A-i{)}^{-1}-(A-z{)}^{-1}=-(z-i)(A-i{)}^{-1}(A-z{)}^{-1}$) writes $(A-z{)}^{-1}=(A-i{)}^{-1}(I+(z-i)(A-z{)}^{-1})$, a composite of a compact with a bounded operator, hence compact by [F4]. This proves (c).