Simple complete spectrum of the separated Robin realisation

[Compact resolvent and infinite dimension] By [F1], $(A-i{)}^{-1}$ is compact and $i$ lies in the resolvent set of the self-adjoint operator $A$. By [F4], $L^2(0,L;\mathbb{C})$ is infinite dimensional.

[Boundedness below] Put $V_{\mathrm{\infty }}:={\max }_{[0,L]}\mathrm{\mid }V\mathrm{\mid }$ and $\delta :=\min \{L,(4(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }+1){)}^{-1}\}$, so that $2\delta (\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })\le \frac{1}{2}$. For $u\in D(A)$, [F5] gives $⟨Au,u⟩=\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2+{\int }_0^{L}V\mathrm{\mid }u{\mathrm{\mid }}^2+\alpha \mathrm{\mid }u(0){\mathrm{\mid }}^2-\beta \mathrm{\mid }u(L){\mathrm{\mid }}^2\ge (1-2\delta (\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }))\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2-(V_{\mathrm{\infty }}+2{\delta }^{-1}(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }))\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2\ge -C\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2,$ where $C:=V_{\mathrm{\infty }}+2{\delta }^{-1}(\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid })$: the integral term is at least $-V_{\mathrm{\infty }}\mathrm{\parallel }u{\mathrm{\parallel }}^2$, each of the two boundary terms is at least $-\mathrm{\mid }\alpha \mathrm{\mid }$ resp. $-\mathrm{\mid }\beta \mathrm{\mid }$ times the trace bound of [F5], and the coefficient of $\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}^2$ is at least $\frac{1}{2}\ge 0$. In particular every eigenvalue $\lambda$ of $A$ satisfies $\lambda \ge -C$, because $\lambda \mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2=⟨Au,u⟩\ge -C\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2$ for an eigenvector $u\mathrm{\ne }0$.

[Simplicity] Let $A\psi =\lambda \psi$ with $\psi \mathrm{\ne }0$. By [F3] the solution $u_a(\cdot ,\lambda )$ is the unique solution of the initial value problem with data $(1,\alpha )$ at $0$, so the $C^2$ representative of $\psi$ equals ${\psi }^{\ast }(0)u_a(\cdot ,\lambda )$; here ${\psi }^{\ast }(0)\mathrm{\ne }0$, because otherwise the representative vanishes at $0$ together with its derivative $\alpha {\psi }^{\ast }(0)=0$, forcing $\psi =0$ by uniqueness. Hence $\ker (A-\lambda )=\mathbb{C}{u}_a(\cdot ,\lambda )$ is one-dimensional: every nonzero eigenfunction at $\lambda$ is a nonzero multiple of $u_a(\cdot ,\lambda )$. The eigenvalue condition is the right boundary condition evaluated on $u_a(\cdot ,\lambda )$; $u_a(L,\lambda )=0$ would give $u_a^{\mathrm{\prime }}(L,\lambda )=\beta u_a(L,\lambda )=0$ and hence $u_a\equiv 0$ by uniqueness, which is impossible since $u_a(0)=1$.

[A real resolvent point, eigenbasis and increasing list] Put $c=-C-1$ with $C$ from step 1.2. For $u\in D(A)$, that lower bound and Cauchy-Schwarz give $\mathrm{\parallel }(A-c)u\mathrm{\parallel }\ge \mathrm{\parallel }u\mathrm{\parallel }$. The range of $A-c$ is closed: if $(A-c)u_j\to f$, this estimate makes $u_j$ Cauchy, and closedness of the self-adjoint operator gives a limit $u\in D(A)$ with $(A-c)u=f$. By Orthogonal decomposition by a closed subspace, its closed range is all of $L^2$ once its orthogonal complement is zero. A vector orthogonal to it belongs to $\ker ((A-c{)}^{\ast })=\ker (A-c)=\{0\}$ by the adjoint definition and the same estimate. Thus $A-c$ is bijective with inverse norm at most one, so $c\in \rho (A)$. The resolvent identity gives $(A-c{)}^{-1}=(A-i{)}^{-1}[I+(c-i)(A-c{)}^{-1}]$; the compactness for every resolvent in [F1] therefore gives compactness at this real $c$. Now [F2] applies: a complete orthonormal eigenfamily exists, all eigenspaces are finite dimensional, the spectrum is exactly the eigenvalue set, and the essential spectrum is empty, with $\mathrm{\mid }{\mu }_n\mathrm{\mid }\to \mathrm{\infty }$. Step 1.2 gives ${\mu }_n\ge -C$, hence ${\mu }_n\to +\mathrm{\infty }$. By step 2.1 the eigenvalues are pairwise distinct; their increasing enumeration is consequently ${\lambda }_1<{\lambda }_2<\cdots$ with ${\lambda }_n\to \mathrm{\infty }$. This proves (a) and (b).