Bounded perturbations of the interval Dirichlet Laplacian

[Self-adjointness] By [F1], $M_V$ is bounded self-adjoint with norm $\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}$ and is $A_D$-bounded with relative bound $0$ (take $a=0$, $b=\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}$ in Relative boundedness with respect to an operator). Since $A_D$ is self-adjoint on $D(A_D)$, [F2] makes $A=A_D+M_V$ self-adjoint on $D(A_D)$; and $Au=A_{D}u+Vu=-D^{2}u+Vu$ by Identification of the Dirichlet Laplacian with the sine diagonal operator.

[The free resolvent is a norm limit of finite-rank operators] Let $c<{\lambda }_1$ be real and let $R_D:=(A_D-c{)}^{-1}$, the diagonal operator with respect to the orthonormal basis $(e_n)$ of Identification of the Dirichlet Laplacian with the sine diagonal operator: $R_D{e}_n=({\lambda }_n-c{)}^{-1}{e}_n$, from $A_D{e}_n={\lambda }_n{e}_n$. For $N\ge 1$ let $R_D^{(N)}$ be the diagonal operator with the same eigenvectors but coefficients $({\lambda }_n-c{)}^{-1}$ for $n\le N$ and $0$ for $n>N$; it is finite rank. For $x={\sum }_n{c}_n{e}_n\in L^2$ one has $(R_D-R_D^{(N)})x={\sum }_{n>N}({\lambda }_n-c{)}^{-1}{c}_n{e}_n$, so $\mathrm{\parallel }(R_D-R_D^{(N)})x\mathrm{\parallel }\le ({\sup }_{n>N}({\lambda }_n-c{)}^{-1})\mathrm{\parallel }x\mathrm{\parallel }$, and the norm identity $\mathrm{\parallel }{R}_D-R_D^{(N)}\mathrm{\parallel }={\sup }_{n>N}({\lambda }_n-c{)}^{-1}$ follows by testing on unit eigenvectors; this supremum tends to $0$ because ${\lambda }_n=(n\pi \mathrm{/}L{)}^2\to \mathrm{\infty }$. Hence $R_D$ is a norm limit of finite-rank operators, and by [F3] it is compact.

[Compactness of the perturbed resolvent] Choose $c:={\lambda }_1-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}-1<{\lambda }_1$, so that $d:=\mathrm{dist}(c,\sigma (A_D))={\lambda }_1-c=\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ and $\mathrm{\parallel }{R}_D{M}_V\mathrm{\parallel }\le \mathrm{\parallel }{R}_D\mathrm{\parallel }\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}+1)<1$. Then $I+R_D{M}_V$ is boundedly invertible by the Neumann series, and on $D(A_D)$ $(A-c)=(A_D-c)(I+R_D{M}_V),$ because $R_D(A_D-c)=I$ on $D(A_D)$. Hence $(A-c{)}^{-1}=(I+R_D{M}_V{)}^{-1}{R}_D,$ a product of a bounded operator with the compact operator $R_D$, hence compact (Compositions with a compact operator are compact); the inverse exists because $A$ is self-adjoint and $c<{\lambda }_1-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\le \inf \sigma (A)$: indeed $⟨Au,u⟩=⟨A_{D}u,u⟩+⟨Vu,u⟩\ge {\lambda }_1\mathrm{\parallel }u{\mathrm{\parallel }}^2-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }u{\mathrm{\parallel }}^2$, so $A\ge ({\lambda }_1-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }})I$ and $c\in \rho (A)$.

[Eigenbasis and discreteness] The space $L^2(0,L;\mathbb{C})$ is infinite dimensional: the vectors $e_n$, $n\ge 1$, of Identification of the Dirichlet Laplacian with the sine diagonal operator form an infinite orthonormal family (Orthonormal families, complete orthonormal systems and Hilbert bases), while a space with a finite spanning set has no linearly independent subset equinumerous with $\mathbb{N}$, and a finite-dimensional space has a finite basis, hence a finite spanning set (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $\mathbb{N}$, Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). Step 2.1 supplies $c\in \rho (A)$ with compact resolvent, so [F4] gives a complete orthonormal eigenfamily $({\psi }_n{)}_{n\ge 1}$ of eigenvectors with eigenvalues $E_n$ satisfying $\mathrm{\mid }{E}_n\mathrm{\mid }\to \mathrm{\infty }$, finite dimensional eigenspaces, $\sigma (A)$ equal to the eigenvalue set and ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)=\mathrm{\varnothing }$; since $A\ge ({\lambda }_1-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }})I$ the eigenvalues are bounded below, so $E_n\to +\mathrm{\infty }$ along any enumeration, and the eigenvalues may be listed with multiplicity in increasing order. This is (b).

[Min-max comparison] Both $A$ and $A_D$ are bounded below and have empty essential spectrum by step 3.1 (for $A_D$: $\sigma (A_D)=\{{\lambda }_n\}$, all isolated eigenvalues of finite multiplicity by Identification of the Dirichlet Laplacian with the sine diagonal operator and the definition of the discrete spectrum, so ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A_D)=\mathrm{\varnothing }$). By [F5], for $n\ge 1$, $E_n={\inf }_{L^{\mathrm{\prime }}\subseteq D(A),\dim L^{\mathrm{\prime }}=n}{\sup }_{\begin{array}{c}x\in L^{\mathrm{\prime }}\\ \mathrm{\parallel }x\mathrm{\parallel }=1\end{array}}⟨Ax,x⟩,{\lambda }_n={\inf }_{L^{\mathrm{\prime }}\subseteq D(A_D),\dim L^{\mathrm{\prime }}=n}{\sup }_{\begin{array}{c}x\in L^{\mathrm{\prime }}\\ \mathrm{\parallel }x\mathrm{\parallel }=1\end{array}}⟨A_{D}x,x⟩\mathrm{.}$ Since $D(A)=D(A_D)$, and $\mathrm{\mid }⟨M_{V}x,x⟩\mathrm{\mid }\le \mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x{\mathrm{\parallel }}^2$, every trial space gives $⟨A_{D}x,x⟩-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\le ⟨Ax,x⟩\le ⟨A_{D}x,x⟩+\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}$ for $\mathrm{\parallel }x\mathrm{\parallel }=1$; taking suprema over then infima over the same family of $n$-dimensional trial spaces yields ${\lambda }_n-\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }}\le E_n\le {\lambda }_n+\mathrm{\parallel }V{\mathrm{\parallel }}_{\mathrm{\infty }},$ which is (c).