Non-invertible elliptic shifts form a discrete set in the self-adjoint case

The candidate series. Suppose $\lambda \mathrm{\notin }\{{\lambda }_j\}$. Since ${\lambda }_j\to +\mathrm{\infty }$ and the distinct eigenvalues have no finite accumulation point, the set of distinct eigenvalues is closed and its distance $\delta :=\mathrm{dist}(\lambda ,\{{\lambda }_j\})$ to $\lambda$ is positive. Set $c_j:=(f,e_j{)}_{L^2}({\lambda }_j-\lambda {)}^{-1}$. Then $\mathrm{\mid }{c}_j\mathrm{\mid }\le {\delta }^{-1}\mathrm{\mid }(f,e_j{)}_{L^2}\mathrm{\mid }$, and [F1] gives ${\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\le {\delta }^{-2}{\sum }_j\mathrm{\mid }(f,e_j{)}_{L^2}{\mathrm{\mid }}^2={\delta }^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2<\mathrm{\infty }$; hence $u:={\sum }_j{c}_j{e}_j$ converges in $L^2(\mathrm{\Omega })$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}\le {\delta }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}$.

Strong form convergence and the equation. For $M<N$, form orthogonality of the eigenfunctions gives $a_{\mu }(u_N-u_M,u_N-u_M)={\sum }_{M<j\le N}({\lambda }_j+\mu )\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\mathrm{.}$ The ratio $({\lambda }_j+\mu )({\lambda }_j-\lambda {)}^{-2}$ is bounded over $j$ (the denominator is nonzero and quadratic growth dominates the linear numerator), so Parseval [F1] gives ${\sum }_j({\lambda }_j+\mu )\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\le C{\sum }_j\mathrm{\mid }(f,e_j{)}_{L^2}{\mathrm{\mid }}^2=C\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2<\mathrm{\infty }\mathrm{.}$ Its tails tend to zero, so $(u_N)$ is Cauchy in the $a_{\mu }$ norm; by [F3] this norm is complete and equivalent to $H_0^1$, hence $u_N$ converges strongly in $H_0^1$ to some $\tilde{u}$. The continuous inclusion $H_0^1\hookrightarrow L^2$ and the $L^2$ convergence of step 1.1 identify $\tilde{u}=u$, so $u\in H_0^1$ and $u_N\to u$ strongly there. For each $v\in H_0^1(\mathrm{\Omega })$, boundedness of $a$ and the eigenrelations give $a(u,v)={\lim }_{N}a(u_N,v)={\lim }_N{\sum }_{j\le N}[(f,e_j{)}_{L^2}+\lambda c_j](e_j,v{)}_{L^2}=(f,v{)}_{L^2}+\lambda (u,v{)}_{L^2},$ where the last equality uses the $L^2$ basis expansions of $f$, $u$ and $v$. Thus $a(u,v)=(f+\lambda u,v{)}_{L^2}$ for all $v$, so $u\in D(L)$ and $(L-\lambda )u=f$.

Bijectivity. If $\lambda ={\lambda }_k$ for some $k$, the eigenfunction $e_k\mathrm{\ne }0$ satisfies $(L-{\lambda }_k)e_k=0$, so $L-\lambda$ is not injective and hence not bijective. If $\lambda \mathrm{\notin }\{{\lambda }_j\}$, step 2.1 produces a solution of $(L-\lambda )u=f$ for every $f\in L^2(\mathrm{\Omega })$, so $L-\lambda$ is surjective; it is injective, because $(L-\lambda )u=0$ makes $u$ a weak eigenfunction with eigenvalue $\lambda$, forcing $\lambda \in \{{\lambda }_j\}$ by [F2] or $u=0$. The solution estimate of step 1.1 gives $\mathrm{\parallel }(L-\lambda {)}^{-1}f{\mathrm{\parallel }}_2\le {\delta }^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_2$, so $L-\lambda$ is bijective with bounded inverse exactly for $\lambda \mathrm{\notin }\{{\lambda }_j\}$.

The inverse series and its norm. For $\lambda \mathrm{\notin }\{{\lambda }_j\}$ the series of step 1.1 has coefficients $(f,e_j{)}_{L^2}({\lambda }_j-\lambda {)}^{-1}$, so the solution is $R_{\lambda }f={\sum }_j(f,e_j{)}_{L^2}({\lambda }_j-\lambda {)}^{-1}{e}_j$, converging in $L^2$ and, by step 2.1, with $H_0^1$ membership; its $L^2$ norm satisfies $\mathrm{\parallel }{R}_{\lambda }f{\mathrm{\parallel }}_{L^2}^2={\sum }_j\mathrm{\mid }(f,e_j{)}_{L^2}{\mathrm{\mid }}^2\mathrm{\mid }{\lambda }_j-\lambda {\mathrm{\mid }}^{-2}\le {\delta }^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^2}^2$. Choose $k$ with $\mathrm{\mid }{\lambda }_k-\lambda \mathrm{\mid }=\delta$ (attained because the eigenvalue set is closed); testing at $f=e_k$ gives $\mathrm{\parallel }{R}_{\lambda }{e}_k{\mathrm{\parallel }}_{L^2}=1\mathrm{/}\delta$, so the operator norm is exactly $\mathrm{\parallel }{R}_{\lambda }\mathrm{\parallel }=1\mathrm{/}\delta =1\mathrm{/}\mathrm{dist}(\lambda ,\{{\lambda }_j\})$.

Complex spectrum. By [F5], every nonreal scalar is in $\rho (\tilde{L})$. For a real $\lambda \mathrm{\notin }\{{\lambda }_j\}$, step 3.1 gives a bounded inverse for $L-\lambda$ over the base field; if $\mathbb{K}=\mathbb{C}$ this is directly the complex resolvent, while if $\mathbb{K}=\mathbb{R}$ its complexification is a bounded inverse of $\tilde{L}-\lambda$. Conversely, each ${\lambda }_j$ is an eigenvalue, so $\tilde{L}-{\lambda }_j$ is not injective (in the real case, complexify its nonzero real eigenfunction). Therefore $\sigma (\tilde{L})=\{{\lambda }_j\}$, which is discrete with no finite accumulation point because ${\lambda }_j\to +\mathrm{\infty }$, bounded below by ${\lambda }_1\ge -\beta$ from the discrete spectral theorem, and unbounded above because ${\lambda }_j\to +\mathrm{\infty }$.