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As already pointed out in the comments, $V\to\infty$ does imply discrete spectrum in general. In fact, this becomes an equivalence if a somewhat more general version of this condition is used: The …

The spectrum of $(Af)(x)=a(x)f(x)$ is the essential range of $a(x)$. As usual in this context, essential here means basically (resisting a silly pun) "ignore what happens on null sets." More precisely …

In view of Terry's comment on Michael's answer, it's perhaps worth pointing out that this monotonicity also fails if both domains are required to be convex. We can take $M_2=[0,L]^2$ as a square of si …

A general concept that fits this well is strong resolvent convergence. As the name suggests, $T_n\to T$ in this sense means that $(T_n-i)^{-1}\to (T-i)^{-1}$ strongly. In your case, we first of all ha …

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1/2]$. Since the ac spectrum is invariant …

As a general rule of thumb, it's usually most convenient in one-dimensional problems to work with solutions of the ODE $-y''+Vy=Ey$ rather than operator theoretic methods. Here, everything follows fro …

This runs into obvious technical issues. For example, $K_0$ will not be bounded (let alone compact) even for very nice $V$. So one also has to think about what exactly one wants to prove. This paper …

This is a slightly expanded (and slightly more systematic) summary of my comments above. First of all, the equation $$ -\varphi''+\frac{1}{r}\varphi' + (V+\frac{m}{r^2})\varphi=\lambda\varphi \quad\qu …

In general, there won't be a uniform bound on all eigenfunctions simultaneously. If $[a,b]$ is a short interval with Dirichlet boundary conditions $y(a)=y(b)=0$ and constant potential $V=c$, then the …

If $0<s\le 1$, then $D(T^s)\supseteq D(T)$ for any self-adjoint $T$, as is obvious from the description of the domain that you quote at the end of your post. If $s=n+t$ with $n\in\mathbb N$ and $0<t\l …

Your (unbounded) operator is in the limit circle case at $+\infty$. This means that self-adjoint realizations are obtained only when a boundary condition at $\infty$ is imposed, and the eigenvalues wi …

This is false. Take $H_n=P_n$ as the projection onto $\ell^2(\{ x: |x|\le n\})$. Then $P_n\to 1$ strongly, $\dim \chi_{\{ 0\} }(P_n)=\infty$, but $\sigma (1) =\{ 1\}$. What you have in this situation …

This runs into similar problems as before. The Laplacian has spectrum $\sigma(\Delta)=[-4,4]$ in dimension $2$, and any finitely supported potential $V\ge 0$, $V\not\equiv 0$ will give $\Delta+V$ an e …

If I interpret your question correctly, then you want an elementary proof of the fact that if $A$ is normal, then $\sigma(p(A,A^*))=f(\sigma(A))$, where $f(z):=p(z,\overline{z})$. By looking at appro …

Let's rewrite your equation as a Schrödinger equation, as follows: Introduce the new variable $t\in (-\pi/2, \pi/2)$ by $x=\sin t$. Then if $f$ solves your boundary value problem (with periodic bounda …