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This follows indeed from a number of standard results, though a written out proof with all details included would be lengthy. (1) The essential spectrum (this, and not continuous spectrum is the prope …

One such condition (which is weaker, though not dramatically so perhaps) is that $\rho(x)>0$ a.e. on $(-2,2)$. This is usually called the Denisov-Rakhmanov theorem; see here. In fact, it will give the …

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 …

The eigenvalues with even/odd eigenfunctions can equivalently be described as the eigenvalues of the half line problem $$ -y''(x)+V(x)y(x)=Ey(x), \quad 0\le x<\infty , $$ with boundary conditions $y'( …

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 …

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 …

This is not a very explicit answer, just an attempt to put things into context. First of all, by periodic boundary conditions, one usually means $\psi(0)=\psi(a)$, $\psi'(0)=\psi'(a)$ (without the co …

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 …

We can actually do this directly and my comment above is not that relevant. Let $u\in H^1$ and also assume that $u$ is compactly supported, so $\int u'=0$. Approximate $\alpha u'$ in $L^2$ by $v_n\in …

We can use the integral kernel of $R=(A-z)^{-1}$ and nothing much changes when $p\not= 2$. By the variation of constants formula, $(Rf)(x)=\int_0^1 G(x,t;z) f(t)\, dt$. Here $G$ could of course be wor …

Let's first make the interval equal to $(-1,1)$ by the change of variable $$ x = \frac{a+b}{2} + \frac{b-a}{2}\, t ; $$ then the equation becomes $$ \int_{-1}^1 u(s)[c+\log |s-t|]\, ds = g(t) , \quad\ …

This is false. Let $A=S+S^*$ be the free Jacobi matrix on $\ell^2$; I write $S$ for the shift $(Sy)_n = y_{n+1}$. Then $\sigma_{ac}(A)=[-2,2]$. Let $B$ be multiplication by a bounded sequence $b_n\ge …

You wrote "normal" earlier, but I assume (from your use and description of $A_{ac}$) that you want to consider self-adjoint $A$. Then your set (denote it by $T$) is simply $T=\sigma_{ess}(A)$ again. I …

Edit (complete rewrite, my first attempt was utter nonsense): In fact, $\sigma(H)=[0,\infty)$, $H=-(1/\beta)d^2/dx^2$. One quick way to see that there is no negative spectrum is to consider the quadra …