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Yes, this can be done. By spectral representation, $Q_t$ is unitarily equivalent to two copies of multiplication by $e^{-t\lambda}$ in $L^2(0,\infty)$, and now we only need to decompose $(0,\infty)=\b …

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 …

This is a rather lazy version of an answer, but I will also indicate how I think a full answer can be produced. Let $u(x,z)$ be the solution of $-u''+qu=z u$ with $u(0,z)=0$, $u'(0,z)=1$. Then $$ B_L …

What is clear here is that $H_2$ has spectrum above $4$, and that $\lambda\in\sigma_{\textrm{ess}}(H_2)$ also, with $\lambda:=\max\sigma(H_2)=\|H_2\|>4$. What $\lambda$ is actually equal to seems a ra …

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 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 …

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be n …

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 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 …

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 …

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 …

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 …

No, the infimum equals zero. In the spirit of Asaf's comment, try $$ f(x,y)=e^{i(mx+ny)} . $$ Then $\|f\|=1$, $\nabla_{irr} f = i(m+\alpha n) f$, with $\alpha=\sqrt{2}$ (but it'll work for any irratio …

(1) is perhaps better rephrased as: (1') $\lambda\in\sigma_p$ (in words: $\lambda$ is an eigenvalue) because then both (1') and (3) are statements about spectral properties. Also, in more compact nota …

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 …