For $W:R \to R$ is a bounded function, consider the Schrodinger operator $$[H_{W}y](x)=-y''(x)+W(x)y(x), x \in R$$ where the domain is $D(H_{W})=\{y:R \to R| y \in AC_{loc}(R), y' \in AC_{loc}(R), y''\in L^{2}(R)\}$.
How to show the spectrum $\sigma(H_{W})$ is bounded from below but not from above. Thanks!
The operator $\partial f = -if'$ is a selfadjoint linear operator on the domain $$ \mathcal{D}(\partial)=\{ f \in L^2 : f \mbox{ is absolutely continuous with } f' \in L^2 \}. $$ And $\partial^2f=-f''$ is selfadjoint on the domain $\mathcal{D}(H_W)$ that you have described. This can be established with the Fourier transform because $f\in\mathcal{D}(H_W)$ iff $(1+s^2)\hat{f}(s) \in L^2$.
The operator $Mf = W(x)f(x)$ is a bounded selfadjoint operator on $L^2$. So the operator under consideration, $H_W = \partial^2+M$ is a bounded selfadjoint perturbation of a selfadjoint operator, which makes $H_W$ selfadjoint.
The spectrum of $H_W$ is contained in its numerical range $$ \{ (H_Wf,f) : f \in \mathcal{D}(H_W),\;\; \|f\|=1 \}. $$ The numerical range of $\partial^2$ is positive because $(\partial^2f,f)=(\partial f,\partial f) \ge 0$ for $f\in\mathcal{D}(\partial^2)$. This leads to a simple argument that the spectrum is bounded below, because $$ (H_Wf,f) = \|\partial f\|^2 + (Mf,f) \ge (Mf,f) \ge -\|M\|\|f\|^2. $$ Therefore the spectrum of $H_W$ must lie in the interval $[-\|M\|,\infty)$. However, the operator $H_W$ cannot have totally bounded spectrum because that would force $H_W$ to be bounded, leading to the conclusion that $\partial^2 = H_W-M$ is bounded, which it is not. So the spectrum of $H_W$ is an unbounded subset of $[-\|M\|,\infty)$.
