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I mean, $\lambda\in\sigma_{ess}(T)$ if and only if $\lambda$ is accumulation point of the spectrum of $T$ or $\dim\ker(T-\lambda)=\infty$. So $T\equiv 0$ implies that $\sigma(T)=\sigma_{ess}(T)=\{0\}$ because $H$ is infinite dimensional.
I think it also works: If $x\in\mathrm{rg}(E_A(\lambda))$ then $$(Ax,x)=(AE_A(\lambda)x,x)=\left(\int_{\mathbb{R}} t\chi_{(-\infty,\lambda]}(t)dE_A(t)x,x\right)=\int_{\mathbb{R}} t\chi_{(-\infty,\lambda]}(t)d\left(E_A(t)x,x\right)\leq \lambda(x,x)$$
With this definition, for the angle $\theta_k$ between $u_k$ and $u_{k+1}$ I did the following estimate: $$\theta_k=\arccos\left(\frac{|\langle u_k,u_{k+1}\rangle|}{\|u_k\|\|u_{k+1}\|}\right)\leq \arccos\left(\frac{1−\sin\frac{π}{2n}}{1+\sin\frac{π}{2n}}\right)$$ but $$\arccos\left(\frac{1−\sin\frac{π}{2n}}{1+\sin\frac{π}{2n}}\right)>\frac{π}{2n}.$$ Is this correct?
