Self-adjoint operators, projections, and resolutions of the identity.

Question

In my Functional Analysis course, we're discussing the Spectral Theorem and the like. One question from a previous exam states the following:

Let $H$ be Hilbert over $\mathbb C$, let $T \in B(H)$ be selfadjoint. Show that there is a selfadjoint projection $P \in B(H)$ with $TP - 2P$ and $2(I-P) - T(I-P)$ positive (in the sense that $(Ax,x) \geq 0$ for all $x \in H$).

This has to do with its resolution of the identity, I'm sure. I have the feeling $P = E([2, \infty) \cap \sigma(T))$ will do the trick, with $E$ the resolution of the identity (as in, just project onto a part of the spectrum), but cannot prove it. Anybody?

Answer 1

\begin{align} TP-2P&=(T-2I)P,\\ 2(I-P)-T(I-P)&=-(T-2I)(I-P) \end{align} $P=E[2,\infty)$ does the job because $I-P=E(-\infty,2]$.

Answer 2

After some fiddling, I feel I have a satisfactory answer. Define $\omega := [2,\infty) \cap \sigma(T)$, and $P := E(\omega)$. This is a self-adjoint projector by definition of $E$. By symbolic calculus, we have $$((TP - 2P)x,x) = ((T-2I)Px,x) = \int_{\sigma(T)} (\lambda - 2)\chi_{\omega} d(E_\lambda x,x).$$

We know $(E_\lambda x,x)$ is a positive measure (cf Rudin 12.17), and the integrand is nonnegative: it is zero on the part of the spectrum where $\lambda < 2$ by the indicator function, and in the part where $\lambda \geq 2$, we have $\lambda - 2 \geq 0$. So indeed, $TP-2P$ is positive. The proof for the other operator is analogous.