# Spectral integral: verification of my conceptual understanding

Let $$t \in B(H)$$ be a positive operator. Let $$E$$ be the unique spectral measure relative to $$(\sigma(t), H)$$ such that $$t = \int \lambda d E(\lambda)$$ (see e.g. Murphy theorem 2.5.6). Recall that $$E_{\xi, \eta}$$ is by definition a regular Borel measure defined by $$E_{\xi, \eta}(S) = \langle E(S)\xi, \eta\rangle$$ and that $$\left\langle \left(\int f(\lambda) dE(\lambda)\right) \xi, \eta\right\rangle = \int_{\sigma(t)} f(\lambda) d E_{\xi, \eta}(\lambda).$$

In the proof of lemma 1.5 (p62) in Takesaki's book "Theory of operator algebra I", the following is claimed:

If $$\epsilon > 0$$ and $$E(\epsilon):= \int \chi_{[0, \epsilon]}(\lambda)dE(\lambda)$$, then $$\langle t\xi, \xi\rangle \ge \epsilon \|\xi\|^2$$ for every $$\xi \in \operatorname{Im}(1-E(\epsilon))$$.

I want to verify that my proof is correct, and thus verify that my understanding of the spectral integral is correct.

Given $$\xi \in H$$, we have using the fact that $$E(\epsilon)$$ and $$t$$ commute and $$1-E(\epsilon)$$ is a projection: \begin{align*}\langle t(1-E(\epsilon))\xi, (1-E(\epsilon))\xi\rangle &= \langle t(1-E(\epsilon))\xi, \xi\rangle\\ &= \left\langle\left( \int\chi_{]\epsilon, \infty[\cap \sigma(t)}(\lambda) \lambda dE(\lambda)\right)\xi, \xi\right\rangle\\ &= \int_{]\epsilon, \infty[\cap \sigma(t)} \lambda dE_{\xi, \xi}(\lambda)\\ &\ge \epsilon \int_{\sigma(t)}dE_{\xi, \xi}(\lambda) \\ &= \epsilon E_{\xi, \xi}(\sigma(t))\\ &= \epsilon \|\xi\|^2\end{align*} Hence, if $$\xi \in \operatorname{Im}(1-E(\epsilon))$$, we get

\begin{align}\langle t\xi, \xi\rangle &= \langle t (1-E(\epsilon))\xi, (1-E(\epsilon))\xi\rangle \ge \epsilon \|\xi\|^2\end{align} as desired.

Is this correct? Please spare no criticism.

Yes, that's the standard way to do it. The only thing I would mention (or ask you to clarify, if I were evaluating you) is when you write $$t\,(1-E(\epsilon))=\int_{\sigma(t)}\chi^{\phantom{A}}_{(\epsilon,\infty)}\,\lambda\,dE(\lambda)$$ you are using the extremely important fact that the map $$f\longmapsto\int_{\sigma(t)}f(\lambda)\,dE(\lambda)$$ is a $$*$$-homomorphism. One is used to integrals being linear, but being multiplicative is a very particular property that depends on the fact that $$E$$ is projection-valued.