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.
