Let $\mathcal H$ be a Hilbert space. Then $\mathcal K(\mathcal H)$ is the only proper closed two sided ideal of $\mathcal B(\mathcal H).$
I am following Rajendra Bhatia's notes on Functional Analysis. The following hint has been given in the book.
Let $\mathcal I$ be a proper closed two sided ideal of $\mathcal B(\mathcal H).$ If $\mathcal I$ contains a positive operator $A$ that is not compact then there exists $\varepsilon \gt 0$ such that $P(\varepsilon, \infty)$ is infinite-dimensional, where $P$ is the projection valued measure associated with $A$. Let $\mathcal M$ be it's range and let $V$ be the unitary operator from $\mathcal H$ onto $\mathcal M.$ Since $A(\mathcal M) = \mathcal M,$ we have $$V^*AV (\mathcal H) = V^*A(\mathcal M) = V^*(\mathcal M) = \mathcal H.$$ Then for every $x \in \mathcal H$ we have $$\|V^*AV x\| \geq \varepsilon \|x\|.$$ Hence $V^*AV$ is invertible and $V^*AV \in \mathcal I.$ So it follows that $\mathcal I = \mathcal B (\mathcal H).$
In the above proof I don't understand few things. Here they are $:$
$(1)$ Why does there exist $\varepsilon \gt 0$ such that the range of the projection $P(\varepsilon, \infty)$ is infinite-dimensional?
$(2)$ How to guarantee the existence of an unitary operator $V : \mathcal H \longrightarrow \mathcal M\ $?
$(3)$ Why do we have $A(\mathcal M) = \mathcal M\ $?
I have understood the rest of the part in the proof quite clearly. Could anyone please help me understanding the proof?
Thanks for your time.
EDIT $:$ I have managed to answer the first question. Here it is $:$
Suppose for every $\varepsilon \gt 0,$ the range of $P(\varepsilon,\infty)$ is finite dimensional. Then for every $n \in \mathbb N,$ $AP\left (\frac {1} {n}, \infty \right )$ is a finite rank operator. Now we have $$AP \left (\frac {1} {n},\infty \right ) = \left (\int t\ dP \right ) \left (\int \chi_{\left (\frac {1} {n}, \infty \right )}\ dP \right ) = \int t\ \chi_{\left (\frac {1} {n} , \infty \right )}\ dP.$$ Since the integrand $t\ \chi_{\left (\frac {1} {n} , \infty \right )}$ increases to $t,$ by MCT we have $$A P \left (\frac {1} {n}, \infty \right ) \xrightarrow {\text {NORM}} A.$$ But then $A$ becomes compact as it is the norm limit of a sequence of finite rank operators, a contradiction. Hence there should exist some $\varepsilon \gt 0$ such that the range of $P(\varepsilon, \infty)$ is infinite-dimensional.
For $(3)$ we first note that $$\begin{align*}AP \left (\varepsilon,\infty \right ) & = \left (\int t\ dP \right ) \left (\int \chi_{\left (\varepsilon, \infty \right )}\ dP \right ) \\ & = \int t\ \chi_{\left (\varepsilon , \infty \right )}\ dP \\ & = \int \chi_{\left (\varepsilon , \infty \right )}\ t\ dP \\ & = \left (\int \chi_{\left (\varepsilon, \infty \right )}\ dP \right ) \left (\int t\ dP \right ) \\ & = P(\varepsilon, \infty) A \end{align*}$$ Now for any $x \in \mathcal M = \text {ran} \left (P(\varepsilon, \infty) \right )$ we have $$Ax = A P(\varepsilon, \infty) x = P (\varepsilon, \infty) A x \in \mathcal M$$ proving that $A(\mathcal M) \subseteq \mathcal M.$ But I don't know how to show the reverse inclusion i.e. how to show that $\mathcal M \subseteq A(\mathcal M).$
RE-EDIT $:$ For the other part of the inclusion in $(3)$ we first note that the map $t \mapsto \frac {1} {t}$ is a bounded measurable function on $(\varepsilon, \infty)$ and hence we have
$$P(\varepsilon, \infty) = \displaystyle {\int_{(\varepsilon, \infty)} dP = \left (\int_{(\varepsilon, \infty)} t\ dP \right ) \left (\int_{(\varepsilon, \infty)} t^{-1}\ dP \right ) = A E(\varepsilon, \infty) B}$$
where $B = \displaystyle {\int_{(\varepsilon, \infty)} t^{-1}\ dP.}$ This shows that $\mathcal M \subseteq A(\mathcal M).$