TITLE: Norm of the product of a self-adjoint operator and one of its spectral projections
QUESTION [6 upvotes]: Let $A$ be a bounded self-adjoint operator on a Hilbert space $H$ such that $A$ has a cyclic vector. That is, there exist $x \in H$ such that the linear subspace spanned by $\{ x, Ax, A^2x,...\}$ is dense in $H$.
Also let $E$ denote the spectral measure corresponding to $A$, i.e. $E : \Sigma \rightarrow B(H)$ with $E(\Omega)=E_{\Omega}=\chi_{\Omega \cap \sigma(A)} (A)$ given by the functional calculus for $A$. Here $\Sigma $ is the Borel-$\sigma$-Algebra over $\mathbb{R}$ and $\chi_M$ denotes the characteristic function of a set $M$. $\sigma(A)$ is the spectrum of $A$.
Question: Assume that the interval $[0,1]$ is contained in the spectrum $\sigma(A)$ of A. What can we say about the norm $||E_\Omega A||$ if $\Omega$ is, for example, a subinterval of $[0,1]$, say $\Omega=[\tfrac{1}{4}, \tfrac{1}{2})$? What if $\Omega=[\tfrac{1}{4}, \tfrac{1}{3}) \cup ([\tfrac{1}{3}, \tfrac{1}{2}) \cap \mathbb{Q})$?
One thing we know is that $||E_\Omega A|| = ||A E_\Omega||$. So we only have to compute the norm on the range of $E_\Omega$. But I don't really see how the cyclic vector comes into play. It tells us that $H$ is separable. I don't know how that could be helpful.

REPLY [2 votes]: In the second case where $\Omega=[\tfrac{1}{4}, \tfrac{1}{3}) \cup ([\tfrac{1}{3}, \tfrac{1}{2}) \cap \mathbb{Q})$ there is no definite answer even if we assume the existence of a cyclic vector.
First, consider $A$ to be multiplication by $f(x)=x$ on $L^2[0,1]$ with the usual Lebesgue measure. A cyclic vector is given by the constant function 1 since polynomials are dense in $L^2[0,1]$. Then the spectral measure $E_\Omega$ is given explicitly by the functional calculus as the operator $\chi_\Omega (A) \phi (x) = \chi_\Omega(x) \phi$.
In particular, the operator $E_\Omega A$ is given by multiplication with $g(x)=x \chi_\Omega(x)$. Computing its norm amounts to computing its spectrum which amounts to computing its essential range with respect to the Lebesgue measure. Clearly, the essential range of $g$ is $\{ 0 \} \cup [\tfrac{1}{4}, \tfrac{1}{3}]$, so $|| E_\Omega A ||= \tfrac{1}{3}$.
Second, consider $L^2[0,1]$ with the measure given by $\mu = \lambda + \delta_{\tfrac{1}{2}}$, where $\lambda$ denotes the Lebesgue measure. I.e. we give the point $\tfrac{1}{2}$ mass 1. Then, by the same argument as above we have to find the essential range of $g(x)=x \chi_\Omega(x)$ w.r.t. $\mu$. Here, we have that the essential range is $\{ 0 \} \cup [\tfrac{1}{4}, \tfrac{1}{3}] \cup \{ \tfrac{1}{2} \}$, so $|| E_\Omega A ||= \tfrac{1}{2}$.
Note: In the second case, we also have a cyclic vector, namely the constant function 1.  This is consequence of the following: For finite, regular Borel measures on compact subsets $K$ of $\mathbb{R}$ the continuous functions are dense in $L^2(K)$. Now, the polynomials are dense in the continuous functions (w.r.t. to the sup-norm), so particular they are dense w.r.t. to the $L^2$ norm.