Norm of the product of a self-adjoint operator and one of its spectral projections 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.
 A: Whenever you apply a spectral projection $E(S)\ne I$ to $A$, you end up with $0$ in the point spectrum of $E(S)A=AE(S)$ because $\{AE(S)\}E(\sigma\setminus S)=0$ and $E(\sigma\setminus S) \ne 0$. So that special case always requires special attention.
If $(a,b)\subseteq\sigma(A)$, then $E(a,b) \ne 0$; otherwise $\mu \in (a,b)$ would lead to a bounded operator,
$$
         R(\mu)=\int_{\sigma}\frac{1}{\lambda-\mu}dE(\mu),
$$
which would have to be the resolvent $R(\mu)=(A-\mu I)^{-1}$. And that would force $(a,b)\in\rho(A)$, contrary to assumption.
Suppose $[0,1]\subseteq\sigma(A)$. Then $E(a,b) \ne 0$ for $(a,b)\subseteq[0,1]$. I'll first consider your first question where you ask about the spectrum of $AE[1/4,1/2)$. Automatically $0\in\sigma(AE[1/4,1/2))$ because $\{AE[1/4,1/2)\}E(1/2,1)=0$ and $E(1/2,1)\ne 0$. For $\mu\ne 0$ and $\mu\notin [1/4,1/2]$,
$$
    AE[1/4,1/2)-\mu I=(A-\mu I)E[1/4,1/2)-\mu E(\sigma\setminus[1/4,1/2))
$$
has a bounded inverse given by
$$
    (AE[1/4,1/2)-\mu I)^{-1}=\int_{[1/4,1/2)}\frac{1}{\lambda-\mu}dE(\lambda)-\frac{1}{\mu}E(\sigma\setminus[1/4,1/2)).
$$
Therefore, $\sigma(AE[1/4,1/2))\subseteq [1/4,1/2]\cup\{0\}$. Conversely $0\in\sigma(AE[1/4,1/2))$ was noted above, and, for any $\mu\in(1/4,1/2)$, the projections $E(\mu-\delta,\mu+\delta)\ne 0$ for all $\delta > 0$, which gives the existence of a non-zero vector $x_{\delta}$ such that $E(\mu-\delta,\mu+\delta)x_{\delta}=x_{\delta}$ and, hence,
\begin{align}
         \|AE[1/4,1/2)x_{\delta}-\mu x_{\delta}\|
    & = \|(A-\mu I)E(\mu-\delta,\mu+\delta)x_{\delta}\| \\
    & \le \delta \|E(\mu-\delta,\mu+\delta)x_{\delta}\| \\
    & = \delta\|x_{\delta}\|.
\end{align}
So $AE[1/4,1/2)-\mu I$ cannot be continuously invertible, which proves
$$
              \{0\}\cup (1/4,1/2) \subseteq \sigma(AE[1/4,1/2))
$$
Because the spectrum is closed,
$$
                \{0\} \cup [1/4,1/2]\subseteq \sigma(AE[1/4,1/2)).
$$
The opposite inclusion was previously shown. So
$$
                     \sigma(AE[1/4,1/2))=\{0\}\cup[1/4,1/2].
$$
The operator $AE[1/4,1/2]$ is selfadjoint. So its norm is its spectral radius, which gives $\|AE[1/4,1/2)\|=1/2$.
I'll let you consider the other cases. Note for example that $S=[1/3,1/2]\cap\mathbb{Q}$ could be such that $E(S)=0$, or it could give $E(S)=E[1/3,1/2]$, or $E(S)=E(T)$ could hold for a lot of closed subsets $T$ of $[1/3,1/2]$ because the spectrum is closed, and every subset of $[1/3,1/2]\cap\mathbb{Q}$ could consist of eigenvalues.
A: 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.
