A functional Hilbert space $\mathscr H=\mathscr H(\Omega)$ is a Hilbert space of complex valued functions on a (nonempty) set $\Omega$, which has the property that point evaluations are continuous i.e. for each $\lambda\in \Omega$ the map $f\mapsto f(\lambda)$ is a continuous linear functional on $\mathscr H$. The Riesz representation theorem ensure that for each $\lambda\in \Omega$ there is a unique element $k\_{\lambda}\in \mathscr H$ such that $f(\lambda)=\langle f,k\_{\lambda}\rangle$ for all $f\in \mathscr H$. The collection $\{k\_{\lambda} : \lambda\in \Omega\}$ is called the reproducing kernel of $\mathscr H$. For $\lambda\in \Omega$, let $\hat{k\_{\lambda}}=\frac{k\_{\lambda}}{\|k\_{\lambda}\|}$ be the normalized reproducing kernel of $\mathscr H$.


For a bounded linear operator $A$ on $\mathscr H$, we define the following norm:
\begin{align\*}
N(A):=\sup\{\big|\langle T\widehat{k}\_{\lambda},\widehat{k}\_{\lambda}\rangle\big|: \lambda\in\Omega\}.
 \end{align\*}



> 
> It is well-known that $w(A^n)=w^n(A)$ holds for any selfadjoint operator $A$, $w(.)$ denotes the numerical radius. Now, assume that $A$ is a selfadjoint operator. Is
> $$N(A^n)=N^n(A)\;?$$
> 
> 
> 



Is $N(A^n)=N^n(A)\;$ for a self-adjoint operator $A$?
No. If $A=P$ is a projection, then $P^2=P$, but in general you won't have $N(P)=0$ or $1$. In fact, you could just take $\Omega=\{ 1,2\}$, so $\mathcal H\cong\mathbb C^2$, and $P$ as the projection onto $(1,1)$.


