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 norms:
\begin{align\*}
 N(A):=\sup\{\|A\widehat{k}\_{\lambda}\|: \lambda\in\Omega\}.
 \end{align\*}



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> Is $ N(A)= N(A^\*)$?
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Is $\sup\{\|A\widehat{k}_{\lambda}\|: \lambda\in\Omega\}=\sup\{\|A^*\widehat{k}_{\lambda}\|: \lambda\in\Omega\}$? where $A$ is an operator on RKHS
No, this fails already when $\Omega=\{ 1, 2\}$, so $H=\mathbb C^2$ and $\widehat{k}\_j=e\_j$. We can take
$$
A= \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} .
$$
Then $N(A)=\sqrt{2}$, $N(A^\*)=1$.


