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\_1(A):=\sup\{\big|\langle T\widehat{k}\_{\lambda},\widehat{k}\_{\mu}\rangle\big|: \lambda,\mu\in\Omega\} \qquad \textrm{and} \qquad N\_2(A):=\sup\{\|T\widehat{k}\_{\lambda}\|: \lambda\in\Omega\}.
 \end{align\*}



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> Is $N\_1=N\_2$? Clearly $N\_1(A)\leq N\_2(A)$.
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A reproducing kernel Hilbert space
A useful test case for RKHS (which is not like the interesting examples, but does satisfy the definitions) is $\Omega={\mathbb N}$ and $H=\ell^2({\mathbb N})$. Note that $\hat{k\_n}$ is just the usual unit basis vector that is $1$ in position $n$ and $0$ everywhere else.


Viewing $T$ as an ${\mathbb N}\times {\mathbb N}$ matrix, $N\_1(T)$ is the maximum absolute value of all matrix entries, and $N\_2(T)$ is the maximum $\ell^2$-norm of all columns in the matrix.


It is then easy to find examples where $N\_1(T)$ is strictly less than $N\_2(T)$, because this is basically asking for vectors in $\ell^2$ whose sup norm is strictly smaller than their $\ell^2$ norm.


In fact, we could have built a counterexample with $\Omega$ being a 2-element set; then the RKHS is just ${\mathbb C}^2$ and you could take $T$ to be the $2\times 2$ matrix with all entries equal to $1$.


