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*}

Is $N_1=N_2$? Clearly $N_1(A)\leq N_2(A)$.

  • $\begingroup$ Typo: in your definitions of $N_1$ and $N_2$, $T$ should be $A$ $\endgroup$
    – Yemon Choi
    Aug 16, 2022 at 21:48

1 Answer 1


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 $1$.


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