# A reproducing kernel Hilbert space

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 $$N_1(A)\leq N_2(A)$$.

• Typo: in your definitions of $N_1$ and $N_2$, $T$ should be $A$ Aug 16, 2022 at 21:48

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 $$1$$.