Orthonormal bases in RKHSs via interpolating sequences

Question

Definitions and setting

Let $\mathcal{H}$ be a separable, infinite-dimensional, reproducing kernel Hilbert space on a nonempty set $X$. As usual, denote the reproducing kernel on $\mathcal{H}$ by $K$ and let $K_x:=K(x,\cdot)$.

Let $M(\mathcal{H})$ denote the multiplier algebra of $\mathcal{H}$, that is, the functions $f:H\rightarrow \mathbb{R}$ satisfying: for all $h\in \mathcal{H}$ one has $x\mapsto f(x)h(x)$ is a function belonging to $H$ (i.e. pointwise multiplication).

A sequence $(k_i)_{i\in \mathbb{N}}\subseteq X^{\mathbb{N}}$ is said to be an interpolating sequence of $M(\mathcal{H})$ if, for any bounded sequence of reals $(r_i)_{i\in \mathbb{N}} \in \ell^{\infty}$ one can always find a multiplier $\mu\in M(\mathcal{H})$ satisfying: $$ \phi(k_i) = w_i \mbox{ for all i in }\mathbb{N} . $$ As discussed in the recent article Tsikalas - Interpolating sequences for pairs of spaces, many RKHSs have the property that the set of interpolating sequences can be characterized as the set of sequences $\boldsymbol{k}:=(k_i)\in X^{\mathbb{N}}$ for which $T_{\boldsymbol{k}}(\mathcal{H})=\ell^2$ where $$ \begin{aligned} T_{\boldsymbol{k}}:\mathcal{H} & \rightarrow \mathbb{R}^{\mathbb{N}}\\ T_{\boldsymbol{k}}(f) & := \big(\frac{f(k_i)}{\|K_{k_i}\|}\big)_{i\in \mathbb{N}} \end{aligned} $$ for all $f\in \mathcal{H}$.

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I recently came across this very old MO post, claiming that, $(K_{k_i})_{i\in \mathbb{N}}$ is a Schauder, or even a Riesz, basis for $\mathcal{H}$ if $(k_i)_{i\in \mathbb{N}}$ is an interpolating sequence of $M(\mathcal{H})$. If this is correct, why is this this the case?

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Intuition: I expect this somehow follows from the characterization of "good interpolating sequences", by which I mean, those satisfying $T_{\boldsymbol{k}}(\mathcal{H})=\ell^2$; however, I don't really see it….

Answer 1

It is true that if a sequence $(k_n)$ is interpolating for $M(\mathcal{H})$, then the normalized Kernel vectors $g_n:=K_{k_n}/\Vert K_{k_n} \Vert_\mathcal{H} $ form a Riesz system in $\mathcal{H}$. Of course it does not have to be a complete system, so it is not always a Riesz basis.

To see this, let $g_n$ be the normalized kernel vectors, and for $t\in \mathbb{R}, j\in \mathbb{N}$ define $w_j = e^{2\pi i j t} $. Then by hypothesis, for every $t$ there exists a multiplier $\varphi_t$ such that $\varphi_t(z_j) = w_j,\, j=1,2,\dots$. Furthermore by the open mapping theorem we can choose $\varphi_t$ such that $\Vert \varphi_t \Vert_{M(\mathcal{H})} \leq M < +\infty$. Then consider the adjoint of the multiplication operator by $\varphi_t$, $M_{\varphi_t}^*:\mathcal{H} \to \mathcal{H}$. By the bound on the multiplier norms of $\varphi_t$, \begin{equation}\label{eq1} \Vert M^*_{\varphi_t} f \Vert_\mathcal{H} ^2 \leq M^2 \Vert f \Vert_\mathcal{H}^2,\,\,\, \forall t\in \mathbb{R}, \,\,\, \forall f \in \mathcal{H}. \end{equation}

Now consider $(a_n)$ a sequence of complex numbers with only finite non zero terms, and $f=\sum_{n}a_n g_n $. Applying the above inequality for this choice of $f$ and using the fact that $M^*_{\varphi_t} g_n = \overline{\varphi_t(k_n)} g_n $ we arrive at the inequality, $$ \sum_{n,m} a_n \overline{a_m} e^{2\pi i (n-m) t} \langle g_n , g_m \rangle \leq M^2 \Vert \sum_{n} a_n g_n \Vert^2, $$ and integrating with respect to $t$ in $(0,1)$ we find that $$ \sum_n |a_n|^2 \leq M^2 \Vert \sum_{n} a_n g_n \Vert^2 $$ By replacing $a_n$ by $e^{-2\pi i tn} a_n $ and using the same ineqaulity and integrating with respect to $t$ we find that $$\Vert \sum_{n} a_n g_n \Vert^2 \leq M^2 \sum_n |a_n|^2. $$ This proves that $g_n$ is a Riesz system.