Finiteness of "novel variance" from a kernel on a compact space [closed]

Question

Let $c(i,i')$ be a kernel function on a reasonable index space $I$. Choose a dense sequence of points $\{i_1, i_2, \cdots \} \subseteq I$, and define the one-point kernel functions $k_n := c(\cdot, i_n)$. Let $U$ be the Hilbert span of these functions; by the reproducing kernel property, $\langle k_n, k_{n'} \rangle := c(i_n, i_{n'})$.

Using the Gram-Schmidt procedure, we convert the sequence $\{k_n\}$ into an orthogonal basis $\{\hat k_n\}$, by subtracting away the projection onto the previous axes (without normalization). Define the "novel variance" $\hat \sigma_n^2 := \langle \hat k_n, \hat k_n \rangle$, which represents the new contribution to variance from data located at $i_n$.

Now, define $\Sigma^2 := \sum \hat \sigma_n^2$, which represents all the possible novel variance to be found in $I$. It is not difficult to show that $\Sigma^2$ does not depend on the choice of basis.

Question: If $I$ is compact, then is $\Sigma^2 < \infty$?

Answer 1

You haven't specified what role is played by the topology of $I$, so my default assumption will be that the kernel is continuous, in which case the answer to your question is negative.

For a counterexample, let $I$ be the one-point compactification of $\mathbb{N}$, that is, $\{1,2,\dots,\infty\}$, and take $k_{nm} := \delta_{nm} n^{-1}$. Then the basis is already orthogonal, so $\hat\sigma_n^2 = k_{nn}$, and $\sum_n \hat\sigma_n^2 = \infty$.