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 $\Sigma^2 < \infty$?