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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 $\Sigma^2 < \infty$?

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closed as unclear what you're asking by Tom LaGatta, Chris Godsil, Stefan Kohl, user9072, Noah Stein Apr 16 '14 at 18:50

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  • $\begingroup$ What do you mean by "doesn't depend on the choice of basis"? It seems that you are not choosing a basis. $\endgroup$ – Alexander Shamov Apr 3 '14 at 0:20
  • $\begingroup$ On the other hand, it does depend on the order in which you take the $i_n$'s. For instance, even for 2 vectors, say, $i_1$ and $i_2$, your sum of squared norms will be $\Vert i_1 \Vert^2 + \Vert i_2 \Vert^2 - \frac{\langle i_1,i_2 \rangle}{\Vert i_1 \Vert^2}$, which is not symmetric in $i_1,i_2$. $\endgroup$ – Alexander Shamov Apr 3 '14 at 0:22
  • $\begingroup$ Thanks, Alexander. I clearly have screwed something up in the definition. The 2 vector case is supposed to be $\|i_1\|^2 + \|i_2\|^2 - \langle i_1, i_2 \rangle$, which is symmetric. $\endgroup$ – Tom LaGatta Apr 3 '14 at 0:40
  • $\begingroup$ Then I don't have a guess what you are computing in the case of more than 2 vectors. $\endgroup$ – Alexander Shamov Apr 3 '14 at 0:46
  • $\begingroup$ Voting to close my own question. $\endgroup$ – Tom LaGatta Apr 3 '14 at 1:02
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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 $\sum_n \hat\sigma_n^2 = \infty$.

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  • $\begingroup$ Thanks for the counterexample. As your comments show the question was ill-formed. I'm marking this as an answer to give you the points in gratitude for your observation. $\endgroup$ – Tom LaGatta Apr 3 '14 at 1:27

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