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Suppose I have a RKHS $\mathcal{H}$ on a set $X$ with kernel $K$. Moreover, othere are $\{x_1,\dots,x_n\}\subset X$ distinct points. We denote by $Q=(K(x_i,x_j))\in\mathbb{R}^{n\times n}$. In An Introduction to the Theory of Reproducing Kernel Hilbert Spaces, by V.I. Paulsen and M. Raghupathi they say for each $f\in\mathcal{H}$ there is a vector $w\in\mathbb{C}^n$ such that

$$Qw = (f(x_1),\dots,f(x_n)) $$

How can this conclusion be drawn? I was trying to work with the reproducing kernel property but didn't manage to get.

$$f(x_i) = \langle f, K(\cdot, x_i)\rangle $$

and trying to invoke the adjoint, without any success. Any hint / help would be much appreciated

swissy
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1 Answers1

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Define $k_i(x):=K(x,x_i)$ for $i=j,\ldots,n$. Project $\bar f\in\mathscr{H}$ orthogonally onto the span of the $k_i$ to obtain $$ \bar f=\sum_{i=1}^n \alpha_i k_i + g$$ for suitable $\alpha_i\in\mathbb{C}$ and a $g\in\mathscr{H}$ orthogonal to the span. Now for all $i=1,\ldots,n$ $$ \bar f(x_j)=\langle \bar f,k_j\rangle = \sum_{i=1}^n \alpha_i \langle k_i, k_j\rangle=\sum_{i=1}^n \alpha_i K_{ji}$$ where we used the fact that $\langle g, k_j\rangle=0$ due to orthogonality.

With $v=\big(f(x_1),\ldots, f(x_n)\big)$, $w=\bar \alpha$ and $\bar Q=K(x_j,x_i)$ this is $$ \bar v_j = \big( \bar Q\alpha\big)_j$$ or $$ v=Qw.$$

g g
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  • thanks for the comment. I totally agree that your mentioned Theorem 3.4 allows this. However, check Theorem 3.8, I quote: "Let \mathcal{H} be an RKHS on $X$ with Kernel K, ${x_1,\dots,x_n}\subset X$ be a finite set of distinct points, let $v=(\lambda_1,\dots,\lambda_n)\in\mathbb{C}^n$ and let $Q=(K(x_i,x_j))$. Then there exists a vector $w=(\alpha_1,\dots,\alpha_n)$ s.t. $(v-Qw)\in\mathcal{N}(Q)$ \dots". There is no mentioning of the range of $Q$ or invertible $K$. – swissy Jun 08 '21 at 12:47
  • Yes, you are right I was a bit quick and did not find this respective citation. I hope it is clear now. – g g Jun 08 '21 at 14:10
  • Thanks for the update. Could you share a resource why this is true: "Note that Q is a symmetric, square matrix. This means Rn is the direct sum of R the range of Q and N its null space. " – swissy Jun 08 '21 at 16:27
  • The point which still puzzles me is that it seems to imply that $Q$ is surjective, isn't it? That would mean every symmetric positive semi-definite matrix is surjective – swissy Jun 09 '21 at 05:56
  • This is what confused me in the beginning as well! But notice that you only demand that $(f(x_1),\ldots,f(x_n))$ is in the range of $Q$ not all vectors. Take as extreme example $K=0$ such that $Q=0$. Then $Q$ is definitely not surjective but the statement still holds, since the only $f\in\mathscr{H}$ is the zero function. But you are right I should give arguments that $r=(f(x_1),\ldots,f(x_n))$. – g g Jun 09 '21 at 07:06
  • @swissy: See my revised approach. This should now be complete. – g g Jun 09 '21 at 09:22
  • Many thanks for the additional detail! – swissy Jun 10 '21 at 05:58