In my research I'm dealing with the following question.

Let $E$ set, $K:E \times E \to \mathbb R$ a positive type function, and $\mathcal H := \mathcal H(1+K)$ (in the sense of the Moore theorem). Now let $\xi: H \to \mathbb R$ a continuous, linear functional with $\xi(1) = 1$ (where the first $1$ denotes the constant function $1$). Now I know from the Aronszajn paper that $\operatorname{ker}(\xi)$ has a r.k. $\eta$ with $\eta \ll 1 + K$.

Now I'm interested in the following question: Is there always a functional $\xi$, so that I find a positive $\eta$, i.e. $\eta(x,y) \geq 0$ for all $x,y \in E$?

I could show the following statements already:

  1. You can calculate $\eta$ as $\eta(x,y) = 1 + K(x,y) - \frac{e(x) e(y)}{\left<e, e \right>}$, where $e \in \mathcal H$ is the Riesz representant from $\xi$, i.e. $\xi(f) = \left<f, e \right>$ for all $f \in \mathcal H$.

So you can ask equivalently, if you can choose an $e \in \mathcal H$, so that $\eta$ is positive.

  1. If $K$ is positive and $\mathcal H(1) \cap \mathcal H(K) =\{0\}$, then you can choose $e = 1$ and you get $\eta(x,y) = K(x,y) \geq 0$ for all $x,y \in E$.

  2. If $K(x,y) \geq \frac{1}{\Vert 1 \Vert_{\mathcal H(K)}}$ and $\mathcal H(1) \cap \mathcal H(K) \neq \{0\}$, you can choose $e=1$ again and you get $\eta(x,y) = K(x,y) - \frac{1}{\Vert 1 \Vert_{\mathcal H(K)}} \geq 0$ for all $x,y \in E$.

I got the great problem, that there are very few elements in $\mathcal H$ for that I can calculate $\eta$ directly. And all my results depend on the positivity of $K$.


The answer is no, and a simple counterexample can be obtained by taking, say, $E = \{1,\dots,n\}$, where $n \ge 3$, and $K(x,y) = C (\delta_{xy} - \frac{1}{n})$, where $C > n$. In fact, what we will show is that no rank one perturbation of $1 + K$ is positive pointwise.

Normalizing your $\xi$ to have norm one (which we can do, since the statement only depends on $\ker \xi$, and the condition $\langle \xi, 1 \rangle = 1$ won't play any role), we can represent $\xi$ as a vector in our RKHS, i.e. a function. Since $\Vert \xi \Vert^2 = 1$, the kernel $\xi \otimes \xi$ is exactly the "complement" to $\eta$, i.e. we may write the condition as follows:

$1 + K(x,y) = \xi(x) \xi(y) + \eta(x,y)$

where $\eta \ge 0$, so in particular, $\xi(x) \xi(y) \le 1 + K(x,y)$.

For $x \neq y$ we have $K(x,y) = -C/n < -1$, so we get $\xi(x) \xi(y) < 0$, so all $\xi(x)$ have different signs, which is impossible as long as $n \ge 3$.

  • $\begingroup$ Thanks, that helps me very much! Can I cite your counterexample in my thesis? Further is there any practical use for this kernel? $\endgroup$ – 3Matrolod Jun 17 '16 at 19:03
  • $\begingroup$ And are you sure that $K$ is a positive type function? $\endgroup$ – 3Matrolod Jun 17 '16 at 19:26
  • $\begingroup$ It's not hard to check that this kernel is the reproducing kernel of the space of functions on $\{1,\dots,n\}$, such that $\sum_x f(x) = 0$, with the standard inner product $\langle f,g \rangle = \operatorname{const} \cdot \sum_x f(x) g(x)$, so it's certainly of positive type. I don't claim any credit, feel free to use it wherever you like. $\endgroup$ – Alexander Shamov Jun 17 '16 at 19:43

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