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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.

2. 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 $x,y \in E$.

3. 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 $K$.

I would really appreciate remarks, ideas, sources and results to this topic. Thank you very much in advance!

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.

2. 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 $x,y \in E$.

3. 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 $K$.

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.

2. 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 $x,y \in E$.

3. 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 $\eta(x,y) \geq K(x,y) > 0$ 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 $K$.

I would really appreciate remarks, ideas, sources and results to this topic. Thank you very much in advance!

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.

2. 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 $x,y \in E$.

3. 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 $K$.

I would really appreciate remarks, ideas, sources and results to this topic. Thank you very much in advance!

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 $e$ 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.

2. 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 $x,y \in E$.

3. 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 $\eta(x,y) \geq K(x,y) > 0$ 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 $K$.

I would really appreciate remarks, ideas, sources and results to this topic. Thank you very much in advance!

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.

2. 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 $x,y \in E$.

3. 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 $\eta(x,y) \geq K(x,y) > 0$ 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 $K$.

I would really appreciate remarks, ideas, sources and results to this topic. Thank you very much in advance!