Skip to main content
deleted 23 characters in body
Source Link
Hans
  • 3k
  • 13
  • 17

A function $f:\mathbb{R}\to\mathbb{R}$ is called positive definite (in the semigroup sense) if for all $n\geq 1$ and $x_1,\ldots,x_n\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^n$ is positive definite. Such functions have a nice characterization, see e.g. Thm.6.2 in by https://arxiv.org/pdf/1301.2449v3.pdf(a version of) Bochner's theorem.

Now I am interested in the following relaxation of this notion: Fix $k\in\mathbb{N}$. We say that a function $f:\mathbb{R}\to\mathbb{R}$ is $k$-positive if for all $x_1,\ldots,x_k\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^k$ is positive definite.

For instance $1$-positive means that $f(x)>0$ for all $x\in\mathbb{R}$. Being $2$-positive means additionally that $\log(f)$ is strictly midpoint convex.

In general, it is clear that $k$-positive definite implies $(k-1)$-positive definite, and the above example shows that the converse is not always true. My question is the following: Is it true that for every $k\in\mathbb{N}$ there is a $k$-positive function which is not $(k+1)$-positive?

A function $f:\mathbb{R}\to\mathbb{R}$ is called positive definite (in the semigroup sense) if for all $n\geq 1$ and $x_1,\ldots,x_n\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^n$ is positive definite. Such functions have a nice characterization, see e.g. Thm.6.2 in https://arxiv.org/pdf/1301.2449v3.pdf.

Now I am interested in the following relaxation of this notion: Fix $k\in\mathbb{N}$. We say that a function $f:\mathbb{R}\to\mathbb{R}$ is $k$-positive if for all $x_1,\ldots,x_k\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^k$ is positive definite.

For instance $1$-positive means that $f(x)>0$ for all $x\in\mathbb{R}$. Being $2$-positive means additionally that $\log(f)$ is strictly midpoint convex.

In general, it is clear that $k$-positive definite implies $(k-1)$-positive definite, and the above example shows that the converse is not always true. My question is the following: Is it true that for every $k\in\mathbb{N}$ there is a $k$-positive function which is not $(k+1)$-positive?

A function $f:\mathbb{R}\to\mathbb{R}$ is called positive definite (in the semigroup sense) if for all $n\geq 1$ and $x_1,\ldots,x_n\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^n$ is positive definite. Such functions have a nice characterization by (a version of) Bochner's theorem.

Now I am interested in the following relaxation of this notion: Fix $k\in\mathbb{N}$. We say that a function $f:\mathbb{R}\to\mathbb{R}$ is $k$-positive if for all $x_1,\ldots,x_k\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^k$ is positive definite.

For instance $1$-positive means that $f(x)>0$ for all $x\in\mathbb{R}$. Being $2$-positive means additionally that $\log(f)$ is strictly midpoint convex.

In general, it is clear that $k$-positive definite implies $(k-1)$-positive definite, and the above example shows that the converse is not always true. My question is the following: Is it true that for every $k\in\mathbb{N}$ there is a $k$-positive function which is not $(k+1)$-positive?

Source Link
Hans
  • 3k
  • 13
  • 17

Relaxation of notion of positive definite function

A function $f:\mathbb{R}\to\mathbb{R}$ is called positive definite (in the semigroup sense) if for all $n\geq 1$ and $x_1,\ldots,x_n\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^n$ is positive definite. Such functions have a nice characterization, see e.g. Thm.6.2 in https://arxiv.org/pdf/1301.2449v3.pdf.

Now I am interested in the following relaxation of this notion: Fix $k\in\mathbb{N}$. We say that a function $f:\mathbb{R}\to\mathbb{R}$ is $k$-positive if for all $x_1,\ldots,x_k\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^k$ is positive definite.

For instance $1$-positive means that $f(x)>0$ for all $x\in\mathbb{R}$. Being $2$-positive means additionally that $\log(f)$ is strictly midpoint convex.

In general, it is clear that $k$-positive definite implies $(k-1)$-positive definite, and the above example shows that the converse is not always true. My question is the following: Is it true that for every $k\in\mathbb{N}$ there is a $k$-positive function which is not $(k+1)$-positive?