Relaxation of notion of positive definite function

Question

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?

Answer 1

$\newcommand\R{\mathbb R}$For real $c$ and $x$, let $$f_c(x):=f(x)-c,$$ where $f(x):=e^{x^2/2}$. Note that $f$ is the moment generating function of the standard normal distribution and thus a mixture of exponential functions. Since the exponential functions are positive semidefinite (in the semigroup sense), $f$ is also positive semidefinite. With some further effort, one should be able to show that $f$ is positive definite.

By Theorem 2.5 on p. 55, Theorem 5.3 on p. 65, and Theorem 8.1 on p. 78 of Karlin - Total positivity, vol. I, for $f_c$ to be $r$-positive semidefinite it is sufficient that the Hankel determinant $$d_{k,c}(x):=d_{f;k,c}(x):=\det((f_c^{(i+j)}(x))_{0\le i,j\le k-1})$$ be $>0$ for all $k\in[r]:=\{1,\dots,r\}$ and necessary that this determinant be $\ge0$ for all $k\in[r]$. Note that $$d_{k,c}(x)=d_{k,0}(x)-c\tilde d_k(x),$$ where $$\tilde d_k(x):=d_{f'';k-1,0}(x)=\det((f^{(i+j)}(x))_{1\le i,j\le k-1});$$ note also that, similarly to $d_{k,0}(x)\ge0$ for real $x$, we have $\tilde d_k(x)\ge0$ for real $x$ (and, likely, $\tilde d_k(x)>0$ for real $x$).

Let
$$c_k:=\sup\{c\colon d_{k,c}(x)\ge0\ \forall x\in\R\} =\inf_{x\in\R}\frac{d_{k,0}(x)}{\tilde d_k(x)}. $$ We have $$c_1=1>c_2=\frac{\sqrt e}2=0.82\ldots>c_3=\frac23=0.66\ldots.$$ Moreover, it appears that $c_k>c_{k+1}$ for all natural $k$.

It will then follow that $f_c$ is $r$-positive semidefinite but not $(r+1)$-positive semidefinite if $c_{r+1}<c<c_r$. (One may also note here that, for $f_c$ to be $r$-positive semidefinite but not $(r+1)$-positive semidefinite, it is necessary that $c_{r+1}\le c\le c_r$.)

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Remark: It appears that $$d_{k,0}(x)\overset{\text{(?)}}=(k-1)\\\$\,e^{k x^2/2},$$ where $n\\\$:=\prod_{j=1}^n j!$, the superfactorial. It also appears that $$\tilde d_k(x)\overset{\text{(?)}}=P_{k-1}(x^2)e^{(k-1)x^2/2},$$ where $P_{k-1}$ is a polynomial of degree $k-1$. I do not know much more about the polynomials $P_{k-1}$.

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The conjecture that $c_k>c_{k+1}$ for all natural $k$ and the above remark are illustrated by the following image of a Mathematica notebook (click on the image to magnify it):