A function $f: \mathbb R \to \mathbb R$ is called totally positive if for every $N \in \mathbb N$, every $a_1< a_2< \dotsb < a_N \in \mathbb R$ and every $b_1 < b_2 < \dotsb < b_N \in \mathbb R$ we have $$ \det [f(a_j-b_k)]_{j,k} \geq 0.\tag{*}\label{star} $$ Assuming further that $f$ is integrable yields a so-called Pólya frequency function. I was wondering if there exists a study of functions where \eqref{star} is replaced by the different assumption that $$ \det [f(a_j-b_k)]_{j,k} \neq 0 $$ for all $N$, all $a_j$ and all $b_j$. Do such functions have a special name or does there exist literatur on them?
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1$\begingroup$ The statement is not clear at all. I do not see the role of index $k$, as if the entries were constant on each row. Is the matrix considered $(f(a_j-b_k)]_{j,k}$? But in this case, I am also surprised to find no assumptions on $a_1,\ldots,a_N,b_1,\ldots,b_N$, since exchanging $a_1$ and $a_2$ would have the effect to exchange rows $1$ and $2$ and to change the detetminant into its opposite. $\endgroup$– Christophe LeuridanCommented Jan 19, 2023 at 21:46
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$\begingroup$ I fixed the statement. There were some typos. $\endgroup$– J. SwailCommented Jan 19, 2023 at 21:54
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$\begingroup$ You can't ask for the determinant to be nonzero for ALL $a_j$ and $b_j$ since, if $a_i = a_j$ or $b_i = b_j$, there are two identical columns. So you have to impose that the determinant is nonzero for $a_i$ and $b_i$ distinct. In this case, by continuity, there is some $\epsilon_n \in \{ \pm 1 \}$ such that $\epsilon_n \det(f(a_j - b_k)) >0$ for all $a_1 < a_2 < \cdots < a_n$ and $b_1 < b_2 < \cdots < b_n$. The standard case is that $\epsilon_n = 1$ for all $n$. One could ask whether there are any other sequences of signs $\epsilon_j$ for which there is a reasonable theory. $\endgroup$– David E SpeyerCommented Jan 20, 2023 at 19:34
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$\begingroup$ Basic observations: Replacing $f(x)$ with $-f(x)$ replaces $\epsilon_n$ by $(-1)^n \epsilon_n$; replacing $f(x)$ by $f(-x)$ replaces $\epsilon_n$ by $(-1)^{\binom{n}{2}} \epsilon_n$. $\endgroup$– David E SpeyerCommented Jan 20, 2023 at 19:36
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$\begingroup$ Have a look at Karlin's Total positivity. The functions you are interested in are called "sign-regular" there. $\endgroup$– Mateusz KwaśnickiCommented Jan 23, 2023 at 23:58
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