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$\textbf{Definition:}$ 1. A function $h : (a,b)\rightarrow\mathbb{R}$ is exponentially convex if it is continuous and $$\sum _{i, j=1}^n\xi_i\xi_jh(x_i+x_j)\geq 0,$$ for all $n\in\mathbb{N}$ and all choices of $\xi_i,x_i\in\mathbb{R}$, $i = 1,\ldots ,n$, such that $x_i+x_j\in(a,b)$, $1 \leq i, j \leq n$.

$\textbf{Proposition:}$ Let $h:(a,b)\rightarrow\mathbb{R}$. The following propositions are equivalent.

(i) $h$ is exponentially convex.

(ii)$ h$ is continuous and $$\sum _{i, j=1}^n\xi_i\xi_jh(\frac{x_i+x_j}{2})\geq 0,$$ for all $n\in\mathbb{N}$ and all choices of $\xi_i,x_i\in\mathbb{R}$, $i = 1, \ldots,n$, such that $x_i\in(a,b)$, $1 \leq i \leq n$.

The proof of above stated proposition is shown in this thread. This proposition is further followed by a corollary.

$\textbf{Corollary:}$ If $h$ is exponentially convex then $$det\big[h\big(\frac{x_i+x_j}{2}\big)\big]_{i,j=1}^n\geq 0.$$ for all $n\in\mathbb{N}$ and $x_i\in(a,b)$, $1 \leq i \leq n$.

Can Someone help me to prove it?

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closed as off-topic by Suvrit, Chris Godsil, Ricardo Andrade, Karl Schwede, Stefan Waldmann Feb 11 '14 at 7:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Suvrit, Chris Godsil, Ricardo Andrade, Karl Schwede, Stefan Waldmann
If this question can be reworded to fit the rules in the help center, please edit the question.

up vote 2 down vote accepted

Hint: From the (ii), we will get the matrix $(h(\frac{x_i+x_j}{2}))_{i,j=1}^{n}$ is positive-definite matrix.

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yes but I am looking for a formal proof. – Shinning Star Feb 10 '14 at 4:59
@ShinningStar A matrix is positive is – gaoxinge Feb 10 '14 at 5:02
you mean positive definite? Do we have some results the determinant of a positive definite matrix is positive? If you can give me some reference? – Shinning Star Feb 10 '14 at 5:24
@ShinningStar You will find in the wiki about characterizations's forth statement which I have given to you. The whole theorem is that A matrix is positive-definite if and only if its leading principal minors are all positive. So back to your question, the matrix in your corollary is a special leading principle minor. – gaoxinge Feb 10 '14 at 5:36
Yes I got that. thank you ! :) – Shinning Star Feb 10 '14 at 5:43

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