$\textbf{Definition:}$ 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](https://mathoverflow.net/questions/156730/exponential-convexity). 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?



Exponential Convexity Results

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> **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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