$\textbf{Definition:}$ 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 $x\_i+x\_j\in(a,b)$, $1 \leq i, j \leq n$.


There is a proposition in many research papers, but I can not find any clear proof of it.


$\textbf{Proposition:}$ 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 $x\_i\in(a,b)$, $1 \leq i \leq n$.


Can Someone give me its formal proof? 



Exponential Convexity
You just multiply or divide by 2, no?


To show that (i) implies (ii), take any $\xi\_i,x\_i\in\mathbb{R}$ such that $x\_i\in(a,b)$ for $i=1,\ldots,n$. Since the interval $(a,b)$ is convex, the midpoints $\frac{x\_i+x\_j}{2}$ are also all in $(a,b)$. Now set $y\_i=\frac{x\_i}{2}$ for $i=1,\ldots,n$. Then we have $y\_i+y\_j=\frac{x\_i+x\_j}{2}$ in $(a,b)$ for all $1\le i,j\le n$, so we can apply (i) and we get
$$\sum\_{i,j=1}^n \xi\_i \xi\_j h\left(\frac{x\_i+x\_j}{2}\right)=\sum\_{i,j=1}^n \xi\_i \xi\_j h\left(y\_i+y\_j\right)\ge 0.$$


For $(ii)\Rightarrow (i)$, let $\xi\_i,x\_i\in\mathbb{R}$ such that $x\_i+x\_j\in(a,b)$ for any $1\le i,j\le n$ and define $y\_i=2x\_i$ for $i=1\ldots,n$. The $y\_i$ are in $(a,b)$, since they are equal to $x\_i+x\_i$, so we can apply (ii), and we get
$$\sum\_{i,j=1}^n \xi\_i \xi\_j h\left(x\_i+x\_j\right)=\sum\_{i,j=1}^n \xi\_i \xi\_j h\left(\frac{y\_i+y\_j}{2}\right)\ge 0.$$


