$\textbf{Definition:}$ 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?

  • $\begingroup$ What is the relationship between $\xi_i$ and $x_i$? $\endgroup$ Feb 5, 2014 at 8:48
  • $\begingroup$ its mentioned in definition. there is not any relationship between these two. $x_i\in(a,b)\subset\mathbb{R}$ and $\xi_i\in\mathbb{R}$. for all $1\leq i\leq n$. $\endgroup$ Feb 5, 2014 at 9:02
  • $\begingroup$ Okay, your phrasing is confusing because you refer to "all choices of $\xi_i$… such that" something is true about the $x_i$. $\endgroup$ Feb 5, 2014 at 9:05
  • $\begingroup$ In any case, something has to be wrong or missing here. In (ii) of the Proposition, $(x_i + x_j)/2$ isn't necessarily in the domain of $h$. $\endgroup$ Feb 5, 2014 at 9:07
  • $\begingroup$ Yes, it was mistyping. Its fine now. $\endgroup$ Feb 5, 2014 at 9:14

1 Answer 1


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 $$\sum_{i,j=1}^n \xi_i \xi_j h\left(x_i+x_j\right)=\sum_{i,j=1}^n \xi_i \xi_j h\left(\frac{y_i+y_j}{2}\right)\ge 0.$$

  • $\begingroup$ yes; actually both versions are the same; just a notational confusion some authors on this topic have caused, I think. $\endgroup$
    – Suvrit
    Feb 5, 2014 at 15:02

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