1
$\begingroup$

$\textbf{Definition:}$ 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?

$\endgroup$
7
  • $\begingroup$ 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

2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.