Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I quote the proof here from Applebaum's Lévy Processes and stochastic calculus (and the things before it to present the full picture)

We say $\phi:\mathbb{R}^d\rightarrow\mathbb{C}$ is conditionally positive definite if for all $n\in\mathbb{N}$ and $c_1,\ldots,c_n\in\mathbb{C}$ for which $\sum^n_{j=1}c_j=0$ we have $\sum^n_{j,k=1}c_j\bar{c_k}\phi(u_j-u_k) \geq 0$ for all $u_1, ... u_n\in\mathbb{R}^d$. The mapping $\phi:\mathbb{R}^d\rightarrow\mathbb{C}$ is said to be hermitian if $\bar{\phi(u)}=\phi(-u)$ for all $u\in\mathbb{R}^d$

Theorem (Schoenberg correspondence). The mapping $\phi:\mathbb{R}^d \to \mathbb{C}$ is hermitian and conditionally positive definite if and only if $e^{t\phi}$ is positive definite for each $t>0$.

Proof. We give the easy part here.

Suppose that $e^{t\phi}$ is positive definite for all $t>0$. Fix $n\in\mathbb{N}$ as above and choose $c_1,\ldots,c_n$ and $u_1,\ldots,u_n$ as above. We then find that, for each $t>0$. $$ \frac{1}{t}\sum^{n}_{j,k=1}c_j\bar{c}_k[e^{t\phi(u_j-u_k)}-1]\geq 0 $$ and so $$ \sum^{n}_{j,k=1} c_j\bar{c}_k\phi(u_j-u_k)=\lim\limits_{t\rightarrow 0}\frac{1}{t}\sum^{n}_{j,k=1}c_j\bar{c}_k[e^{t\phi(u_j-u_k)}-1]\geq 0. $$

My question is how do you arrive at the first equation? $\sum^{n}_{j,k=1}c_j\bar{c}_ke^{t\phi(u_j-u_k)}\geq 0$ I accept because it is just the property of positive definiteness for a characteristic function, but I cannot see why this assertion is true.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

$\sum_{j,k=1}c_j\bar{c}_k=\sum_{k=1}^n\bar{c}_k\sum_{j=1}^nc_j=0$ since $\sum_{j=1}^nc_j=0$ by assumption.

share|improve this answer
    
thank you. I feel ridiculously stupid right now. –  Lost1 Oct 18 '12 at 18:13
    
@Yufan: nah, these high brow theorems play tricks on anyone's mind. –  Alex R. Oct 18 '12 at 18:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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