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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.

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up vote 1 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.

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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

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