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A function $f: \mathbb R\to \mathbb C$ is of positif type if, for all $t_1, \dots, t_n \in \mathbb R$ and for all $a_1, \dots, a_n \in \mathbb C$ we have: $$\sum_{i,j=1}^{n}f(t_i -t_j) a_i \overline{a}_j\geq 0.$$ I would like to prove that the $f(x,y)=e^{-\frac{|x-y|}{2}}$ is of positif type on $\mathbb R\times \mathbb R$.

Thank you in advance

Z. Alfata
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    Then you have to define "positive type" for functions on $\mathbb R^2$. – Mundron Schmidt Mar 28 '18 at 19:17
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    As pointed out by Mundron Schmidt, you only defined "positive type" for functions $\mathbb{R} \to \mathbb{C}$ but now you are asking positive type on $\mathbb{R}^2$. I guess that what you have in your mind is the condition $$ \sum_{i,j=1}^{n} f(t_i, t_j) a_i \bar{a}_j \geq 0 $$ for all $t_i$'s in $\mathbb{R}$ and $a_i$'s in $\mathbb{C}$, but definitely it is up to you to clarify this. – Sangchul Lee Mar 29 '18 at 00:55

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This answer is based on the assumption that the definition of "positive type" for a function $f : \mathbb{R} \times \mathbb{R} \to \mathbb{C}$ is as follows:

$$ \sum_{i,j=1}^{n} f(t_i, t_j) a_i \bar{a}_j \geq 0 $$

for all $t_i$'s in $\mathbb{R}$ and $a_i$'s in $\mathbb{C}$.

Notice that, for $\alpha \geq 0$ and $\mathrm{i}=\sqrt{-1}$, we have

$$ \forall t \in \mathbb{R} \ : \quad e^{-\alpha|t|} = \frac{1}{\pi} \int_{\mathbb{R}} \frac{e^{\mathrm{i}\alpha t x}}{1+x^2} \, dx $$

(See this answer, for instance.) Then

\begin{align*} \sum_{i,j = 1}^{n} e^{-\alpha |t_i - t_j|} a_i \bar{a}_j &= \frac{1}{\pi} \int_{\mathbb{R}} \frac{1}{1+x^2} \left( \sum_{i,j = 1}^{n} e^{\mathrm{i}\alpha (t_i - t_j) x} a_i \bar{a}_j \right) \, dx \\ &= \frac{1}{\pi} \int_{\mathbb{R}} \frac{1}{1+x^2} \left| \sum_{j=1}^{n} e^{\mathrm{i}\alpha t_j x} a_j \right|^2 \, dx \\ &\geq 0. \end{align*}

Alternatively, this follows by noting that the function $f(s, t) = e^{-\alpha|s-t|}$ can be realized as the covariance kernel of an Ornstein-Uhlenbeck process.

Sangchul Lee
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