We say a positive symmetric $n\times n$ matrix $M$ over $\mathbb{R}^n$ is semi-definite if $v^{\intercal}Mv\geq 0$ for all nonzero $v\in\mathbb{R}^n$.

We say a function $f:\mathbb{T}^2\longrightarrow\mathbb{R}$ to be positive semi-definite if $\Big(f(t\_k, t\_j)\Big)\_{k,j=1}^n$ is a positive semi-definite matrix for all $(t\_k)\_{k=1}^n\in\mathbb{T}^n$

With this definition, I am working on an exercise asking me to show

> 
> $(1)$ the function $f:(s,t)\mapsto t\wedge s$ defined for $s,t\geq 0$ is positive semi-definite;
> 
> 
> $(2)$ the function $c:(s,t)\mapsto e^{-|t-s|}$ is positive semi-definite.
> 
> 
> 

For the first one, I tried to use the fact that $$\int\_{\mathbb{R}^{+}}\mathbb{1}\_{[0,t]}\mathbb{1}\_{[0,s]} \,d\mu = t\wedge s,$$ so that each term in the matrix is of the form $b\_{i,j}:=f(t\_i,t\_j)=\int\_{\mathbb{R}^{+}}\mathbb{1}\_{[0,t\_{j}]} \mathbb{1}\_{[0,t\_{}} \, d\mu$, for $t,j=1,\cdots,n$.

Let $v=(v\_1,\ldots, v\_n)\in\mathbb{R}^n$, then $$v^{\intercal} Mv=a\_1 \sum\_{i=1}^n a\_i b\_{i,1}+a\_2\sum\_{i=1}^n a\_i b\_{i,2} + a\_3 \sum\_{i=1}^n a\_i b\_{i,3}+\cdots+a\_n \sum\_{i=1}^n a\_i b\_{i,n}.$$

**But then I don't know what to do next..**

For the second one, the exercise gives an hint: using an auxiliary Hilbert space $H$ and a $h\_t\in H$ such that $\langle h\_t, h\_s\rangle=c(s,t)\ldots$ I don't really know how to use this hint...

I really need an answer with some details, since this is an exercise in Stochastic Process, instead of functional analysis and so forth, so I don't have enough background of this...

Thank you so much!

---

**Edit 1: (Proof of the first one)**

Following **MaoWao**'s suggestion, I think I proved the first one.

Firstly let me claim that if $H$ is a Hilbert space, then its corresponding inner product $\langle\cdot, \cdot\rangle\_H:H\times H\longrightarrow\mathbb{R}$ is positive semi-definite. 

Indeed, we have for any $n\in\mathbb{N}$, $x\_1,\ldots, x\_n \in H$ and $c\_1,\ldots, c\_n \in\mathbb{R}$ that $$\sum\_{i,j=1}^n c\_i c\_j \langle x\_i,x\_j\rangle\_H = \left<\sum\_{i=1}^n c\_i x\_i,\sum\_{j=1}^n c\_j x\_j \right>\_H =\Big\|\sum\_{i=1}^n c\_i x\_i\Big\|\_H^2\geq 0.$$

In fact, the above result also holds for pre-Hilbert space, since the notion of completeness was not involved in the above argument. 

Thus, we only need to find a specific (pre-)Hilbert Space $H$ and a $h\_{t}\in H$ such that $\langle h\_t, h\_s\rangle\_H = t\wedge s$.

But this is easy, let's consider $H:=L^2(\mathbb{R}\_{+})$, and $h\_t:=\mathbb{1}\_{[0,t]}$. It is clear that $h\_t\in H$, and for any $t,s\geq 0$, we have $$\langle h\_t, h\_s\rangle\_{L^2(\mathbb{R}\_{+})} = \int\_{\mathbb{R}\_{+}}\mathbb{1}\_{[0,t]} \mathbb{1}\_{[0,s]} \, d\mu=t\wedge s,$$ and thus we are done. 

---

**Edit 2: (Proof of the second one)**

I've searched all over the places. The function in $(2)$ is Abel kernel, but it is rarely discussed since Abel kernel is closely related to Poisson kernel, and most of the discussions are on the latter. 

Later, I found a really close one: the Gaussian kernel and here is a link about the proof of Gaussian kernel is really a kernel. That is, it is positive semidefinite. 

<https://stats.stackexchange.com/questions/35634/how-to-prove-that-the-radial-basis-function-is-a-kernel>

In this link, one answer used the characteristic function. This greatly inspired me. I also found a characteristic function, which is the one for Cauchy distribution. 

Below is the proof:

Recall the Cauchy Distribution $(x\_{0},\gamma)$ with $\gamma>0$ has the characteristic function $$\varphi(t)=e^{ix\_{0}t-\gamma|t|}.$$

Using this, we can write $c(s,t)=h(s-t)$ where $h(t):=e^{-|t|}=\mathbb{E}e^{itZ}$ is the characteristic function of a random variable $Z$ with Cauchy $(0,1)$ distribution.

Then for real numbers $x\_{1},\cdots, x\_{n}$ and $a\_{1},\cdots, a\_{n}$, we have 
\begin{align\*}
\sum\_{j,k=1}^{n}a\_{j}a\_{k}h(x\_{j}-x\_{k})&=\sum\_{j,k=1}^{n}a\_{j}a\_{k}\mathbb{E}e^{i(x\_{j}-x\_{k})Z}\\
&=\mathbb{E}\Big(\sum\_{j,k=1}^{n}a\_{j}e^{ix\_{j}Z}a\_{k}e^{-ix\_{k}Z}\Big)\\
&=\mathbb{E}\Big(\Big|\sum\_{j=1}^{n}a\_{j}e^{ix\_{j}Z}\Big|^{2}\Big)\geq 0.
\end{align\*}

Thus, $c$ is positive semi-definite.

**It seems that I did not use the hint at all for part $(2)$, so I believe there must be another way.**

**I do need someone to check if my proof in the edit 1 and edit 2 is correct. I am gonna open a bounty in 19 hours later, for proof checking and possible new proof. Thank you!**
['https://math.stackexchange.com/questions/3550591', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/502975/']
As **MaoWao** suggested, I am gonna post my own proof here.

---

*Proof of (1):*

Firstly let me claim that if $H$ is a Hilbert space, then its corresponding inner product $\langle \cdot, \cdot\rangle\_{H}:H\times H\longrightarrow\mathbb{R}$ is positive semi-definite. 

Indeed, we have for any $n\in\mathbb{N}$, $x\_{1},\cdots, x\_{n}\in H$ and $c\_{1},\cdots, c\_{n}\in\mathbb{R}$ that $$\sum\_{i,j=1}^{n}c\_{i}c\_{j}\langle x\_{i}, x\_{j}\rangle\_{H}=\Big<\sum\_{i=1}^{n}c\_{i}x\_{i},\sum\_{j=1}^{n}c\_{j}x\_{j}\Big>\_{H}=\Big\|\sum\_{i=1}^{n}c\_{i}x\_{i}\Big\|\_{H}^{2}\geq 0.$$

In fact, the above result also holds for pre-Hilbert space, since the notion of completeness was not involved in the above argument. 

Thus, we only need to find a specific (pre-)Hilbert Space $H$ and a $h\_{t}\in H$ such that $\langle h\_{t}, h\_{s}\rangle\_{H}=t\wedge s$.

But this is immediate. Let's consider $H:=L^{2}(\mathbb{R}\_{+})$, and $h\_{t}:=\mathbb{1}\_{[0,t]}$. It is clear that $h\_{t}\in H$, and for any $t,s\geq 0$, we have $$\langle h\_{t}, h\_{s}\rangle\_{L^{2}(\mathbb{R}\_{+})}=\int\_{\mathbb{R}\_{+}}\mathbb{1}\_{[0,t]}\mathbb{1}\_{[0,s]}d\mu=t\wedge s,$$ and thus we are done. 

---

*Proof of $(2)$:*

Recall the Cauchy Distribution $(x\_{0},\gamma)$ with $\gamma>0$ has the characteristic function $$\varphi(t)=e^{ix\_{0}t-\gamma|t|}.$$

Using this, we can write $c(s,t)=h(s-t)$ where $h(t):=e^{-|t|}=\mathbb{E}e^{itZ}$ is the characteristic function of a random variable $Z$ with Cauchy $(0,1)$ distribution.

Then for real numbers $x\_{1},\cdots, x\_{n}$ and $a\_{1},\cdots, a\_{n}$, we have 
\begin{align\*}
\sum\_{j,k=1}^{n}a\_{j}a\_{k}h(x\_{j}-x\_{k})&=\sum\_{j,k=1}^{n}a\_{j}a\_{k}\mathbb{E}e^{i(x\_{j}-x\_{k})Z}\\
&=\mathbb{E}\Big(\sum\_{j,k=1}^{n}a\_{j}e^{ix\_{j}Z}a\_{k}e^{-ix\_{k}Z}\Big)\\
&=\mathbb{E}\Big(\Big|\sum\_{j=1}^{n}a\_{j}e^{ix\_{j}Z}\Big|^{2}\Big)\geq 0.
\end{align\*}

Thus, $c$ is positive semi-definite.

---

Also, as **MaoWao** pointed out, the proof of $(2)$ can be equivalently written into something similar to what we did in the $(1)$, by taking the Hilbert space $L^{2}(\mathbb{P})$ and $h\_{t}=\exp(it\cdot)$

I am really grateful to the help, suggestion and proof verification from **MaoWao**. I'd like also to express my appreciation to **Michael Hardy**, for the discussion and help. 

Thank you guys :)
This will be a partial answer.

The standard Wiener process $\{B(t)\}\_{t\,\ge\,0},$ also called the standard Brownian motion, assigns to each $t\ge0$ a random variable $B(t)$ in such a way that every increment $B(t)-B(s)$ for $0\le s\le t$ is distributed as $\operatorname N(0,t-s)$ (the standard normal distribution with expected value $0$ and variance $t-s,$ so standard deviation $\sqrt{t-s},$ and for **pairwise disjoint** intervals $(s\_i,t\_i),$ $i=1,\ldots,n,$ with $0\le s\_i\le t\_i,$ the increments $B(t\_i) - B(s\_i)$ are **independent** random variables.

Then for $0\le s\le t$ one can show that
$$
\operatorname{cov}(B(s), B(t)) = s = s\wedge t.
$$
That is done by writing
$$
\operatorname{cov}\Big(B(s), \Big(B(t) - B(s)\Big) + B(s)\Big)
$$
and then using the independence of $B(t) - B(s)$ and $B(s).$

Since all the entries in the matrix of covariances of $B(t\_1),\ldots, B(t\_n)$ are of the form $t\_i\wedge t\_j,$ and since every matrix of covariances is positive-semi-definite, the function $(s,t)\mapsto s\wedge t$ must be positive-semi-definite.