prove negative definiteness of $\mathbf{1}_{(0, \infty)}(x)$

Question

I would like to show the negative definiteness of $\mathbf{1}_{(0, \infty)}(x)$ with certain condition.

Let $h\left(x_{i}, x_{j}\right)=\mathbf{1}_{(0, \infty)}\left(|x_{i}-x_{j}|\right)$.

if $c_{1}+\cdots+c_{n}=0,c_{i}\in \mathbb{R} $

then $\sum_{i, j=1}^{n} c_{i} c_{j} h\left(x_{i}, x_{j}\right) \leq 0$.

I tried to count the summation in many ways (e.g. dividing summation in $\sum_{i, j=1}^{n}=\sum_{i, j=1, x_{i}=x_{j}}+\sum_{i, j=1, x_{i}\neq x_{j}}$) but couldn't solve this and started to think that there might be a
counterexample. I reckon this problem is linked to a variogram of nugget effect model. thank you very much in advance.

Answer 1

Without loss of generality, assume $X=\mathbb{R}$. We prove that

for any $n\in\mathbb{B}$, distinct points $x_1,\ldots,x_n\in \mathbb{R}$ and real (or complex) numbers $c_1,\ldots,c_n$ with $\sum^n_{j=1}c_j=0$, $$\sum_{1\leq j,k\leq n}c_j\overline{c_k}h(x_j,x_k)\leq0$$

For $n=1$ the statement is trivial. Suppose $n\geq2$ and let $x_1,\ldots,x_n$ and $c_1,\ldots,c_n$ as in the statement above. Define the discrete measure $\mu$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ as $$\mu(dx)=\sum^n_{j=1}\delta_{x_j}(dx)$$ where $\delta_{x_j}$ is the unit mass measure at $x_j$, and for each $j$, define $$f=\sum^n_{j=1}c_j\mathbb{1}_{\{x_j\}}$$ Since $m=\sum^n_{j=1}c_j=0$, $f=f-m=\sum^n_{j=1}c_j(\mathbb{1}_{\{x_j\}}-1)$ and so, $$\begin{align} \sum^n_{j=1}|c_j|^2 &=\int|f|^2\,d\mu=\int|f-m|^2\,d\mu=\int\sum_{1\leq j,k\leq n}c_j\overline{c_k}(1-\mathbb{1}_{\{x_j\}})(1-\mathbb{1}_{\{x_k\}})\,d\mu\\ &=\sum_{1\leq j,k\leq n}c_j\overline{c_k}\int(1-\mathbb{1}_{\{x_j\}})(1-\mathbb{1}_{\{x_k\}})\,d\mu\\ \end{align} $$

If $j=k$, $$\int(1-\mathbb{1}_{\{x_j\}})\,d\mu=(n-1)+(n-2)h(x_j,x_k)$$ whereas if $j\neq k$, $$\int(1-\mathbb{1}_{\{x_j\}})(1-\mathbb{1}_{\{x_k\}})\,d\mu=(n-2)h(x_j,x_k)$$

Hence $$\begin{align} \sum_{1\leq j,k\leq n}c_j\overline{c_k}\int(1-\mathbb{1}_{\{x_j\}})(1-\mathbb{1}_{\{x_k\}})\,d\mu&=(n-2)\sum_{\stackrel{1\leq j,k\leq n}{j\neq k}}c_j\overline{c_k}h(x_j,x_k)\\ &\qquad\qquad +(n-1)\sum^n_{j=1}|c_j|^2\\ &=(n-2)\sum_{1\leq j,k\leq n}c_j\overline{c_k}h(x_j,x_k)\\ &\qquad\qquad+ (n-1)\sum^n_{j=1}|c_j|^2 \end{align}$$

Putting things together gives $$ (n-2)\sum^n_{1\leq j,k\leq n}c_j\overline{c_k}h(x_j,x_k)=-(n-2)\sum^n_{j=1}|c_j|^2\leq0 $$