Let $p:\mathbb{R}\rightarrow \mathbb{R}$ be a piecewise continous, nonnegative function with compath support. Define $K:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{C}$ by $$K(x,y)=\int\limits_{-\infty}^{\infty}p(t)e^{i(x-y)t}\mathrm{d}t$$.
I want to show that this is a kernel function. Therefore I have to proof that for every $n \in \mathbb{N}$ and all sets $\{x_{1},\ldots,x_{n}\}\subset \mathbb{R}$, $\{\alpha_{1},\ldots,\alpha_{n}\}\subset \mathbb{C}$ $$\sum\limits_{i,j}\alpha_{i}\overline{\alpha_{j}}K(x_{i},x_{j})\geq 0$$ is satisfied.
My first idea is the following:
$$\sum\limits_{k,l}\alpha_{k}\overline{\alpha_{l}}K(x_{k},x_{l}) = \sum\limits_{k,l}\alpha_{k}\overline{\alpha_{l}}\int\limits_{-\infty}^{\infty}p(t)e^{i(x_{k}-x_{l})t}\mathrm{d}t\\ = \int\limits_{-\infty}^{\infty} p(t) \sum\limits_{k,l} \alpha_{k}\overline{\alpha_{l}}e^{itx_{k}}\overline{e^{itx_{l}}} \mathrm{d}t\\ = \int\limits_{-\infty}^{\infty} p(t) \left\langle \sum\limits_{k} \alpha_{k}e^{itx_{k}},\sum\limits_{l} \alpha_{l}e^{itx_{l}}\right\rangle \mathrm{d}t\\ = \int\limits_{-\infty}^{\infty} p(t) \left|\left|\sum\limits_{k} \alpha_{k}e^{itx_{k}}\right|\right|^{2} \mathrm{d}t\geq0$$
Is this correct? Since I've only used that $p(t)\geq0$ for all $t\in\mathbb{R}$ I have the feeling I'm missing something.