In Hilbert space theory, one usually defines positive operators $T:H\rightarrow H$ to be \begin{align*} \left<Tf,f\right>\geq 0,~~~~f\in H. \end{align*} In Riesz space theory, or certain area operator theory, positive operators $T$ are such that \begin{align*} T(f)\geq 0,~~~~f\geq 0. \end{align*} Of course, a space need no to have the order property, for $f\geq 0$ needs to be defined properly.
Consider now $H=\mathcal{S}(\mathbb{R}^{n})$, the Schwarz class, but we restrict to those of real-valued, and a pairing $\left<\cdot,\cdot\right>:H\times H\rightarrow\mathbb{R}$ to be the canonical integral representation \begin{align*} \left<f,g\right>=\int_{\mathbb{R}^{n}}f(x)g(x)dx,~~~~f,g\in\mathcal{S}(\mathbb{R}^{n}). \end{align*} We define $f\geq 0$ to be the canonical meaning that $f(x)\geq 0$ for all $x\in\mathbb{R}^{n}$.
Let $T:H\rightarrow H$ be such that \begin{align*} T(f)(x)=\int_{\mathbb{R}^{n}}K(x,y)f(y)dy, \end{align*} where $K$ is some measurable function and assume that $K$ is sufficient regular to have $Tf\in H$, is there any necessary and sufficient condition such that \begin{align*} \left<Tf,f\right>\geq 0,~~~~f\in H~~~~\text{if and only if}~~~~T(f)\geq 0,~~~~f\geq 0. \end{align*} An uninteresting sufficient condition would be $K(x,y)\geq 0$ a.e., but I wonder if anything can be said more about it?
A general and rough question would be, when do the Hilbertian-positivity and the Rieszian-positivity coincide?
If $K$ is given by $K(x,y)=k(x-y)$, where $k\in L^{1}$, then we can write $Tf=k\ast f$ and hence $T\varphi_{\epsilon}(x)=k\ast\varphi_{\epsilon}(x)\rightarrow k(x)$ a.e. for a standard mollifier $(\varphi_{\epsilon})$, where we choose $\varphi\geq 0$.
If $\left<Tf,f\right>\geq 0$ for $f\in H$, then $K(x,y)\geq 0$ a.e., so a necessary and sufficient condition is obtained.
But for general kernel $K$ of non-convolution type, I cannot figure out anything to say about it.