Let $f$ be a nonnegative function on the $d$-dimensional torus $\mathbb{T}^d$, which you can take to be smooth. Let $\bar{f}:=\int_{\mathbb{T}^d}fdx$. I am interested in whether the following positivity inequality is true: for $1\leq q<\infty$ and $0\leq \alpha\leq 2$, $$\int_{\mathbb{T}^d}|f(x)-\bar f|^{q-2} (f(x)-\bar f)f(x)(-\Delta)^{\alpha/2}(f-\bar f)(x)dx\geq 0, \tag{1}$$ wgere $(-\Delta)^{\alpha/2}$ is the fractional Laplacian of order $\alpha$, that is the Fourier multiplier with symbol $(2\pi |\xi|)^{\alpha}$.
If $\alpha=0$, then (1) is true since the integrand becomes $$|f(x)-\bar f|^{q-2}(f(x)-\bar f) f(x) (f(x)-\bar f) = |f(x)-\bar f|^q f(x)\geq 0,$$ by assumption that $f\geq 0$. If $\alpha=2$ (the other local case), then it's not clear to me (1) is true. Integrating by parts
\begin{multline} \int_{\mathbb{T}^d}|f(x)-\bar f|^{q-2} (f(x)-\bar f)f(x)(-\Delta)(f-\bar f)(x)dx \\\\ =(q-1)\int_{\mathbb{T}^d} |f(x)-\bar f|^{q-2}f(x)|\nabla f(x)|^2 dx + \int_{\mathbb{T}^{d}} |f(x)-\bar f|^{q-2}(f(x)-\bar f) |\nabla f(x)|^2 dx\\\\ = \int_{\mathbb{T}^d}|f(x)-\bar f|^{q-2}(qf(x)-\bar f)|\nabla f(x)|^2dx. \end{multline}
Does inequality (1) hold true when $\alpha \in (0,2)$?
Note that if I replace $\bar f$ by zero, then (1) is true for $\alpha\in (0,2]$ by the Cordoba-Cordoba inequality (a convexity inequality) for the fractional Laplacian.
