Let
- $c_0>0$ and $\ell\in[0,1]$;
- $(E,\mathcal E,\lambda)$ be a measure space;
- $\mu$ be a probability measure on $(E,\mathcal E)$;
- $p:E\to[0,\infty)$ be $\mathcal E$-measurable with $$p_\lambda:=\int p\:{\rm d}\lambda>0$$;
- $f\in L^2(\pi)$ and $$\mu f:=\int f\:{\rm d}\mu.$$
Are we able to show that $$\Gamma(f):=c_0\langle f,f-\mu f\rangle_{L^2(\mu)}-\frac\ell{2p_\lambda}\int\lambda({\rm d}x)\int\mu({\rm d}y)\min(p(x),p(y))|f(x)-f(y)|^2$$ is nonnegative?
If this is too hard or not possible to show in general, I'm especially interested in the case where $E=[0,1)^d$ for some $d\in\mathbb N$, $\mathcal E$ is the Borel $\sigma$-algebra on $E$, $\lambda$ is the restriction of the $d$-dimensional Lebesgue measure on $E$ and $\mu$ is the uniform distribution on $E$. In particular, $\lambda=\mu$. I would also be willing to assume that $c_0\ge1$ or even $c_0=1$.
The right-hand side is clearly bounded below by $$c_0\langle f,f-\mu f\rangle_{L^2(\mu)}-\frac\ell2\int\pi({\rm d}x)\int\mu({\rm d}y)|f(x)-f(y)|^2,\tag1$$ where $\pi$ is the measure with density $\frac p{p_\lambda}$ with respect to $\lambda$. However, I also wasn't able to show that $(1)$ is nonnegative so far.