Does $\int f_1\le\int f_2$ imply $\sup_g\int f_1g\le\sup_g\int f_2g$?

Question

Let $(E,\mathcal E,\mu)$ be a finite measure space and $f_i:E^2\to[0,\infty)$ be $\mathcal E^{\otimes2}$-measurable with $$\int f_1\:{\rm d}\mu^{\otimes2}\le\int f_2\:{\rm d}\mu^{\otimes2}\tag1.$$ I would like to conclude that $$\sup_{\substack{g\in\mathcal L^2(\mu)\\g\ge0\\\int g\:{\rm d}\mu=0\\\int g^2\:{\rm d}\mu\le1}}\int\mu^{\otimes2}({\rm d}(x,y))f_1(x,y)|g(x)-g(y)|^2\le\sup_{\substack{g\in\mathcal L^2(\mu)\\g\ge0\\\int g\:{\rm d}\mu=0\\\int g^2\:{\rm d}\mu\le1}}\int\mu^{\otimes2}({\rm d}(x,y))f_2(x,y)|g(x)-g(y)|^2\tag2$$ Is that possible or is there a counterexample? The Cauchy-Schwarz inequality is not enough to answer this question. So, what can we do?