$L^2$ norm of $f(x) - f(y)$

Question

Let $W(x,y) \in L^\infty(\Omega^2)$.

Is there a way to get a good lower bound on $\int_{\Omega^2} W(x,y) (f(x) - f(y))^2 dydx$ in terms of $\int_\Omega W(x,y) f(x)^2 dx$?

We can say \begin{align} \int_{\Omega^2} W(x,y) & (f(x) - f(y))^2 dydx \\ &= \int_{\Omega^2} W(x,y)\big( f(x)^2 - 2f(x)f(y) + f(y)^2 \big) dydx \\ &= 2\int_{\Omega^2} W(x,y) f(x)^2 dydx - 2\int_{\Omega^2} W(x,y)f(x) f(y) dydx \end{align}

and using Cauchy-Schwarz we know the first term $$ 2\int_{\Omega^2} W(x,y) f(x)^2 dydx \leq 2\|W\|_\infty \|f\|^2_2$$ and the second term $$2\int_{\Omega^2} W(x,y)f(x) f(y) dydx 2 \leq 2\|W\|_\infty \|f\|^2_2 \,.$$

I'm not sure if this problem reduces to asking whether we can bound $\|f(x) - f(y)\|^2_2 = \int_{\Omega^2} (f(x) - f(y))^2dydx$ using $\|f\|^2_2$, or how to do that either.

Answer 1

I don't think you can get a good lower bound in general: if $f$ is constant (and $\Omega$ is bounded, probably), then $$\int_{\Omega^2}W(x,y)(f(x)-f(y))^2dxdy=0.$$