Let $f:\mathbb R^n\to\mathbb R$ be a smooth function. Let $k>0$ and consider the cutoff function $T_k:\mathbb R\to\mathbb R$ defined for any $t\in\mathbb R$ as$$T_k(t)=\begin{cases}t-k & t>k\\t+k & t<-k\\0 & |t|\le k\\\end{cases}.$$As an exercise (calc II), I need to find a lower bound for the double integral$$\iint_{\mathbb R^{2n}} (f(x)-f(y)) \big(T_k(f)(x)-T_k(f)(y)\big) dx dy.$$
The lecturer suggested (because I think it will be useful in the proof of a subsequent theorem) to show that$$\iint_{\mathbb R^{2n}} \big|T_k(f)(x)-T_k(f)(y)\big|^2 dx dy\le \iint_{\mathbb R^{2n}} (f(x)-f(y)) \big(T_k(f)(x)-T_k(f)(y)\big) dx dy.$$The inequality sounds reasonable, but I can not prove it.
I started considering the integral on RHS and I write $\mathbb R^{2n}$ in this way:$$\mathbb R^{2n}=\{ x: |f(x)|\le k, y\in\mathbb R^n\} \cup \{ x: |f(x)|> k, y\in\mathbb R^n\}$$and considering all the possible cases.
The computations is a mess, but I found, for example, that in the set$$\{ x: |f(x)|> k, y: |f(y)|> k\},$$it is$$(f(x)-f(y)) \big(T_k(f)(x)-T_k(f)(y)\big)= 4|f(x)-f(y)|^2.$$This suggests that to get a lower bound, I can forget about $\{ x: |f(x)|> k, y: |f(y)|> k\}$. Nonetheless, I can not get the desired inequality.
Anyone could help with that? Is there any "faster" way to get it?