I was reading this post , where the following question is discussed:
Let $h:[0,1] \rightarrow[0,1]$ be a $C^{1}$ function such that $h^{\prime}(x)<0$ for all $x \in(0,1)$. Then, $$ \inf _{f \in \mathcal{H}}\left(\int_{0}^{1} \int_{0}^{1} h(|x-y|)(f(x)-f(y))^{2} d x d y\right)<2 \int_{0}^{1} h(x) d x $$ where $\mathcal{H}=\left\{f \in L^{2}([0,1]): \int_{0}^{1} f(t) d t=0\right.$ and $\left.\int_{0}^{1} f^{2}(t) d t=1\right\}$
Someone in the answers proved this by explicitly finding an $f \in \mathcal{H}$. Their proof starts by arguing that $$ \int_{0}^{1} \int_{0}^{1} h(|x-y|)(f(x)-f(y))^{2} d x d y-2 \int_{0}^{1} h(x) d x=2 D_{f}(h) $$ where $$ D_{f}(h):=\int_{0}^{1} d x \int_{x}^{1} d y h(y-x)(f(x)-f(y))^{2}-\int_{0}^{1} h . $$
This raised a couple of questions for me, since I'm working on something similar:
why does this hold? $$D_{f}(h):=\int_{0}^{1} d x \int_{x}^{1} d y h(y-x)(f(x)-f(y))^{2}-\int_{0}^{1} h$$ I checked it for a few functions and I'm convinced that it is in fact true, but I can't prove it.
If $h(x,y)=h(|x-y|)+E(x,y)$, what assumptions do we need on $E(x,y)$ for the same statement to hold for $h$? For example, would it hold if $E$ was "small enough"?
I would deeply appreciate any help!