upper bound for infimum of integral

Question

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:

1. 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.

2. 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!

Answer 1

We have \begin{equation} I:=\int_0^1\int_0^1 h(|x-y|)(f(x)-f(y))^2\,dx\,dy =I_1+I_2, \end{equation} where \begin{equation} \begin{aligned} I_1&:=\int_0^1\int_0^1 1(x>y)h(|x-y|)(f(x)-f(y))^2\,dx\,dy, \\ I_2&=\int_0^1\int_0^1 1(x<y)h(|x-y|)(f(x)-f(y))^2\,dx\,dy. \end{aligned} \end{equation} Next, \begin{equation} \begin{aligned} I_1&=\int_0^1\int_0^1 1(x>y)h(|x-y|)(f(x)-f(y))^2\,dx\,dy \\ &=\int_0^1\int_0^1 1(x>y)h(x-y)(f(x)-f(y))^2\,dx\,dy \\ &=\int_0^1\int_0^1 1(v>u)h(v-u)(f(v)-f(u))^2\,du\,dv \\ &=\int_0^1\int_0^1 1(u<v)h(v-u)(f(u)-f(v))^2\,du\,dv \\ &=\int_0^1\int_0^1 1(x<y)h(y-x)(f(x)-f(y))^2\,dx\,dy \\ &=\int_0^1\int_0^1 1(x<y)h(|y-x|)(f(x)-f(y))^2\,dx\,dy=I_2. \end{aligned} \end{equation} Also, reading the above multiline display backwards, we see that \begin{equation} \begin{aligned} I_2&=\int_0^1\int_0^1 1(x<y)h(y-x)(f(x)-f(y))^2\,dx\,dy \\ &= \int_0^1 dx\,\int_x^1 dy\,h(y-x)(f(x)-f(y))^2. \end{aligned} \end{equation} So, \begin{equation} \begin{aligned} I&=I_1+I_2 \\ &=2I_2=2 \int_0^1 dx\,\int_x^1 h(y-x)(f(x)-f(y))^2\,dx\,dy, \end{aligned} \end{equation} which immediately implies \begin{equation} \int_0^1\int_0^1 h(|x-y|)(f(x)-f(y))^2\,dx\,dy-2\int_0^1 h(x)dx \\ =I-2\int_0^1 h=2D_f(h), \end{equation} as desired.