Infimum upper bound Let $h:[0,1]\to[0,1]$ be a $\mathcal{C^1}$ function such that $h'(x)<0$ for all $x\in(0,1)$. I am trying to show that (not sure if it is true):
$$
\inf_{f\in \mathcal{H}}\left(\int_0^1\int_0^1 h(|x-y|)(f(x)-f(y))^2dxdy\right) < 2\int_0^1 h(x)dx
$$
where $\mathcal{H} = \left\{f\in L^2([0,1]) : \int_0^1 f(t)dt = 0 \text{ and } \int_0^1 f^2(t)dt = 1\right\}$.
 A: Note that 
\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=2D_f(h),
\end{equation}
where 
\begin{equation}
 D_f(h):=\int_0^1 dx \int_x^1 dy\,h(y-x)(f(x)-f(y))^2-\int_0^1 h. 
\end{equation}
So, it is enough to find, for each $h$ as in the OP, a function $f\in\mathcal H$ such that $D_f(h)<0$. 
We shall do a bit more than this -- by letting $h$ be any continuous function on $[0,1]$ such that 
\begin{equation}
 h(0)>h(1). 
\end{equation}
Take any $t\in(0,1/2)$, and for all $x\in[0,1]$ let 
\begin{equation}
 f(x):=f_t(x):=\tfrac1{\sqrt{2t}}(I\{x<t\}-I\{x>1-t\}),
\end{equation}
where $I$ is the indicator. Then $f\in\mathcal H$. 
Also, 
\begin{equation}
 2t(f(x)-f(y))^2=I\{x<t\}I\{y\ge t\}+I\{x\le1-t\}I\{y>1-t\}+2I\{x<t\}I\{y\ge1-t\}
\end{equation}
if $0<x<y<1$, whence, letting $t\downarrow0$, we have
\begin{equation}
 \int_0^1 dx \int_x^1 dy\, h(y-x)(f(x)-f(y))^2=\frac{J_1+J_2+2J_3}{2t}, 
\end{equation}
where 
\begin{align}
 J_1&:=\int_0^t dx \int_t^1 dy\, h(y-x) \\ 
&= \int_0^t dx \int_{t-x}^{1-x} du\, h(u) \\ 
&= t\int_0^1 h- \int_0^t dx \Big(\int_0^{t-x}h+ \int_{1-x}^1 h\Big) \\ 
&= t\int_0^1 h- \int_0^t dx \Big(\int_0^{t-x}[h(0)+o(1)]+ \int_{1-x}^1 [h(1)+o(1)]\Big) \\ 
&= t\int_0^1 h- \frac{t^2}2\,[h(0)+h(1)+o(1)];  
\end{align}
similarly, 
\begin{align}
 J_2&:=\int_{1-t}^1 dy\,\int_0^{1-t} dx\,  h(y-x) \\    
&= t\int_0^1 h- \frac{t^2}2\,[h(0)+h(1)+o(1)];  
\end{align}
and 
\begin{align}
 J_3&:=\int_0^t dx \int_{1-t}^1 dy\, h(y-x) \\ 
&= t^2[h(1)+o(1)].   
\end{align}
Collecting all the pieces, we have 
\begin{equation}
 D_f(h)=\frac{J_1+J_2+2J_3}{2t}-\int_0^1 h=[h(1)-h(0)+o(1)]t/2<0
\end{equation}
for small enough $t>0$ (depending on $h$). 
Thus indeed, for each continuous function $h$ on $[0,1]$ such that $h(0)>h(1)$, 
we have constructed a function $f\in\mathcal H$ such that $D_f(h)<0$.
A: Based on the previous comment, it is enough to find a function $f\in \mathcal{H}$ such that
$$
\int\int_{|x-y| < u} (f(x)-f(y))^2 dxdy < 2u
$$
for almost every $u\in (0,1)$.
