# 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 $$\mathcal{H} = \left\{f\in L^2([0,1]) : \int_0^1 f(t)dt = 0 \text{ and } \int_0^1 f^2(t)dt = 1\right\}$$.

Note that $$$$\int_0^1\int_0^1 h(|x-y|)(f(x)-f(y))^2\,dx\,dy-2\int_0^1 h(x)dx=2D_f(h),$$$$ where $$$$D_f(h):=\int_0^1 dx \int_x^1 dy\,h(y-x)(f(x)-f(y))^2-\int_0^1 h.$$$$ So, it is enough to find, for each $$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 $$$$h(0)>h(1).$$$$

Take any $$t\in(0,1/2)$$, and for all $$x\in[0,1]$$ let $$$$f(x):=f_t(x):=\tfrac1{\sqrt{2t}}(I\{x1-t\}),$$$$ where $$I$$ is the indicator. Then $$f\in\mathcal H$$. Also, $$$$2t(f(x)-f(y))^2=I\{x1-t\}+2I\{x if $$0, whence, letting $$t\downarrow0$$, we have
$$$$\int_0^1 dx \int_x^1 dy\, h(y-x)(f(x)-f(y))^2=\frac{J_1+J_2+2J_3}{2t},$$$$ where \begin{align} J_1&:=\int_0^t dx \int_t^1 dy\, h(y-x) \\ &= \int_0^t dx \int_{t-x}^{1-x} du\, h(u) \\ &= t\int_0^1 h- \int_0^t dx \Big(\int_0^{t-x}h+ \int_{1-x}^1 h\Big) \\ &= t\int_0^1 h- \int_0^t dx \Big(\int_0^{t-x}[h(0)+o(1)]+ \int_{1-x}^1 [h(1)+o(1)]\Big) \\ &= t\int_0^1 h- \frac{t^2}2\,[h(0)+h(1)+o(1)]; \end{align} similarly, \begin{align} J_2&:=\int_{1-t}^1 dy\,\int_0^{1-t} dx\, h(y-x) \\ &= t\int_0^1 h- \frac{t^2}2\,[h(0)+h(1)+o(1)]; \end{align} and \begin{align} J_3&:=\int_0^t dx \int_{1-t}^1 dy\, h(y-x) \\ &= t^2[h(1)+o(1)]. \end{align} Collecting all the pieces, we have $$$$D_f(h)=\frac{J_1+J_2+2J_3}{2t}-\int_0^1 h=[h(1)-h(0)+o(1)]t/2<0$$$$ for small enough $$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$$.

• Thanks @IosifPinelis for you answer. I am not sure if the function $g(x) = I_{\left\{x<1/2\right\}}- I_{\left\{x>1/2\right\}}$ is the solution for the problem. In fact if you take $h(x) = 1-x^2$, then $D_g(h) = 1/12>0$. I think the problem in your solution is in the calculation of $D_g(h_u)$, in fact for $u\in (0,1/2)$ your formula is correct. Nevertheless when $u\in (1/2,1)$ we have $D_g(h_u) = 2-4(1-u)^2-2u>0$! – Samovem Mar 4 at 14:21
• @Samovem : Thank you for your comment. I have fixed the error. – Iosif Pinelis Mar 5 at 6:48

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 $$u\in (0,1)$$.

• I think this is conjecture is false. – Iosif Pinelis Mar 5 at 6:50