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

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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 $D_f(h)<0$.

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  • $\begingroup$ 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 $D_g(h_u) = 2-4(1-u)^2-2u>0$! $\endgroup$ – Samovem Mar 4 at 14:21
  • $\begingroup$ @Samovem : Thank you for your comment. I have fixed the error. $\endgroup$ – Iosif Pinelis Mar 5 at 6:48
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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 $u\in (0,1)$.

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

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