Let $p(u,x):=(4 \pi u)^{-1/2}e^{-\frac{x^2}{4u}},u>0,x \in \mathbb{R}.$

Let $\mathcal{E}:=\{\phi \in C_c^\infty (\mathbb{R}),\operatorname{supp}(\phi) \subset B(0,1),\|\phi\|_\infty \leq 1\}.$

Prove or disprove that for all $U>0,\beta>0,$ there exist $\epsilon>0,C>0$ such that for all $\lambda \in \left]0,1\right],u,v \in [0,U],$ $$\sup_{x \in \mathbb{R}} \sup_{\phi \in \mathcal{E}}\left(\int_0^{|v-u|} \int_{\mathbb{R}} \left(\int_{\mathbb{R}} \phi_x^\lambda(y_1)p(r,y_1-y_2) \, dy_1 \right)^2 \,dy_2 \, dr\right)^{1/2}\leq C|v-u|^\varepsilon \lambda^{1/2-\beta},$$ where $\phi_x^\lambda(y) = \lambda^{-1} \phi(\lambda^{-1}(y-x)).$

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    $\begingroup$ The conspicuous typographical difference between $\lambda\in]0,1]$ and $\lambda\in\left]0,1\right]$ results from the use of \left and \right in \lambda\in\left]0,1\right], so this is another example of why \left and \right are not only about sizes of delimeters but also about proper horizontal spacing. $\endgroup$ Jan 26, 2023 at 19:09
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    $\begingroup$ Isn't the integral on the LHS the $L^2$ norm (both in space and time) of the solution of the heat equation with initial datum $\phi_x^\lambda$? Why do you think it is true? $\endgroup$ Jan 26, 2023 at 22:08
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    $\begingroup$ Why is it true then? Are you refering to a theorem? $\endgroup$
    – mathex
    Jan 26, 2023 at 22:11
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    $\begingroup$ Call $T=|v-u|$. The integral on the LHS is $\int_0^T \|e^{t \Delta} \phi^\lambda\|_2^2 dt$, where I fixed $x=0$, the $L^2$ norm refers to the space variable and $e^{t \Delta}$ is the heat semigroup with your kernel $p$. If $I_\lambda (\phi)(x)=\phi (x/\lambda)$ then $e^{t \Delta} \phi_\lambda=\lambda^{-1}I_\lambda (e^{t\lambda^2 \Delta} \phi)$ and then its $L^2$ norm coincides with that of $e^{t \lambda^2 \Delta} \phi$ and the initial integral equals $\int_0^T \|e^{t \lambda^2 \Delta} \phi\|_2^2dt$ which tends to $T\|\phi\|_2^2$ when $\lambda \to 0$. $\endgroup$ Jan 26, 2023 at 22:36
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    $\begingroup$ @MichaelHardy, re, in this case \left and \right are something in the way of distractions; the same effect can be achieved without any sizing using $\lambda \in \mathopen]0, 1\mathclose]$ \lambda \in \mathopen]0, 1\mathclose], and I suspect (but have not checked) that \left and \right use \mathopen and \mathclose in addition to their sizing effects. $\endgroup$
    – LSpice
    Jan 31, 2023 at 21:36

1 Answer 1


$\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$What you wanted us to prove is not true.

Indeed, take any $\phi\in\EE$ such that $\phi\ge1_{[-1/2,1/2]}$. Write $A\gg B$ for $A\ge cB$, where $c$ is a universal positive real constant.

Then, for $w:=x-y_2$,
\begin{equation} \begin{aligned} &\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1 \\ &\gg\frac1\la\,\int_\R dy_1\, 1(|y_1-x|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(y_1-y_2)^2}{4r} \\ &=\frac1\la\,\int_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(w+z)^2}{4r} \\ &\ge\frac1\la\,\int_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{w^2+z^2}{2r} \\ &\ge\exp\Big(-\frac{\la^2}{8r}\Big)\frac1{\sqrt r}\,\exp-\frac{w^2}{2r}. \end{aligned} \end{equation}
So, \begin{equation} \begin{aligned} &\int_\R dy_2\,\Big(\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1\Big)^2 \\ &\gg \exp\Big(-\frac{\la^2}{4r}\Big) \int_\R dw\,\frac1r\,\exp-\frac{w^2}r \\ &\gg \frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \end{aligned} \end{equation} and hence \begin{equation} \begin{aligned} I&:=\int_0^{|v-u|}dr\,\int_\R dy_2\,\Big(\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1\Big)^2 \\ &\gg \int_0^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\ &\ge \int_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\ &\ge \int_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{2|v-u|}\Big) \\ &\gg |v-u|^{1/2}\exp\Big(-\frac{\la^2}{2|v-u|}\Big). \end{aligned} \end{equation}

Letting now, for instance, $U=1$, $v=1$, and $u=0$, for all $\la\in(0,1]$ we get \begin{equation} I\gg1. \end{equation} So, if $\beta<1/2$, then there is no real $\ep>0$ and $C>0$ such that $I^{1/2}\le C|v-u|^{\ep} \la^{1/2-\beta}$ for all $\la\in(0,1]$. $\quad\Box$

  • $\begingroup$ What if $\beta>1/2$? $\endgroup$
    – mathex
    Jan 31, 2023 at 2:51
  • $\begingroup$ I think this would work, but I have not checked the details. If you post this version of the question separately and let me know, I will consider it. $\endgroup$ Jan 31, 2023 at 3:14
  • $\begingroup$ The question is here: mathoverflow.net/questions/439814/…, after checking computation this version is correct $\endgroup$
    – mathex
    Jan 31, 2023 at 21:31
  • $\begingroup$ Hi, any good ideas for $\beta>1/2$? $\endgroup$
    – mathex
    Feb 1, 2023 at 15:54
  • $\begingroup$ @mathex : I saw the answer at mathoverflow.net/a/439822/36721 . Are you not satisfied with it? $\endgroup$ Feb 1, 2023 at 16:41

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