# Integral with inequality

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 $$\phi_x^\lambda(y) = \lambda^{-1} \phi(\lambda^{-1}(y-x)).$$

• 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. Jan 26, 2023 at 19:09
• 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? Jan 26, 2023 at 22:08
• Why is it true then? Are you refering to a theorem? Jan 26, 2023 at 22:11
• 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 $\lambda \to 0$. Jan 26, 2023 at 22:36
• @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. Jan 31, 2023 at 21:36

$$\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$$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{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}
So, \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} and hence \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}

Letting now, for instance, $$U=1$$, $$v=1$$, and $$u=0$$, for all $$\la\in(0,1]$$ we get $$$$I\gg1.$$$$ So, if $$\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$$

• What if $\beta>1/2$? Jan 31, 2023 at 2:51
• 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. Jan 31, 2023 at 3:14
• The question is here: mathoverflow.net/questions/439814/…, after checking computation this version is correct Jan 31, 2023 at 21:31
• Hi, any good ideas for $\beta>1/2$? Feb 1, 2023 at 15:54
• @mathex : I saw the answer at mathoverflow.net/a/439822/36721 . Are you not satisfied with it? Feb 1, 2023 at 16:41