Integral with inequality

Question

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

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 $\la\in(0,1]$. $\quad\Box$