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 that for all $U>0,\beta>0,$ there exist $\epsilon>0,C>0$ such that for all $\lambda \in \left]0,1\right],u,v \in [0,U],$ $$u\leq v \implies \sup\_{x \in \mathbb{R}} \sup\_{\phi \in \mathcal{E}}\left(\int\_u^{v} \int\_{\mathbb{R}} \left(\int\_{\mathbb{R}} \phi\_x^\lambda(y\_1)p(v-r,y\_1-y\_2) \, dy\_1 \right)^2 \,dy\_2 \, dr+\int\_0^u\int\_{\mathbb{R}}\left(\int\_{\mathbb{R}}\phi\_x^\lambda(y\_1)(p(v-r,y\_1-y\_2)-p(u-r,y\_1-y\_2))dy\_1\right)^2dy\_2dr\right)\leq C|v-u|^\varepsilon \lambda^{1-2\beta},$$
where $\phi\_x^\lambda(y) = \lambda^{-1} \phi(\lambda^{-1}(y-x)).$


How can we prove this inequality?



Integral and inequality
At least when (say) $U=1$, $v=1$, and $u=0$, the integral
$$\int\_u^{v} \int\_{\mathbb{R}} \left(\int\_{\mathbb{R}} \phi\_x^\lambda(y\_1)p(v-r,y\_1-y\_2) \, dy\_1 \right)^2 \,dy\_2 \, dr$$
coincides (in view of the variable change $r\leftrightarrow v-r$) with the integral
$$\int\_0^{|u-v|} \int\_{\mathbb{R}} \left(\int\_{\mathbb{R}} \phi\_x^\lambda(y\_1)p(r,y\_1-y\_2) \, dy\_1 \right)^2 \,dy\_2 \, dr$$
in [your previous post](https://mathoverflow.net/q/439437/36721).


So, in view of the [previous answer](https://mathoverflow.net/a/439456/36721), your current inequality is not true either.


