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



Integral with inequality
$\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$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 $\la\in(0,1]$. $\quad\Box$


