Integral and 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 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?

Answer 1

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.

So, in view of the previous answer, your current inequality is not true either.