Let $E$ be a normed $\mathbb R$-vector space and $$\left\|f\right\|_{0+\alpha}:=\sup_{\stackrel{x,\:y,\:x',\:y'\:\in\:E}{x\:\ne\:x',\:y\:\ne\:y'}}\frac{\left|f(x,y)-f(x',y)-f(x,y')+f(x',y')\right|}{\left\|x-x'\right\|_E^\alpha\left\|y-y'\right\|_E^\alpha}$$ for $f:E\times E\to\mathbb R$ and $\alpha\in(0,1]$. Now, let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $T>0$
- $I:=(0,T]$
- $A:\Omega\times\overline I\times E\times E\to\mathbb R$
I'll write $A_t$ instead of $A(\;\cdot\;,t,\;\cdot\;,\;\cdot\;)$ for $t\in\overline I$ and $A(x,y)$ instead of $A(\;\cdot\;,\;\cdot\;,x,y)$ for $x,y\in E$. Fix $\alpha\in(0,1]$. Assume that
- $A_t(x,y)$ is $\mathcal A$-measurable for all $t\in\overline I$ and $x,y\in E$
- $A(x,y)$ is continuous almost surely for all $x,y\in E$
- $\left\|A_t\right\|_{0+\alpha}<\infty$ almost surely for all $t\in\overline I$
Let $C>0$ and $$\tau:=\inf\left\{t\in\overline I:\left\|A_t\right\|_{0+\alpha}\ge C\right\}.$$
How can we show that $$A^\tau_t(x,x)-2A^\tau_t(x,y)+A^\tau_t(y,y)\le C\left\|x-y\right\|_E^{2\alpha}\;\;\;\text{for all }t\in\overline I\text{ almost surely}\tag1$$ for all $x,y\in E$?
It's clear to me that $$A_t(x,x)-2A_t(x,y)+A_t(y,y)\le\left\|A_t\right\|_{0+\alpha}\left\|x-y\right\|_E^{2\alpha}\tag2$$ for all $x,y\in E$ almost surely for all $t\in\overline I$. So, my problem is somehow the selection of a null set.