Let $c>0$ and $T>0$ be fixed. Denote by $F$ the Gaussian CDF, i.e. $F:\mathbb R\to\mathbb R$ is defined by


$$F(x):=\int\_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz.$$


For every $a\in [0,1)$, does there exist $C\equiv C\_a>0$ s.t.


$$\left|\int\_0^\infty \frac{1}{s-t}\left(F\left(\frac{y}{\sqrt{c(s-t)}}\right)-F\left(\frac{y-x}{\sqrt{c(s-t)}}\right)\right) dy\right| \le \frac{C|x|^a}{(s-t)^{(1+a)/2}}$$


holds for all $0\le t<s\le T$ and $x\in\mathbb R$?



On an integral of Gaussian CDFs
No. Indeed, let
$$u:=\frac x{\sqrt{c(s-t)}}.$$
Then, with the substitution $z:=\dfrac y{\sqrt{c(s-t)}}$, the inequality in question can be rewritten as
$$|I(u)|\le C|u|^a \tag{1}\label{1}$$
for all real $u$, where
$$I(u):=\int\_0^\infty dz\,(F(z)-F(z-u)).$$


For $u>0$,
$$I(u)=\int\_0^\infty dz\,\int\_{z-u}^z dF(t)
=\int\_{-u}^\infty dF(t)\int\_0^\infty dz\,1(t<z<t+u) \\ 
\ge\int\_0^\infty dF(t)\,u=\frac12\,u. $$
Similarly, $|I(u)|\le|\int\_{-\infty}^\infty dz\,(F(z)-F(z-u))|=|u|$ for all real $u$.


So,
\eqref{1} holds for all real $u$ if and only if $a=1$.
(This conclusion actually holds for any cdf $F$ such that $F(0-)<1$.)


