$ \newcommand{\bR}{\mathbb{R}} \newcommand{\diff}{\mathop{}\!\mathrm{d}} $ We fix $\alpha \in (0, 1)$ and $c>0$. Let $f : \bR^d \to \bR$ and $\ell : \bR^d \to \bR_+$ be measurable such that $\ell$ is a probability density function and that $$ \| f \|_\infty + \|\ell\|_{\infty} + \sup_{\substack{x, y \in {\bR}^d \\ x \neq y}} \frac{| \ell (x)-\ell (y)|}{|x-y|^\alpha} + \int_{\bR^d} |y| \ell (y) \diff y \le c. $$

Let $(p_t)_{t>0}$ be the standard Gaussian heat kernel on $\bR^d$, i.e., $$ p_t (x) := \frac{1}{(4 \pi t)^{\frac{d}{2}}} \exp \left ( -\frac{|x|^2}{4 t} \right ). $$

We define $$ I_t := \int_{\bR^d \times \bR^d} |f(x)-f(y)| (1+|y|) \ell (x) p_t (x-y) \diff x \diff y. $$

Is there a constant $c_1 >0$ (depending only on $d,\alpha, c$) such that $I_t \le c_1 t^{\frac{\alpha}{2}}$ for $t>0$?

Other upper bounds are also welcome. Thank you for your elaboration.