Upper bound $I_R := \int_{B_R^c} |x| (P_t \ell_\nu) (x) \, \mathrm d x$ in terms of $R, \nu, t$?

Question

Let $(p_t)_{t >0}$ be the Gaussian heat kernel on $\mathbb R^d$ and $(P_t)_{t >0}$ its induced semi-group, i.e., $$ \begin{align} p_t (x) &:= (4\pi t)^{-\frac{d}{2}} e^{-\frac{|x|^2}{4t}}, \quad t>0, x \in \mathbb R^d, \\ P_t f (x) &:= \int_{\mathbb R^d} p_t ( x- y) f(y) \, \mathrm d y. \end{align} $$

Let $\nu$ be a Borel probability measure on $\mathbb R^d$ with finite first moment. We denote by $\ell_\nu$ the probability density function of $\nu$. For $R>0$, let $B_R$ be the open ball centerted at $0$ with radius $R$. Let $B_R^c := \mathbb R^d \setminus B_R$ and $$ I_R := \int_{B_R^c} |x| (P_t \ell_\nu) (x) \, \mathrm d x, \quad R >0. $$

Is there an upper bound of $I_R$ in terms of $R, \nu, t$?

Thank you so much for your elaboration!

Answer 1

Let $Z$ be a standard normal random vector in $\mathbb R^d$. Let $Y$ be a random vector in $\mathbb R^d$, independent from $Z$, with distribution $\nu$. Then, for each $s\in(0,1)$, $$I_R=E|\sqrt{2t}\,Z+Y|\,1(|\sqrt{2t}\,Z+Y|\ge R) \\ \le E(\sqrt{2t}\,|Z|+|Y|) \,[1(\sqrt{2t}\,|Z|\ge sR)+1(|Y|\ge(1-s)R)] \\ =\sqrt{2t}\,E|Z|1(|Z|\ge sR/\sqrt{2t}) \\ +\sqrt{2t}\,E|Z|\,P(|Y|\ge(1-s)R) \\ +E|Y|\,P(|Z|\ge sR/\sqrt{2t}) \\ +E|Y|\,1(|Y|\ge(1-s)R). $$

In turn, you can upper-bound the latter four summands using the following applications of Markov's inequality: $$ E|Z|1(|Z|\ge sR/\sqrt{2t})\le \frac{E|Z|^2}{sR/\sqrt{2t}}=\frac n{sR/\sqrt{2t}}, \\ E|Z|\,P(|Y|\ge(1-s)R)\le\sqrt{E|Z|^2}\,\frac{E|Y|}{(1-s)R} =\frac{\sqrt n\,E|Y|}{(1-s)R} \\ P(|Z|\ge sR/\sqrt{2t})\le\frac{E|Z|^2}{(sR/\sqrt{2t})^2} =\frac{n}{(sR/\sqrt{2t})^2} \\ E|Y|\,1(|Y|\ge(1-s)R)\le\frac{E|Y|^2}{(1-s)R}. $$