Let $\kappa\_1>0$, $\beta\in [0, 1]$ and $b: \mathbb R\_+ \times \mathbb R^d \to \mathbb R^d$ such that for all $t\ge0$ and $x,y \in \mathbb R^d$ we have $|b(t, 0)| \le \kappa\_1$ and $|b(t, x) - b(t, y)| \le \kappa\_1 ( |x-y| \vee |x-y|^\beta )$ for all $x,y \in \mathbb R^d$. I'm reading section 1.2 in the paper [Density and gradient estimates for non degenerate Brownian SDEs with unbounded measurable drift](https://doi.org/10.1016/j.jde.2020.09.004).




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To state our main result, we need to prepare some deterministic regularized flow associated with the drift $b$. Let $\rho$ be a nonnegative smooth function with support in the unit ball of $\mathbb{R}^d$ and such that $\int\_{\mathbb{R}^d} \rho(x) \mathrm{d} x=1$. For $\varepsilon \in(0,1]$, define
$$
\rho\_{\varepsilon}(x):=\varepsilon^{-d} \rho\left(\varepsilon^{-1} x\right), \quad b\_{\varepsilon}(t, x):=b(t, \cdot) \* \rho\_{\varepsilon}(x)=\int\_{\mathbb{R}^d} b(t, y) \rho\_{\varepsilon}(x-y) d y,
$$
i.e. $\*$ stands for the usual spatial convolution. Then for each $n=1,2, \dotsc$, it is easy to see that
\begin{align}
\left|\nabla\_x^n b\_{\varepsilon}(t, x)\right| & =\left|\int\_{\mathbb{R}^d}(b(t, y)-b(t, x)) \nabla\_x^n \rho\_{\varepsilon}(x-y) \mathrm{d} y\right| \tag{1}\label{1} \\
& \leqslant \int\_{\mathbb{R}^d}|b(t, y)-b(t, x)|\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y \\
& \leqslant \kappa\_1 \varepsilon^\beta \int\_{\mathbb{R}^d}\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{\beta-n} .\tag{2}\label{2}
\end{align}




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**My understanding** It seems from (\ref{1}) that we need $\int\_{\mathbb{R}^d}b(t, x) \nabla\_x^n \rho\_{\varepsilon}(x-y) \mathrm{d} y=0$. A sufficient condition is that $\rho$ is symmetric, i.e., $\rho (y) = \rho(-y)$ for all $y \in \mathbb R^d$. We have
\begin{align\*}
\int\_{\mathbb{R}^d}\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y &= \int\_{\mathbb{R}^d}\left|\nabla\_{y}^n \rho\_{\varepsilon}\right|(y) \mathrm{d} y \\
&= \int\_{\mathbb{R}^d}\left|\nabla\_{y}^n \rho \right|(y) \mathrm{d} y.
\end{align\*}



> 
> Could you explain how to get the inequality
> $$
> \int\_{\mathbb{R}^d}\left|\nabla\_x^n \rho\_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{-n}
> $$
> in (\ref{2})?
> 
> 
> 


Thank you so much for your help!



An estimate of the integral of the higher order derivative of a bump function
$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\na}{\nabla}$Note that
\begin{equation}
 (\na^n\rho\_\ep)(x)=\ep^{-d-n}(\na^n\rho)(\ep^{-1}x). 
\end{equation}
So,
\begin{equation}
 \int\_{\R^d}|(\na^n\rho\_\ep)(x-y)|\,dy
 =\ep^{-d-n}\int\_{\R^d}|(\na^n\rho)(\ep^{-1}(x-y))|\,dy \\ 
 =\ep^{-n}\int\_{\R^d}|(\na^n\rho)(\ep^{-1}x-v))|\,dv
 =c\_{n,\rho} \ep^{-n}, 
\end{equation}
where $c\_{n,\rho}:=\int\_{\R^d}|(\na^n\rho)(z)|\,dz$. $\quad\Box$


