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.

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}

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 $$ \int_{\mathbb{R}^d}\left|\nabla_x^n \rho_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{\beta-n} $$ in (\ref{2})?

Thank you so much for your help!

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.

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}

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!

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.

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}

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 $$ \int_{\mathbb{R}^d}\left|\nabla_x^n \rho_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{\beta-n} $$ in (\ref{2})  ?

Thank you so much for your help!

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.

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}

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 $$ \int_{\mathbb{R}^d}\left|\nabla_x^n \rho_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{\beta-n} $$ in (\ref{2})?

Thank you so much for your help!

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.

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}

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 $$ \begin{align*} \int_{\mathbb{R}^d}\left|\nabla_x^n \rho_{\varepsilon}\right|(x-y) \mathrm{d} y &= \int_{\mathbb{R}^d}\left|\nabla_{y}^n \rho_{\varepsilon}\right|(y) \mathrm{d} y \\ &= \varepsilon^{-d-1} \int_{\mathbb{R}^d}\left|\nabla_{y}^n \rho \right|(y) \mathrm{d} y. \end{align*} $$

Could you explain how to get the inequality $$ \int_{\mathbb{R}^d}\left|\nabla_x^n \rho_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{\beta-n} $$ in (\ref{2}) ?

Thank you so much for your help!

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.

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}

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 $$ \int_{\mathbb{R}^d}\left|\nabla_x^n \rho_{\varepsilon}\right|(x-y) \mathrm{d} y \leqslant c \varepsilon^{\beta-n} $$ in (\ref{2}) ?

Thank you so much for your help!