An estimate of the integral of the higher order derivative of a bump function

Question

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.

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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!

Answer 1

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\na}{\nabla}$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 $c_{n,\rho}:=\int_{\R^d}|(\na^n\rho)(z)|\,dz$. $\quad\Box$