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For a compactly support smooth function $\rho \in C_0^\infty(\mathbb{R}^n),$ clearly $\rho$ belongs to the Schwartz space. Given a parameter $0<\epsilon <1,$ consider the following Talyor expansion for $\xi,\eta \in \mathbb{R}^n$ \begin{equation*} \hat{\rho}(\epsilon\xi)-\hat{\rho}(\epsilon\eta)=\sum_{0<|\alpha|<N} (\xi-\eta)^\alpha \frac{\epsilon^{|\alpha|}}{\alpha!} \partial_\xi^\alpha \hat{\rho}(\epsilon\eta)+r_N(\xi,\eta,\epsilon). \end{equation*} Now I want to estimate the remainder term $r_N,$ to be exact, I want to prove that for any integer $k \geq 0,$ there exists a constant number $C,$ such that \begin{equation*} |r_N(\xi,\eta,\epsilon)| \leq C\,\epsilon^N \,|\xi-\eta|^N\,(1+\epsilon |\xi|)^{-k}, \ \text{if}\ |\xi-\eta|\leq \frac{|\xi|}{2}. \end{equation*}

Clearly, the remainder has integral expression \begin{equation*} r_N(\xi,\eta,\epsilon)=N\int_{0}^1 (1-t)^{N-1} \sum_{|\alpha|=N} \partial_{\xi}^\alpha \hat{\rho}(\epsilon \eta+\epsilon t(\xi-\eta)) \,\frac{\epsilon^N (\xi-\eta)^\alpha}{\alpha!}\,\mathrm dt. \end{equation*}

My attempt is to multiply $(\epsilon \eta+\epsilon t(\xi-\eta))^{\beta}$ for multi-index $|\beta|\leq k$ to both side of the integral expression of $r_N,$ and use $\hat{\rho}$ belongs to the Schwartz space. But I cannot achieve the estimate. Can anyone help me?

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You were on the good way. So first since for any multi-index $|x^\alpha|\leq |x|^{|\alpha|}$, it holds \begin{equation*} |r_N| \leq C_N \,\epsilon^N\, |\xi-\eta|^N \int_0^1 (1-t)^{N-1} \sup_{|\alpha|=N} |\partial_{\xi}^\alpha \hat{\rho}(\epsilon\, \eta+\epsilon \,t\,(\xi-\eta))|\,\mathrm d t \end{equation*} where $C_N = \sum_{|\alpha|=N} \frac{N}{\alpha!} = N\,\frac{n^N}{N!}$ by the multinomial theorem. Then, we can multiply and divide by $(1+|\epsilon\, \eta+\epsilon \,t\,(\xi-\eta)|)^k$, inside the supremum and use the definition of the Schwartz space to deduce that there is a constant $C_{\rho}$ depending on the size of $\rho$ such that \begin{equation*} |r_N| \leq C_{\rho}\,C_N\,\epsilon^N\, |\xi-\eta|^N \int_0^1 (1-t)^{N-1} (1+|\epsilon\, \eta+\epsilon \,t\,(\xi-\eta)|)^{-k}\,\mathrm d t. \end{equation*} Since $|\xi-\eta|\leq |\xi|/2$, it follows by the triangle inequality that for any $t\in[0,1]$, $$ |\xi| \leq (1-t)|\xi-\eta| + |(1-t)\,\eta+t\,\xi| \\ \leq (1-t)|\xi|/2 + |\eta+t\,(\xi-\eta)| $$ so that $$ (1+t)|\xi|/2 \leq |\eta+t\,(\xi-\eta)| $$ and so since $(1+t)/2\leq 1$, $$ 1+|\epsilon\, \eta+\epsilon \,t\,(\xi-\eta)| \geq \frac{1+t}2 \,(1+ \epsilon\,|\xi|) $$ and the estimate for $r_N$ becomes \begin{equation*} |r_N| \leq C\,\epsilon^N\, |\xi-\eta|^N (1+ \epsilon\,|\xi|)^{-k} \end{equation*} with $$ C= 2^k\,C_{\rho}\,C_N \int_0^1 (1-t)^{N-1} (1+t)^{-k}\,\mathrm d t \\ \leq \frac{2^k\,C_{\rho}\,C_N}{N} = \frac{2^k\,C_{\rho}\,n^N}{N!}. $$

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  • $\begingroup$ Thank you so much for your excellent answer! $\endgroup$ Aug 20 at 15:43

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