# Estimate of remainder of Taylor expansion

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 $$\xi,\eta \in \mathbb{R}^n$$ $$\begin{equation*} \hat{\rho}(\epsilon\xi)-\hat{\rho}(\epsilon\eta)=\sum_{0<|\alpha| Now I want to estimate the remainder term $$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?

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 $$C= 2^k\,C_{\rho}\,C_N \int_0^1 (1-t)^{N-1} (1+t)^{-k}\,\mathrm d t \\ \leq \frac{2^k\,C_{\rho}\,C_N}{N} = \frac{2^k\,C_{\rho}\,n^N}{N!}.$$