Let $f\in C_c^n(\Bbb{R})$ ($n$ times continuously differentiable compactly supported). Show that, $|\hat{f}(\xi)|\le \frac{C}{1+|\xi|^n}$

Question

Let $f\in C_c^n(\Bbb{R})$ ($n$ times continuously differentiable compactly supported). Show that, there exists $C>0$ such that $$|\hat{f} (\xi)|\le \frac{C}{1+|\xi|^n}\ \forall \xi\in\Bbb{R}$$

I define $g=\frac{d^n}{dx^n}(f)$. Then $g\in L^1(\Bbb{R})$. I have proved by induction that $\hat{g}(\xi)=(2\pi i)^n \hat{f}(\xi)$.

By Riemann Lebesgue Lemma, $|\hat{g}(\xi)|\to0$ as $|\xi|\to\infty$.

So, there is $M>0$ such that $|\hat{g}(\xi)|\le 1\ \forall |\xi|>M$. Thus $|\hat{f}(\xi)|\le \frac{1}{|\xi|^n}\ \forall |\xi|>M$.

But I'm not able to extract the exact expression of $1+|\xi|^n$ in the bound.

Can anyone help complete the proof? Thanks for your help in advance.

Answer 1

\begin{align*} |\widehat{f}(\xi)|&=|\widehat{f}(\xi)|\chi_{|\xi|\leq M}+|\widehat{f}(\xi)|\chi_{|\xi|>M}\\ &=|\widehat{f}(\xi)|\frac{1+|\xi|^{n}}{1+|\xi|^{n}}\chi_{|\xi|\leq M}+|\widehat{f}(\xi)|\frac{1+|\xi|^{n}}{1+|\xi|^{n}}|\chi_{|\xi|>M}\\ &\leq|\widehat{f}(\xi)|\frac{1+M^{n}}{1+|\xi|^{n}}\chi_{|\xi|\leq M}+\frac{1}{|\xi|^{n}}\frac{1+|\xi|^{n}}{1+|\xi|^{n}}|\chi_{|\xi|>M}\\ &\leq\left(\sup_{|\xi|\leq M}|\widehat{f}(\xi)|\right)\frac{1+M^{n}}{1+|\xi|^{n}}+\left(1+\frac{1}{|\xi|^{n}}\right)\chi_{|\xi|>M}\frac{1}{1+|\xi|^{n}}\\ &\leq\left((1+M^{n})\left(\sup_{|\xi|\leq M}|\widehat{f}(\xi)|\right)+\left(1+\frac{1}{M^{n}}\right)\right)\frac{1}{1+|\xi|^{n}}. \end{align*}