Let $f$ be a $C^{1}$ function and $\left | f(x) \right |+\left | f'(x) \right |\leq \frac{C}{(1+\left | x \right |)^{2}} $ for all $x\in \mathbb{R}$.
$\hat{f}(\xi )=\int f(x)e^{-2\pi ix\xi }dx$
Prove that
(a)$\hat{f}$ is continuous and $\lim_{\left | \xi \right |\rightarrow 0}\hat{f}(\xi )=0$
(b)$\hat{f}$ is differentiable and $\frac{d\hat{f}(\xi )}{d\xi }=\hat{g}(\xi )=$where $ g(x)=-2\pi ixf(x)$.
(My attempt)
(a) $\hat{f}=\int f(x)\frac{1}{-2\pi i\xi }\frac{\mathrm{d} (e^{-2\pi ix\xi })}{\mathrm{d} x}dx $
Integrating by parts, $\hat{f}=e^{-2\pi ix\xi }f(x)-\int e^{-2\pi ix\xi }f'(x)dx$
Since $\int \frac{C}{1+\left (| x \right |)^{2}}dx\leq \int \frac{C}{\left (| x \right |)^{2}}dx$ and $\int \frac{C}{\left (| x \right |)^{2}}dx$ is bounded,
$\left | f(x) \right |+\left | f'(x) \right |$ is bounded.
I'm stuck here and don't know how to show the continuity.
(b)$ \frac{\mathrm{d}(\hat{f}(\xi )) }{\mathrm{d} \xi }=\int f(x)\frac{\mathrm{d} (e^{-2\pi ix\xi })}{\mathrm{d} \xi }dx=\int f(x)(-2\pi ix)e^{-2\pi ix\xi }dx=\hat{g}(\xi )$
I don't know how to show the differentiability here.
b) is false. Let $f(x)=\frac 1 {1+x^{2}}$. Then $ \hat {f} (\xi) =\pi e^{-2 \pi |\xi|}$ which is not differentiable at $0$. Note that $|f(x)|+|f'(x)| \leq \frac 2 {1+x^{2}} \leq \frac 4 {(1+|x|)^{2}}$ (because $|-2x| \leq 1+x^{2}$). For part a) all you have to do is apply Dominated Convergence Theorem: if $\xi _n \to \xi$ then $f(x)e^{-2i\pi x \xi _n} \to f(x)e^{-2i\pi x \xi}$ and $|f(x)e^{-2i\pi x \xi _n}| \leq |f(x)|$ so $\hat {f} (\xi_n) \to \hat {f} (\xi )$. The second part of a) is the Riemann Lebesgue Lemma.