Suppose $f\in L^1 (\mathbb{R})$ and that for any $n\in \mathbb{N}$ there is $C_n > 0$ such that its Fourier transform satisfies $$ |\hat{f}(\xi )| \le C_n(1+|\xi |^2)^{-n}. $$ Let $$g(x) = \frac 1{2\pi}\int_{\Bbb R}\widehat f(\xi)e^{i\xi x}d\xi$$ and suppose we want to show that $g$ is smooth.
The decay condition allows us to show that the function $$k_d(x)=\frac 1{2\pi}i^d\int_\Bbb R\widehat f(\xi)e^{i\xi x}\xi^dd\xi$$ is well defined for each $x \in \mathbb{R}$.
How does it then follow that the $d$-th derivative of $g(x)$ is in fact $k_d(x)$?
This question is related to the answer provided in Decaying Fourier transform and smoothness. It says it follows by the dominated convergence theorem but I was not sure how this was the case... I would greatly appreciate some explanation. Thank you very much!