$g$ is differentiable and $g'(y)=\int_{\mathbb{R}}ixf(x)e^{iyx}dm(x)$

Question

Let $f \in \mathcal{L}(\mathbb{R},\mathfrak{M},\mathbb{R})$ where $\mathfrak{M}$ measurable Lebesgue. Asumme that $x\to f(x)$ is measurable.

For $y \in \mathbb{R}$ define: $g(y)=\int_{\mathbb{R}}f(x)e^{iyx}dm(x)$

Shows that $g$ is differentiable and $g'(y)=\int_{\mathbb{R}}ixf(x)e^{iyx}dm(x)$.

As I can apply the dominated convergence theorem to solve this fact or a suggestion to solve it thanks

Answer 1

If we assume that the function $x\mapsto x f(x)$ is integrable, and $(h_n)_{n\geqslant 0}\subset\mathbf R\setminus \{0\}$ a sequence converging to $0$ then , then define $$f_n(x):=f(x)\frac{e^{i(y+h_n)x}-e^{iyx}}{h_n}.$$ Then $$\left|f_n(x)\right|\leqslant\left|f(x)\right|\frac{\left|e^{ixh_n}-1 \right| }{ \left|h_n\right|}\leqslant\left|xf(x)\right|.$$ The pointwise limit of the sequence $(f_n)_{n\geqslant 0}$ is not hard to identify.