Derivative of Fourier transform: $F[f]'=F[-ixf(x)]$

Question

Let us define the Fourier transform of the Lebesgue-summable function $f\in L_1(\mathbb{R},\mu_x)$ as $F[f](\lambda)=\int_{\mathbb{R}}f(x) e^{-i\lambda x} d\mu_x$, where $\mu_x$ is the Lebesgue linear measure. Kolmogorov-Fomin's (p. 430 here) states that, if $x\mapsto xf(x)$ is Lebesgue-summable too, then the Fourier transform is differentiable and$$\frac{d}{d\lambda}F[f](\lambda)=-i\int_{\mathbb{R}}x f(x) e^{-i\lambda x} d\mu_x$$

EDIT: I would be very grateful to the people having voted to close the question, and to anybody else, if they explained how to use the question linked as a duplicate to prove this equality, which I am not able to do. I am trying to teach myself and cannot afford university for the moment and have not got the sufficient mathematical ability to understand how that result implies this equality. I heartily thank you all!!!

Previous trial of mine which may or may not be useful to prove the given equality: In order to prove the statement to myself I had thought that the fact that if the sequence $\{\varphi_n\}\subset L_1(\mathbb{R},\mu_x)$ converges to $\varphi$ with respect to the metric of $L_1(\mathbb{R},\mu_x)$, then $F[f_n]\xrightarrow{n\to\infty} F[f]$, but I cannot apply it. In fact I see that the derivative $F[f]'(\lambda)$ can be seen, if we define $\varphi_n(x)=f(x)\frac{e^{-ih_n x}-1}{h_n}$, where $\{h_n\}$ converges to 0, as the limit $$\lim_n \int_{\mathbb{R}}f(x)\frac{e^{-i(\lambda+h_n)x}-e^{-i\lambda x}}{h_n}d\mu_x=\lim_n \int_{\mathbb{R}} f(x) \frac{e^{-ih_n x}-1}{h_n} e^{-i\lambda x} d\mu_x$$$$=\lim_n F[\varphi_n](\lambda)$$but I cannot prove that $\int_{\mathbb{R}} f(x) \frac{e^{-ih_n x}-1}{h_n} e^{-i\lambda x} d\mu_x$ converges to $-i\int_{\mathbb{R}}x f(x) e^{-i\lambda x} d\mu_x$. I thought about Lebesgue's dominated convergence theorem, but I am not convinced at all that we can majorate $|\varphi_n(x)|$ with a summable function and I am not sure this is the right path to prove the desired equality...

P.S.: I assume that the equality holds for any $f\in L_1(\mathbb{R},\mu_x)$ because the book does not specify any other conditions, but it is perfectly in the style of the book to state theorems valid only under certain conditions without explicitly declare them. So, if you noticed that some other condition is needed...

Answer 1

You know, thanks to Riemann-Lebesgue theorem, that $\,\hat{f}$ and $\,\hat{g}$ are continuos and vanishing at $\infty$, where $g=\widehat{-ixf(x)}$. I prove that $$\hat{f}(y)-\hat{f}(0)=\int_0^y g(t)dt,$$ the result will follow from fundamental calculus theorem.

\begin{align*} \int_0^y g(t)dt&=\int_0^y\int_{\mathbb{R}}-ixe^{-ixt}dt\, f(x)dx=\\ &=\int_{\mathbb{R}} \left.e^{-ixt}\right|_{0}^y\,\, f(x)dx=\int_{\mathbb{R}}f(x)\left[e^{-ixy}-e^{ix0}\right]dx=\hat{f}(y)-\hat{f}(0). \end{align*}

The hypothesis $xf(x)\in L^1$, in addiction with $|e^{i \theta}|=1$ for all $\theta\in\mathbb{R}$, is used to apply Fubini's theorem.

Answer 2

If we have mean value theorem for complex functions, then we can appeal to the dominated convergence theorem and things can be done easily. Unfortunately there is no mean value theorem for complex functions. However, although there is no such thing for complex functions, we have a similar estimate. (c.f. Mean Value Theorem for complex functions?)

Also, note that calculating derivatives is actually taking limits over segments, which are definitely convex. So by this estimate and the dominated convergence theorem, we are done.