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

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...

• possible duplicate of Problem on Big Rudin about Fourier Transform Commented Nov 21, 2014 at 9:42
• @DavidZena : I'm sorry. There's an extra transform in the other problem. I associated the two because the proof that works is identical to the other case. You start with the function you believe to be the derivative, which is $-iF[xf]$, and integrate that function with respect to $\lambda$ over a finite interval to get $\int_{0}^{\mu}(-iF[xf])(\lambda)\,d\lambda = F[f](\mu)+C$. Then you argue that $-iF[xf]$ is continuous in $\lambda$ by dominated convergence, and that gives the existence of $F[f]'(\mu)=-iF[xf](\mu)$. Commented Nov 21, 2014 at 11:48
• $\int_{0}^{t}-ixe^{-i\lambda x}d\lambda = \int_{0}^{t}\frac{d}{d\lambda}e^{-i\lambda x}d\lambda= e^{-itx}-1$. So when you integrate $\int_{0}^{t}F[-ixf](\lambda)\,d\lambda$, you get $F[f](t)-C$ where $C$ is a constant. Then differentiate both sides with respect to $t$ and use the continuity of $F[-ixf](\lambda)$ and the fundamental theorem of calculus. Commented Nov 21, 2014 at 16:42
• Actually, this is classical pointwise everywhere using the Fundamental Theorem of Calculus for the Riemann integral because $F[-ixf](\lambda)$ is continuous in $\lambda$ by the Lebesgue dominated convergence theorem. :) Commented Nov 21, 2014 at 18:00
• Ouch: I didn't realise that Lebesgue=Riemann integral on $[0,t]$! I tend to be so careful not to use lemmas from Riemann integral calculus when dealing with Lebesgue ones that I forget that they can sometimes be the same. $\infty$-ly thanks! Commented Nov 21, 2014 at 18:03

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.

• Very clear: I heartily thank you! Just one thing more: Commented Nov 21, 2014 at 17:33
• what do you want to know? Commented Nov 21, 2014 at 17:34
• @T.A.E. and @ Felice: ...I know that the absolute continuity of $\hat{f}(\lambda)$ as a function of $\lambda$ guarantees that it's differentiable almost everywhere. Nevertheless, we cannot say anything in general about the existence of $\frac{d}{d\lambda}\hat{f}$ everywhere: am I right? Thank you both so much again!!! Commented Nov 21, 2014 at 17:35
• From my proof you can deduce it. Because $\hat{f}$ is the primitive of a continuous function, so it will be differentiable. Commented Nov 21, 2014 at 17:37
• (Fondamental calculus theorem) Commented Nov 21, 2014 at 17:38

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.