# Question about proof of Fourier transform of derivative

If $f\in L^1(\mathbb{R})$, $f'(x)$ exists and is continuous, and $f'\in L^1(\mathbb{R})$, then $\widehat{f'}(t)=2\pi i \widehat{f}(t)$.

I've stated the above theorem from a textbook that I'm reading. The author uses the Sobolev inequality in the proof to show that $f(x)\to 0$ as $x\to 0$. (And Sobolev inequality, as stated in the textbook requires continuity of $f'$.)

However, it seems to me that this theorem holds under the weaker condition that $f'\in L^1(\mathbb{R})$. The proof that I have in mind uses integration by parts and also the fundamental theorem of calculus (to show $f(x)\to 0$ as $x\to 0$).

Can someone, please, let me know whether my proposed proof and the weaker condition are in fact correct for this theorem. And if yes, I still wonder why the author has used the Sobolev theorem to prove this since the FTC seems easier to apply.

Because you are assuming that $f'\in L^1(\mathbb{R})$, then the following limits exist $$\lim_{x\rightarrow\pm\infty}f(x)=\lim_{x\rightarrow\pm\infty}\int_{0}^{x}f'(t)dt+f(0).$$ The above limits must be $0$ because $f \in L^1(\mathbb{R})$; otherwise, for example, if $\lim_{x\rightarrow\infty}f(x)=L \ne 0$, then $|f(x)| \ge L/2$ for large enough $x$, which would contradict the absolute integrability of $f$ on $[0,\infty)$. Because these limits must be $0$, then the evaluation terms below vanish: \begin{align} \int_{-\infty}^{\infty}f'(t)e^{-ist}dt & = \left.f(t)e^{-ist}\right|_{t=-\infty}^{\infty}-\int_{-\infty}^{\infty}f(t)\frac{d}{dt}e^{-ist}dt \\ & =is\int_{-\infty}^{\infty}f(t)e^{-ist}dt. \end{align}
$$\widehat{f'}(t)=2\pi i t \widehat{f}(t).$$
And yes, I don't know why your textbook would use Sobolev inequality as your idea is correct and $f' \in L^1$ is enough (as you need the Fourier transform of the derivative to exist). The proof, which is the one you describe, goes like this: Fourier Transform of Derivative.