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.