In Rudin's book, the author says that the Fourier transform of $f'$ is $it\hat{f}(t)$, provided that $f\in L^1,$ $f'\in L^1,$ and if $f$ is the indefinite integral of $f'$. Thus, using the fact that $f$ is absolutely continous on any compact interval and the dominated convergence theorem, I derived the following:
$\displaystyle\int_{\mathbb{R}} (f'g)\mathrm{d}m = \lim_{\substack{ a \rightarrow -\infty \\ b \rightarrow \infty }} [f(b)g(b) -f(a)g(a)] + it\hat{f}(t),$ where $g(x) = e^{-itx}$ and $m$ is Lebesgue measure on $\mathbb{R}.$
However, it seems that there should be another condition that $f$ tends to zero as $x$ goes to $\displaystyle \pm \infty$ in order to derive the statement. Can I derive this condition using the already mentioned conditions, or should the condition be mentioned independently? There are counterexamples where $f$ is continous and integrable, but does not vanish at infinity. Thus, I am thinking that the new condition should be added.
If $f \in L^{1}$ and $f' \in L^{1}$ and $f$ is the indefinite integral of $f'$ then $\lim_{x \to \pm \infty} f(x)=0$. First note that $f(x)-f(0)=\int _0^{x} f'(t) dt$ so $f$ is bounded. Also $\int_x^{y} f'(t)dt \to 0$ as $ x,y \to \infty$ so $|f(x)-f(y)| \to 0$. From these two facts it follows that $\lim_{x \to \pm \infty} f(x)$ exists and the limit has to be $0$ since $f$ is integrable.