i want to prouve that if $f$ is continuously derivable and $f' \in L^1(\mathbb{R})$, then $F(f')=i(yF(f))$. where $F(f')$ denote the Fourier transformate of $f'$.
I try following solution: we have by definition: $F(f')(\xi)= \displaystyle\int_{-\infty}^{+\infty} e^{-ix\xi} f'(x) dx, \forall \xi \in \mathbb{R}$.
Then by integration by parts we have $$ F(f')(\xi)= [e^{-ix \xi} f(x)]_{-\infty}^{+\infty} + i \xi \displaystyle\int_{-\infty}^{+\infty} e^{-i x \xi} f(x) dx. $$ My question is what's the value of $ [e^{-ix \xi} f(x)]_{-\infty}^{+\infty}$? I think that the result is zero but how we justify it? Please.
Thank you in advance.
Indeed, with these assumptions, the term $[e^{-ix \xi} f(x)]_{-\infty}^{+\infty}$ vanishes (but as discussed in the comments and in this thread or this one, this is not true in general). According to this thread, it suffices to prove that $f$ is uniformly continuous on $\mathbb R$. This follows from the fundamental theorem of analysis and the fact that for each positive $\varepsilon$, there exists a $\delta$ for which the integral on an interval of length smaller than $\delta$ of $\left\lvert f'\right\rvert$ is smaller than $\varepsilon.