conditions on Fourier Transform of derivative

Question

At page 445 of Myint-U's Linear Partial Differential Equations (4th Ed), Fourier Tranform of derivative is defined as: Let $f$ be a continuous and piecewise smooth in $(-\infty, \infty)$. Let $f(x)$ approach zero as $|x| \rightarrow \infty$. If $f$ and $f'$ are absolutey integrable, then $$\mathcal{F}[f'(x)] = \imath\, k \, \mathcal{F}[f(x)] = \imath\, k \, F(k). $$ In expansion of LHS we get this step: $$\mathcal{F}[f'(x)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f'(x) e^{- \imath\, k\, x} dx = \frac{1}{\sqrt{2 \pi}} \left[ f(x) e^{- \imath\, k \, x}\; |_{-\infty}^{\infty}\quad + \cdots \right].$$ We can find functions such that $\lim_{x\rightarrow -\infty} f(x) =0$ $\;$ but $\;$ $\lim_{x\rightarrow -\infty} f(x) e^{- \imath\, k \, x} \neq 0$. Example is $f(x) = \frac{1}{x}$.
Obviously for $f(x)$ satisfying rest of the conditions (continuity and smoothness etc) the required limit does go to zero. Example $f(x) = e^{-x^2}$.
My question: How can we prove (generally) that the term $f(x) e^{- \imath\, k \, x}$ goes to zero as $|x| \rightarrow \infty$ when all conditions are met.

Answer 1

Exponential with purely imaginary argument has absolute value equal to 1 so it is sufficient that $f$ vanishes at infinity.