TITLE: derivative of fourier transform
QUESTION [6 upvotes]: Let $f\in C^k$ and $f^{(k)}$ be absolutely integrable. I want to show for the fourier transform: 
$$\widehat{f^{(k)}}(z)=(iz)^k\widehat{f}(z)$$
I want to prove it for $k=1$ and did the following:
partial integration:  $$\begin{align*}
\int_a^bf'(w)\exp(-iwz)dw&=f(b)\exp(-ibz)-f(a)\exp(-iaz)+iz\int_a^b f(w)\exp(-iwz)dw
\end{align*}$$
The last term for $a,b\rightarrow\infty$ is the fourier transform of $f(z)$. 
But why are $\lim_{b\rightarrow\infty}f(b)=0$ and the same for $a$?

REPLY [3 votes]: The reason is that: 

If $|f|$ and $|f'|$ are both integrable, then $f$ must vanish at infinity.

Notice your question pre-suppose that the Fourier transform for $f$ and $f'$ exist, hence they are both integrable. Now 
$$
\int^{\infty}_0 f'(w) \,dw = \lim_{b\to \infty}f(b) - f(0).
$$ 
By integrability of $f'$, the limit exists. Moreover by integrability of $f$, the only possible limit is $0$. The same argument applies for $-\infty$.