I have been recently studing Fourier transform and there is a proposition that says: If $\lim \limits_{x \to\infty}xf(x)=\lim\limits_{{x}\to -\infty}xf(x)=0$ then $$\hat{f'}(z)=-iz\hat{f}(z)$$
and this is proven like this:
$(f(x)e^{izx})'=f'(x)e^{izx}+izf(x)e^{izx}$
and intergrating by parts we have
$\displaystyle\int_{-\infty}^{\infty}(f(x)e^{izx})'dx=\int_{-\infty}^{\infty}f'(x)e^{izx}dx+\int_{-\infty}^{\infty}izf(x)e^{izx}dx$
or
$f(x)e^{izx}\vert_{x=-\infty}^{x=+\infty}=\hat{f'}(z)+iz\hat{f}(z)$
But $f(x)e^{izx}\vert_{x=-\infty}^{x=+\infty}=0$ , so we get the desired equality.
My quetion here is why we should suppose that $\lim\limits_{{x}\to \infty}xf(x)=\lim\limits_{{x}\to -\infty}xf(x)=0$ and not just that $$\lim\limits_{{x}\to \infty}f(x)=\lim\limits_{{x}\to -\infty}f(x)=0$$ ? Because then we would have $f(-\infty)e^{iz(-\infty)}=f(\infty)e^{iz(\infty)}=0$
Thanks in advance for your time and effort.