Asymptotic behaviour of Fourier transform: $|F[f]|=|\lambda^{-k}F[f^{(k)}]|$ for absolutely continuous $f$ I read in Kolmogorov-Fomin's (p. 429 here) that if function $f:\mathbb{R}\to\mathbb{C}$ is such that $f^{(k-1)}$ [the $(k-1)$-th order derivative] on any finite interval and if $f,...,f^{(k)}\in L_1(-\infty,\infty)$ [i.e. they are Lebesgue-summable on $\mathbb{R}$], then$$F[f^{(k)}]=(i\lambda)^k F[f]$$where $F[f]$ is the Fourier transform defined by $F[f](\lambda)=\int_{\mathbb{R}} f(t)e^{-i\lambda t}d\mu_t$, where $\mu_t$ is the linear Lebesgue measure on $\mathbb{R}$.
I know, from the proof given in the same page, that such an equality holds for $k=1$ (lemma (3)). Provided that $f$ is absolutely continuous on any finite interval, induction guarantees that $F[f^{(k)}]=(i\lambda)^k F[f]$, as far as I understand. I do not think we may drop the hypothesis of the absolutely continuity of $f$.
Then the book, talking as if such a statement held in general, in another lemma bearing another number, (4), says that if $f^{(k)}\in L_1(-\infty,\infty)$ then $$\lim_{|\lambda|\to\infty}|F[f]|=\lim_{|\lambda|\to\infty}\frac{|F[f^{(k)}]|}{|\lambda|^k}=0.$$I think it is still intended that $f,...,f^{(k-1)}\in L_1(-\infty,\infty)$ and $f$ is absolutely continuous on any finite interval: am I right? I do not think that the hypothesis may be dropped.
This asymptotic estimate is then used to prove the following lemma (5), which says that if $f''$ exists and belongs to $L_1(-\infty, \infty)$, then $F[f]$ is [Lebesgue] absolutely integrable, but, again, I think that we must consider it under the hypothesis that $f$ is absolutely continuous on any finite interval and $f,f'\in L_1(-\infty,\infty)$. Is not that so? I should admit that this much celebrated textbook is not so easy for me to do exegesis on...
Thank you so much!!!
 A: The asymptotic estimate $|F[f](\lambda)|\to 0$ holds under much weaker conditions. I can't read Kolmogorov-Fomin, so I don't know what they're claiming, but here's a proof assuming only that $f$ is in $L^1(-\infty,\infty)$.
The idea is to approximate $f$. First, we know how to show that $F[g'](\lambda) = i\lambda F[g](\lambda)$ if $g$ is $C^\infty$ and compactly supported. So for $g\in C_c^\infty$ we have 
$$
|F[g](\lambda)| = \frac{|F[g'](\lambda)|}{|\lambda|}\to 0
$$
as $\lambda\to\infty$.
Since $C_c^\infty$ is dense in $L^1$ (w.r.t $L^1$-norm), given $\varepsilon>0$ we may find $g\in C_c^\infty$ with $\Vert f-g \Vert_{L^1}<\varepsilon$. Then
$$
\left|\int_{-\infty}^\infty f(x)e^{-i\lambda x}~dx\right| \leq \left|\int_{-\infty}^\infty (f(x)-g(x))e^{-i\lambda x}~dx\right| + \left|\int_{-\infty}^\infty g(x)e^{-i\lambda x}~dx\right|.
$$
So we have the bounds
$$
\left|F[f](\lambda)\right| \leq \Vert f-g \Vert_{L^1} + \left|F[g](\lambda)\right| < \varepsilon + \left|F[g](\lambda)\right|.
$$
As $|\lambda|\to\infty$ the second term goes to $0$, and since $\varepsilon>0$ was arbitrary this shows that $|F[f](\lambda)|\to 0$ as $|\lambda|\to\infty$.
Now you can do this for any number of derivatives on $g$ since $g$ is compactly supported, which gives you many more asymptotic estimates in terms of $g$ and its derivatives. Notice that we did not even use the derivative of $f$ in this proof.
On the other hand the identity $i\lambda F[f](\lambda) = F[f'](\lambda)$ is of course going to need some more regularity on $f$ than just integrability.
