In C.D.Sogge's *Fourier Integrals in Classical Analysis* pp.128-129, he proved Lemma4.2.3(Tauberian Lemma):


**Lemma.** Let$g(\lambda)$ be a piece-wise continuous tempered function of $\mathbb{R}$. Assume that for $\lambda>0$
$$|g(\lambda+s)-g(\lambda)|\leq C(1+\lambda)^a,0<a\leq1$$
Then if $\hat{g}(t)=0$ when $|t|\leq1$(The hat means Fourier transform) we have:
$$|g(\lambda)|\leq C(1+\lambda)^a$$#


by using an identity:


$$|G(\lambda)|=|(G'\*\psi)(\lambda)|\leq C(1+\lambda)^a\int|\psi(s)|(1+|s|)^ads\leq C(1+\lambda)^a$$


My confusion is that when the $\hat{\psi}(t):=\frac{\eta(t)}{it}$ where $\eta\in\mathcal{S}$ satisfying $\eta(t)=0$ when $|t|\leq\frac{1}{2}$ and $\eta(t)=1$ when $|t|>1$. It could be that $\psi$ is not integrable at all since $\frac{1}{t}$ is not necessarily integrable at infinity. *So how could this identity holds?*


What is more, in the same book, pp.127 in the proof to Theorem4.2.1 the same argument arise. Altough this can be argued using oscillatory integral in Chap1.


**Remarks from Professor** It could be found in Hormander's PDO Vol.3 this Lemma holds under a more restrictive assumption, and it suffices for later applications. 


**My question is whether the identity above can be argued correctly?** If so, what technique should be used?(I will be deeply appreciated if a detail explanation is given.)


Thanks.



Ask the validity of Tauberian lemma in Sogge's book
The identity $\hat{\psi}(t)=(it)^{-1}\eta(t)$ you quoted is indeed hold except for the typo that $\eta\in C^{\infty}$ instead of Schwartz functions. Although the function on the right hand side is not integrable as you have noticed, itself can be the Fourier transform of a $L^1$ function. The more proper setting here is to use the Fourier transform of tempered distribution, then the problem you concerned would not exist. In fact, the smoothness of the function on the RHS and vanishing near origin allow one to do integration by parts arbitrary times, so $\psi$ is rapidly decay at $\infty$. 


Another simple way to justify that $(it)^{-1}\eta(t)\in \mathcal{F}L^1$ is through the so called Bernstein theorem which says that 
$$
H^{k}(\mathbb{R}^n)\hookrightarrow\mathcal{F}L^1, \quad k>\frac{n}{2},
$$
where $H^{k}(\mathbb{R}^n)$ is the usual Sobolev space. The proof is not hard and you can find it in many standard text books (in fact you can prove it by breaking the integral of the Fourier transform into two parts, and just use Schwartz inequality). Now in this particular case, all you need to do is to check that $(it)^{-1}\eta(t)\in L^2$, and $\frac{d}{dt}(\frac{\eta(t)}{it})\in L^2$, which I believe is obvious to you.


