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.)


  • $\begingroup$ By PDO I mean The Analysis of Linear Partial Differential Operators $\endgroup$
    – Henry.L
    Commented Jun 3, 2015 at 10:18
  • $\begingroup$ And Sogge provided a flawed proof for Theorem 3.1.1 on pp.95-96.(According to my Prof.) $\endgroup$
    – Henry.L
    Commented Jun 3, 2015 at 10:39

1 Answer 1


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.

  • $\begingroup$ Thanks. Subtle is the lord. And Thm 3.1.1 seems to be of the same flaw. :-) $\endgroup$
    – Henry.L
    Commented Jun 6, 2015 at 3:42
  • $\begingroup$ After a while, I found it Not clear at all why itself can be the Fourier transform of a L1 function...So I cancel my selection and hope for a better explanation. $\endgroup$
    – Henry.L
    Commented Jun 13, 2015 at 10:44
  • $\begingroup$ Hi, Henry, I edited the answer, hope it's clear enough now. $\endgroup$
    – Tomas
    Commented Jun 13, 2015 at 13:36
  • $\begingroup$ What about your original way? How can I view them as L1 functions' Fourier transform? It may be better if a reference is provided. I may walk through this again now. $\endgroup$
    – Henry.L
    Commented Jun 14, 2015 at 14:35
  • $\begingroup$ Well, it seems to me that once you observed that it is L^2, we resort to the Plancherel theorem and figure out that the distribution has no singularity near the origin. Fair. Thanks a lot! $\endgroup$
    – Henry.L
    Commented Jun 24, 2015 at 23:54

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