I'm reading Fourier Analysis by Rami Shakarchi and Elia.M.Stein. When I started reading the chapter about Fourier Transform on R I came across very difficult inequalities. One of them take me a long time to prove it. Now I am with one of these. I have tried a lot of things like triangule inequality, Newton's Binomial, etc. I would appreciate some help. The problem is the following:

In the context of prooving that the convolution of two functions in the Schwartz Space lies again in the Schwartz Space it is neccesary to demonstrate that if a function $g$ belongs to the space then the following inequality holds:

$$|x|^l |g(x-y)| \leq A_l(1 + |y|)^l \ \forall \ l \geq 0$$ Remember, $S(\mathbb{R})$, the Schwartz Space, is the set of functions $f$ such that $sup |x|^l|f^{(k)}(x)| < \infty \ \forall \ k,l$

The book gives a hint, it says that consider the cases $|x|> 2|y|$ and $|x|<2|y|$.


  • $\begingroup$ In addition to the direct inequality in the answer below, the fact that Fourier transform converts convolution to multiplication (under various mild hypotheses), and the somewhat easier (?) assertion that products of Schwartz functions are Schwartz... $\endgroup$ Jul 18, 2018 at 22:25
  • $\begingroup$ Yeah but for prove the fact that Fourier maps convolutions to multiplication you need to show before that te convolution lies in S (R). $\endgroup$
    – HFKy
    Jul 19, 2018 at 3:14
  • $\begingroup$ well, there are many possible logical lines here, for sure, but just showing that the convolution is in $L^1$ is sufficient for existence of Fourier transform, etc. Anyway, yes, some inequalities must be proven... $\endgroup$ Jul 19, 2018 at 12:18

1 Answer 1


Recall the following trivial inequality; $$\lvert x\rvert ^{l} \leq \left(\, \lvert x-y \rvert \, + \lvert y \rvert \, \right)^{l}\leq (2\max(\, \lvert x-y\rvert, \, \lvert y \rvert \, )\, )^{l} \leq\, 2^{l} \, \lvert x-y\rvert^{l} + 2^{l}\, \lvert y \rvert ^{l} \qquad \forall l\geq 0$$

Set $$C_{l} := \sup_{x\in\mathbb{R} } \, \lvert x\lvert ^{l} \,\lvert g(x) \, \rvert <\infty \qquad \forall l\geq 0$$

Now using the above estimate, we get $$\lvert x \rvert^{l} \, \lvert g(x-y) \rvert \leq 2^{l} \, \lvert x-y\rvert^{l} \, \rvert g(x-y)\rvert + 2^{l}\, \lvert y \rvert^{l} \, \lvert g(x-y) \rvert \leq \\ 2^{l}\, C_{l}+ 2^{l} \, C_{0} \, \lvert y \rvert^{l} \leq \underbrace{2^{l}(C_{l}+C_{0})}_{=A_{l}}\, (1+\lvert y\rvert)^{l}$$

This establishes the desired inequality. To finish off the proof entirely we may write $$ \lvert x \rvert^{l} \, \lvert (g\ast f)(x) \rvert \leq \int _{\mathbb{R}} \lvert x \rvert^{l} \, \lvert g(x-y) \rvert \, \lvert f(y) \rvert \, dy \leq A_{l} \int_{\mathbb{R}} (1+\lvert y \rvert )^{l+2} \, \lvert f(y) \rvert \frac{dy}{(1+\lvert y \rvert)^{2}} \leq \\ A_{l} \, C_{l+2} \,\int_{\mathbb{R}} \frac{dy}{(1+\lvert y \rvert)^{2}}< \infty $$ Since the bound is independent of $x\in \mathbb{R}$, the claim now follows.

  • $\begingroup$ I am not too sure about some of the inequalities. $\endgroup$
    – HFKy
    Jul 18, 2018 at 20:29
  • $\begingroup$ Firstable, thanks for the help. I do not understand why $(2 max (|x-y|,|y|)^l \leq 2^l |x-y|^l + 2^l |y|^l$. $\endgroup$
    – HFKy
    Jul 18, 2018 at 20:35
  • 1
    $\begingroup$ Ooooh! I get it!!! $\endgroup$
    – HFKy
    Jul 18, 2018 at 21:10

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