2
$\begingroup$

When reading the proof of this proposition in Stein's Fourier Analysis: An introduction, I cannot follow some of the process.

I wonder why, in the second row of the proof, we have the conclusion that $$\sup_x |x|^l |g(x-y)| \leq A_l(1+|y|)^l$$ for some constant $A_l$. There is a direction for the reader to check but I still cannot figure it out.

The author Stein said that consider separately the two cases $|x| \leq 2|y|$ and $|x| \geq 2|y|$ in both the first and the last paragraph of the proof. However, I cannot see where it will be used.

Could anyone give an explanation that is easier to understand?

Here is the definition of "rapidly decreasing" and the "Schwartz space" from Stein's book for reference:

The Schwartz space on $\mathbb{R}$ consists of all infinitely differentiable functions $f$ so that $f$ and its all derivatives $f', f'',...,f^{(l)},...$, are rapidly decreasing, in the sense that $$\sup_{x \in \mathbb{R}} |x|^k |f^{(l)}(x)| < \infty \quad \text{for all } k,l \geq 0$$ and we denote the Schwartz space by $\mathcal{S}(\mathbb{R})$.

img1 img2

$\endgroup$
1
  • $\begingroup$ why $x^l(f*g)(x)$ is bounded? $\int_{-\infty}^{\infty} |f(y)|(1+|y|)^l dy$ converges? $\endgroup$
    – eraldcoil
    Mar 2, 2020 at 2:23

1 Answer 1

0
$\begingroup$

For $|x|\leq 2|y|$ we can estimate \begin{align*} |x|^l |g(x-y)| \leq 2^l |y|^l ||g||_\infty \leq 2^l||g||_\infty (1+|y|)^l. \end{align*} For $|x|> 2|y|$ we can estimate \begin{align*} |x|^l |g(x-y)| &\leq |x|^l \frac{1}{|x-y|^l} \ \sup_z |z|^l|g(z)|\\ &\leq |x|^l \frac{1}{(|x|-|y|)^l} \ \sup_z |z|^l|g(z)| \\ &\leq |x|^l \frac{1}{(|x|-\frac{1}{2}|x|)^l} \ \sup_z |z|^l|g(z)|\\ &\leq |x|^l \frac{1}{(\frac{1}{2}|x|)^l} \ \sup_z |z|^l|g(z)|\\ &\leq 2^l \sup_z |z|^l|g(z)| \leq 2^l \sup_z \big(|z|^l|g(z)|\big)\,(1+|y|)^l. \end{align*}

$\endgroup$
1
  • $\begingroup$ Thank you very much. $\endgroup$
    – ScienceAge
    Feb 4, 2020 at 13:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .