A step in a proof that $\mathcal S(\mathbb R)$ is closed under convolution

Question

This answer to a question on why the Schwartz functions are closed under convolution seems to be making the claim that for any $x\in \mathbb R$

$$\sup_{y\geq x/2} g(y)\leq \frac {C_n} {1+|x|^n}$$

where $C_n$ is a positive constant such that

$$\forall x\in\mathbb R,\ g(x)\leq \frac {C_n} {1+|x|^n}$$

What justifies this? Or am I misunderstanding the proof? Also, can this proof be generalized to higher dimensions?

score 1 · Answer 1 · answered Oct 27 '16 at 10:40

1

What is needed (and written) in the linked answer is that $$\sup_{|y| \geqslant |x|/2}\left| g(y)\right|\leqslant \frac {C_n} {1+|x/2|^n}.$$ This is indeed true, since for $|y| \geqslant |x|/2$, $$\left| g(y)\right|\leqslant \frac {C_n} {1+|y|^n}\leqslant \frac {C_n} {1+|x/2|^n}.$$

answered Oct 27 '16 at 10:40

Davide Giraudo

181,608

So we're just appealing to the monotonicity of $1/(1+x^n)$? – Jack M Oct 27 '16 at 11:29
Yes, nothing more. – Davide Giraudo Oct 27 '16 at 15:50

A step in a proof that $\mathcal S(\mathbb R)$ is closed under convolution

1 Answers1