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Why Schwarz space is given by the set of $f\in \mathcal C^\infty (\mathbb R^n)$ s.t. $$\sup_{x\in \mathbb R^n}|(1+|x|^N)\partial _x^\alpha f(x)|<\infty ,$$ where $\alpha \in \mathbb N^n$ and $N\in\mathbb N$.

To have $$\sup_{x\in \mathbb R^n}|x|^N|\partial _x^\alpha f(x)|<\infty,$$ it is not enough ? I have the impression that both definition are equivalent, no ?

MSE
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2 Answers2

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Yes, the two conditions are equivalent as $|x| \to \infty$, but one may not forget that we are interested in dominant functions $g$ such that $$ \int_{\mathbb R^n} |g|\:dx <\infty $$ which is the case for $$ g(x)=\frac1{(1+|x|^N)}, $$ as $N$ is great enough and which is not the case for $$ g(x)=\frac1{|x|^N}. $$

Olivier Oloa
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    Indeed, it's a good observation. I didn't expect that. – MSE Jul 03 '16 at 17:07
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    We have two families of seminorms. What you say here explains why one family is often more convenient to use than the other, even though it looks more complicated at first. But it also gives the impression that they are not equivalent (especially the language "...equivalent as $|x|\to\infty$< but..). In fact the two families of seminorms are equivalent! The OP asked why the one with $|x|^N$ was not enough - it is enough... – David C. Ullrich Jul 03 '16 at 17:41
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In fact using $|x|^N$ is enough.

The seminorm $\sup|x|^N|f(x)|$ is not equivalent to the seminorm $\sup(1+|x|^N)|f(x)|$.

But that's irrelevant to the definition of the Schwarz space, which involves the family of seminorms for all $N$. The family of seminorms defined by $|x|^N$ is equivalent to the family of seminorms defined by $1+|x|^N$. So the two families of seminorms do define the same space.

This is clear, because on the one hand $|x|^N\le1+|x|^N$, while on the other hand $$\sup(1+|x|^N)|f(x)|\le\sup|f(x)|+\sup|x|^N|f(x)|.$$

David C. Ullrich
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  • Maybe there is something simple that eludes me, but if $\sup(1+|x|^N)|f(x)|<\infty$ then of course $\sup|x|^N|f(x)|<\infty$. If $\sup |x|^N|f(x)|<\infty$ then $\sup(1+|x|^N)|f(x)|=\sup(|x|^N|f(x)| \cdot (1+1/|x|^N))<\infty$. The two conditions are equivalent. – Olivier Oloa Jul 03 '16 at 17:47
  • @OlivierOloa Oops. Of course those two conditions are equivalent, it's the two seminorms that are not. – David C. Ullrich Jul 03 '16 at 17:49
  • Then I'm not convinced by your answer. – Olivier Oloa Jul 03 '16 at 17:51
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    @OlivierOloa Not convinced of what? Each seminorm in either family is dominated by finitely many from the other family, so they define the same space with the same topology. Hence $|x|^N$ "is enough". – David C. Ullrich Jul 03 '16 at 17:53
  • I'm not convinced that $\sup_{x\in \mathbb R^n}|x|^N|\partial x^\alpha f(x)|<\infty$ could have been chosen as a pertinent statement regarding that we are willing to deal with dominant functions $g$ verifying $\int{\mathbb R^n} |g|:dx <\infty$. – Olivier Oloa Jul 03 '16 at 18:10
  • @OlivierOloa ??? I've already agreed that using $1+|x|^N$ is often a better idea, for the reason you give. I didn't say anything about what is or is not a "pertinent statement" - I don't even recall the definition. Are you disputing that we could use $|x|^N$ in the definition of the Schwarz space? It's a fact that we can. – David C. Ullrich Jul 03 '16 at 18:19
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    Ok, using $1+|x|^N$ was a better idea, that's why it has been chosen. I'm not disputing something else. – Olivier Oloa Jul 03 '16 at 18:27
  • @OlivierOloa Fine. I did say often a better idea, though; sometimes knowing the family is equivalent to the one with $|x|^N$ is convenient. Or $(1+|x|^2)^{N/2}$, or for that matter just $|x|^{2N}$ (the last two being smooth). For example when you want to show the Fourier transform is an isomorphism you need to show you can switch the derivative and the multiplier; using $1+|x|^N$ seems less convenient there... – David C. Ullrich Jul 03 '16 at 18:34