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I have one problem with Schwartz space. I remind you which definitions which I am using.

Multi-index: $\alpha=(\alpha_{1}, \ldots , \alpha_{n}) \in \mathbb{Z}_{+}^{n}$, length of multi-index: $| \alpha | = \sum_{i=1}^{n} \alpha_{i}$, partial differential operator: $$D^{\alpha} = \frac{\partial ^{|\alpha|}}{\partial x_{1}^{\alpha_{1}} \ldots \partial x_{n}^{\alpha_{n}}}.$$ Moreover $|x|=\sqrt{x_1^2+ \ldots + x_n^2}$, Schwartz space: $$\mathcal{S}(\mathbb{R}^n) = \{f \in C^{\infty}(\mathbb{R}^n) \colon \sup_{x \in \mathbb{R}^{n}} (1+|x|)^{m}|D^{\beta}f(x)| < \infty, \: m \in \mathbb{N}, \: \beta \in \mathbb{Z}_{+}^{n} \}.$$

It is easy to prove that if $f \in \mathcal{S}(\mathbb{R}^n)$ then $D^{\beta}f \in \mathcal{S}(\mathbb{R}^n)$ for every multi-index $\beta$. I need to prove that if $f \in \mathcal{S}(\mathbb{R}^n)$ then $(1+|x|)^{m}f(x) \in \mathcal{S}(\mathbb{R}^n)$ for every integer $m$. Has anyone idea how to prove it?

I found theorem: If $a(x) \in C^{\infty}(\mathbb{R}^n)$ is such that $$\forall_{\beta \in \mathbb{Z}_{+}^{n}} \; \exists_{c_{\beta}>0,\; n_{\beta} \in \mathbb{N}} \colon |D^{\beta} a(x)| \leq {c_{\beta}} \cdot (1+|x|)^{n_{\beta}}$$ then $a(x)\cdot f(x) \in \mathcal{S}(\mathbb{R}^n)$ for all $f \in \mathcal{S}(\mathbb{R}^n)$. So, if we show that $$|D^{\beta} (1+|x|)^{m}| \leq c_{\beta} \cdot (1+|x|)^{n_{\beta}} $$ then it will be proved. I got only $$|D^{\beta} (1+|x|)^{m}| \leq c_{\beta} \cdot (1+|x|)^{k}\cdot |x|^{-|\beta|}$$ for all $k\geq m$, but it doesn't help. I will be gratful for all tips and solutions.

Greetings from Poland, Marcin.

saz
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Marcin
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1 Answers1

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The function $x \mapsto |x|$ is not smooth, so we cannot expect that $x \mapsto (1+|x|) \cdot f(x)$ is differentiable. Let's prove instead that

$$x \mapsto (1+|x|^2)^m \cdot f(x) \in \mathcal{S}(\mathbb{R}^n) \tag{1}$$

By the Binomial theorem, we have

$$(1+|x|^2)^m = \sum_{k=0}^m {m \choose k} \cdot |x|^{2k}$$

Thus, by the product rule,

$$D^{\beta} \big( (1+|x|^2)^m \cdot f(x) \big) = \sum_{\alpha+\gamma=\beta} C_{\alpha,\gamma} \sum_{k=0}^m {m \choose k} \cdot D^{\alpha}(|x|^{2k}) \cdot D^{\gamma}(f(x))$$

Since the Schwartz space is a linear space and since $x \mapsto D^{\alpha} (|x|^{2k})$ is a polynomial (in $x$) for fixed $\alpha \in \mathbb{N}_0^n$, $k \in \mathbb{N}_0$, it suffices to show that functions of the form

$$x \mapsto g(x) := x^{\alpha} \cdot D^{\gamma} f(x))$$

satisfy

$$\sup_{x \in \mathbb{R}^n} (1+|x|)^{\ell} \cdot |g(x)| < \infty \tag{2}$$

for arbitrary $\ell \in \mathbb{N}$. This follows from the fact that

$$|x_j|^{\alpha_j} \leq (1+|x_j|)^{\alpha_j} \leq (1+|x|)^{\alpha_j}$$

i.e.

$$|x^\alpha| \leq (1+|x|)^{\sum_j \alpha_j} = (1+|x|)^{|\alpha|}$$

Consequently,

$$\begin{align*} (1+|x|)^{\ell} \cdot |g(x)| &\leq (1+|x|)^{|\alpha|} \cdot (1+|x|)^{\ell} \cdot |D^{\gamma}f(x)| \\ &= (1+|x|)^{\ell+|\alpha|} \cdot |D^{\gamma} f(x)|. \end{align*}$$

Thus, $(2)$ follows from the definition of the Schwartz space.

saz
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  • Yeah, I proved this like that. But, you wrote:

    Let's prove instead that

    $$x \mapsto (1+|x|^2)^m \cdot f(x) \in \mathcal{S}(\mathbb{R}^n) \tag{1}$$

    Note that this implies in particular

    $$\sup_{x \in \mathbb{R}^n} \bigg|(1+|x|)^{\ell} \cdot D^{\beta} \big((1+|x|)^m \cdot f(x) \big) \bigg|< \infty \qquad (\ell \in \mathbb{N}, \beta \in \mathbb{N}_0^n)$$

    I'm not sure why this is impied.

     – Marcin Sep 01 '13 at 16:01
  • @Marcin I edited my answer, obviously $(1+|x|)^m \cdot f(x)$ has to be smooth, otherwise we can't talk about derivatives at all. Concerning your question: The point is that there exists a constant $c$ such that $$\frac{1}{c} \cdot (1+|x|^2)^{m/2} \leq (1+|x|)^m \leq c \cdot (1+|x|^2)^{m/2} \qquad (x \in \mathbb{R})$$ which means that both functions $x \mapsto (1+|x|^2)^{m/2}$ and $x \mapsto (1+|x|)^m$ grow with the same speed. Since a similar estimate holds for the derivatives, the claim follows. – saz Sep 01 '13 at 16:23
  • Yeah, these inequalities hold, but I am not sure about derivatives. Of which derivatives are you saying? – Marcin Sep 01 '13 at 16:37
  • @Marcin Well, I'm talking about the derivatives of $(1+|x|^2)^{m/2}$ and $(1+|x|)^m$ (of course, not in $x=0$). Consider for example $n=1$ and calculate the derivatives, then you will see that they are still comparable. Alternatively, you can adopt the argumentation from my answer since the derivatives (where existent) $D^{\alpha}(|x|^k)$ are also polynomials. But I don't see why you cling to the factor $(1+|x|)^m$ since -as already mentioned- you cannot expect $(1+|x|)^m \cdot f(x) \in \mathcal{S}$ whereas all works fine for the smooth function $(1+|x|^2)^{m/2}$. – saz Sep 01 '13 at 16:47
  • I can show that $D^{\alpha}(|x|^k) \leq c |x|^{k-|\alpha|}$ but not for every $k$ this is a polynomial. In your inequality $m/2$ isn't integer for odd $m$.

    I'm trying to prove theorem 9.2 from "Lectures on Linear Partial Differential Equations" written by G. Eskin. This book is on Google Books, but sometimes this theorem isn't shown. So, I attache printscreen link. I'm trying to prove equality 9.13, and next inequality. Only these 2 things make problems for me. Next part of proof I proved. Could you please help me with these things?

     – Marcin Sep 01 '13 at 17:49
  • @Marcin With the same argumentation as above, it suffices to show (given the setting in the book) that $y \mapsto h(y) := D^{\alpha}(|y|^m) \cdot D^{\gamma}(f(y))$ is integrable for $m \in \mathbb{N}$ and $\alpha$, $\gamma$ such that $\alpha_j, \gamma_j \in {0,1}$. Now note that $$D^{\alpha}(|y|^m) = c \cdot y^{\alpha} \cdot |y|^{m-2|\alpha|}$$ i.e. $$|D^{\alpha}(|y|^m)| \leq c \cdot |y|^{m-|\alpha|}$$ Since $|\alpha| \leq n$, $m \geq 1$, this implies that $h$ is integrable on $B_{\mathbb{R}^n}(0,1)$. – saz Sep 01 '13 at 19:16
  • On the other hand, for $|y| \geq 1$, we have $$|D^{\alpha}(|y|^m)| \leq c \cdot |y|^m$$ and by the definition of the Schwarz space, we know that $$|y|^m \cdot |D^{\gamma}f(y)| \leq (1+|y|)^m \cdot |D^{\gamma}f(y)|$$ is integrable. Thus, $h$ is integrable. This proves the existence of the integrals in (9.13) and as far as I can see that's all you need. – saz Sep 01 '13 at 19:19
  • @Marcin So did you succeed in solving your problem? – saz Sep 03 '13 at 17:10
  • Yeah, thanks for this tips. You are right, this integral is finite. What can you tell about next inequality? I am close to prove it but still have a problem. Could you take a look at this PDF file? Problem is with $|y|^{-|\alpha|}$ factor. I mean if $|y| \geq 1$ then problem is solved, but if $|y|<1$ i have no idea what to do. Thanks in advance. – Marcin Sep 04 '13 at 05:40
  • @Marcin Sorry, can't help you with this one. Not even sure whether this estimate is correct at all... – saz Sep 07 '13 at 10:00