2
$\begingroup$

In a book (In portuguese, it can be reed in here in the first pages) the author says that it is clear, by the definition, that the test space $\mathcal{D}(\mathbb{R}) \subset \mathcal{S}(\mathbb{R})$ is a subset of the Schwartz space(he uses other notation for the Schwartz space). The definition he uses are below

The Schwartz space is a space formed by the functions in $$C^{\infty}(\mathbb{R}) = \{f \operatorname{is a function}:f \operatorname{is continuous and every derivative}f^{(q)}\operatorname{for every}q\in\mathbb{N}\operatorname{exists and is also continuous}\}$$ and satisfies that, for every $m,q \in \mathbb{N}_0 = \{0,1,2,\dots\}$ $$\lim_{\vert x \vert \rightarrow \infty}(1 + \vert x \vert)^m\vert f^{(q)}(x)\vert = 0$$

and the definition of the test space is

The test space is a space formed by the functions in $C^{\infty}(\mathbb{R})$ such that the suport of the functions are compact, i.e, the set defined by $$\operatorname{supp}f = \overline{\{x \in \mathbb{R}: f(x)\neq 0\}}$$ where the line above means closure, this set must be a compact set.

Finally my question is that, to me, is not as trivial as it sounds that every test function satisfy the conditions to be a Schwartz function. There is a proof of this or it is really so trivial that this fact does not deserve a demonstration?

  • 1
    It is really that trivial as each derivative of a test function vanishes outside of a compact set. Therefore $(1+|x|)^m|f^{(q)}(x)| = 0$ for $x$ big enough. It doesn't only approach $0$ for large $x$, it **is** $0$ for large $x$. – 2017-01-09
  • 0
    By definition if $\varphi \in D(\mathbb{R})$ then $\varphi \in S(\mathbb{R})$. But also, if $\varphi_n \to \varphi \in D(\mathbb{R})$ then $\varphi_n \to \varphi \in S(\mathbb{R})$. So this is an inclusion of sets and of topological spaces (in $S(\mathbb{R})$ there are more elements and more open sets) – 2017-01-09

2 Answers 2

1

The support of any derivative of $f$ is a subset of the support of $f$ itself (can you see that?). Therefore, any compactly supported function has compactly supported derivatives, for which Schwartz's condition is obvious.

  • 0
    You are saying that $\operatorname{supp}f^{(n)} \subset \operatorname{supp}f$? I cannot see that but, worst, I cannot see how this helps.. – 2017-01-09
  • 0
    Based on other answer I understand what you mean, but now the problem is that how it works? Why the limit does not explode caused by the bounded values of the support ? Am I confusing the things? – 2017-01-09
  • 1
    Sorry @Rafael Wagner, but I don't understand what do you mean - "Why the limit does not explode caused by the bounded values of the support ?". Every $f^{(n)}$ is *zero* outside the support, which on the other hand is contained in a sphere of radius $R$. This yields, for an arbitrary $g(x)$: $$g(x)f^{(n)}(x)=0$$ for $|x|>R$, so $$\lim _{|x|\to \infty} g(x)f^{(n)}(x)=0$$ – 2017-01-09
  • 0
    I completely understand now! I was thinking in the limit entering the region of the support but because of the bounded support (compact in real) in some sense the values in the limit get out of the region of non-vanishing $f^{(n)}$ and all the values are zero. I think now I understand, sorry for my confusing English above... – 2017-01-09
0

If you're outside a function's support, it is identically zero. Compact sets in $\mathbb{R}^n$ are bounded, so for each test function $f$, there is a radius $R(f)$ outside which it is identically zero. The derivative of a function that is identically zero is also zero, and so $f$ satisfies all the limit criteria required to be in the Schwartz space.