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The set of smooth compactly supported functions are contained in the Schwartz space. It is somewhat OK to understand the steps in the proof that $\mathcal D$ is dense in $\mathcal S$ but I do even have hard time of understanding compactly supported smooth functions are also Schwartz functions what I did is as follows:

1) I would like to prove any smooth compactly supported function let's say $u(x)$ has rapidly decaying derivatives. especially $|\partial^{\alpha}x^{\beta}u(x)|\le C_{\alpha\beta}$ for any $\alpha$ and $\beta.$

We know that $u(x)$ has compact support i.e. the closure of $(u^{-1}(\{0\}))^{c}$ is compact.

Roughly speaking we know that $u(x)$ has continuous derivatives of infinite order then by product rule; $\big|\partial^{\alpha}x^{\beta}u(x)\big|=\big|\sum_{\eta+\gamma=\alpha}{\partial^{\eta}(x^{\beta})\partial^{\gamma}(u(x))}\big|,$

3) Since we know that $u(x)$ has compact support i.e. the set where $u(x)\ne 0$ is compact and then each derivative of $u(x)$ of any order is continuous, thus we will be always thinking of the compact subsets of $R^n,$ then each of the term in this sum related to $u(x)$ will be bounded.

4) When we take the maximum of each boundary values for the derivatives of the $u(x)$ say $M_{\alpha},$ we will obtain the following result,

$\big|\partial^{\alpha}x^{\beta}u(x)\big|=\big|\sum_{\eta+\gamma=\alpha}{\partial^{\eta}(x^{\beta})\partial^{\gamma}(u(x))}\big|\le \big|\sum_{\eta+\gamma=\alpha}{\partial^{\eta}(x^{\beta})}\big|M_{\alpha} \le C_{\alpha\beta},$

5) Do I need the above statement or by definition of the Schwartz space, since every derivatives of $u(x)\in C_{c}^{\infty}{(R^{n})}\subset C^{\infty}(R^{n}),$

6) then there is an equivalent semi-norm that is $\big|x^{\beta}\partial^{\alpha}u(x)|\le C_{\alpha\beta}$ for any $\alpha$ and $\beta,$ in Schwartz space, then since $\partial^{\gamma} u(x)\in C_{c}^{\infty}(R^n),$ then by definition of compactly supported smooth function that these are bounded on compact sets otherwise we will be getting $0$ , thus $u(x)$ lies in the Schwarz space.

7) Could you tell me if there is another way to explain that $C_{c}^{\infty}(R^n)\subset \mathcal S(\mathbb R^n)$ in a rigorous way?

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1 Answer 1

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I will not answer 1-6, since

7) The easiest way is just to use that every function in $C_c^\infty$ is bounded. Note that $x^\beta u(x)$ is also in $C_c^\infty$ and so is $\partial^\alpha(x^\beta u(x))$. Hence there are constants $C_{\alpha\beta}$ such that $\sup_x |\partial^\alpha(x^\beta u(x))| < C_{\alpha\beta}$.

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  • $\begingroup$ ok, that is exactly what I wrote in 6. Thank you $\endgroup$
    – user92604
    Sep 1, 2013 at 19:05

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