0
$\begingroup$

The Schwartz space $\mathcal{S}(\mathbb R^n)$ is the set of all function $f:\mathbb R^n\longrightarrow \mathbb C$ such that $\displaystyle \sup_{x\in \mathbb R^n}|x^\beta \partial^\alpha f(x)|<\infty$ for all $\alpha, \beta\in\mathbb N_0^n$ ($\mathbb N_0=\mathbb N\cup\{0\}$). Can anyone give me an example why $$\rho_{\alpha, \beta}(f)=\sup_{x\in \mathbb R^n} |x^\beta \partial^\alpha f(x)|,$$ is not a norm on $\mathcal{S}(\mathbb R^n)$ for each $\alpha, \beta\in\mathbb N_0^n$?

$\endgroup$
2
$\begingroup$

The $\rho_{\alpha,\beta}$ are norms on $\mathcal{S}(\mathbb{R}^n)$. That they are seminorms is straightforward to verify, and $\rho_{\alpha,\beta}(f) = 0$ only for $f \equiv 0$ follows since $x^\beta$ is only zero on a nowhere-dense subset (the union of finitely many coordinate hyperplanes $\{x_i = 0\}$), so $\rho_{\alpha,\beta}(f) = 0 \Rightarrow \partial^\alpha f \equiv 0$, and that implies that $f$ is a polynomial, but the only polynomial in $\mathcal{S}(\mathbb{R}^n)$ is $0$.

However, that these seminorms are in fact norms is not important. The topology one is interested in is generated by the family of all these (semi-) norms, so whether or not a single one of these is a norm doesn't matter. The topology induced by a finite subfamily of these seminorms is not as useful.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.