I just learn some distribution theory.

In Friedlander's book, "Introduction to the theory of distributions" (page 97), he said: The dual of $\mathscr{S}(R^n)$ (Schwartz Space) consists of continuous linear forms on $\mathscr{S}(R^n)$; Obviously, such a linear form $u$ is continuous if and only if there is a constant $C\geq 0$ and a nonnegative integer N such that \begin{equation} |<u,\phi>|\leq C \sum_{|\alpha|,|\beta|\leq N} \sup |x^\alpha\partial^\beta \phi| \quad, \forall \phi\in \mathscr{S}(R^n) \end{equation}

This means that all tempered distribution always have a order (right?). I can't prove that "obviously " fact. Can anyone help me or give a hint?

By the way: $\mathscr{S}(R^n)$ means Schwartz space. Here is the wiki link, http://en.wikipedia.org/wiki/Schwartz_space

And linear form is "continuous" means it a linear & it is continuous in the topological sense.


1 Answer 1


It's the standard "continuous implies bounded" argument: Let \begin{equation} \Vert\phi\Vert_N:= \sum_{|\alpha|,|\beta|\leq N} \sup |x^\alpha\partial^\beta \phi|. \end{equation} If your inequality fails then for every $N$ then there is a $0\neq\phi_N\in\mathscr S$ s.t. $$|\langle u,\phi_N\rangle|\geq N\Vert\phi_N\Vert_N.$$ Let $\psi_N=\phi_N/(N \Vert\phi_N\Vert_N)$. Then $\psi_N\to 0$ in $\mathscr S$, but $|\langle u,\psi_N\rangle|\geq 1$, so $\langle u,\psi_N\rangle$ doesn't converge to $0$.

  • $\begingroup$ I think you are right. Thanks. $\endgroup$ Apr 13, 2014 at 21:28

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