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The Schwartz space on $\mathbb{R}^n$ is the function space $$ S \left(\mathbf{R}^n\right) = \left \{ f \in C^\infty(\mathbf{R}^n) : \|f\|_{\alpha,\beta} < \infty,\, \forall \alpha, \beta\in\mathbb{Z}_+^n \right \}, $$ where $\mathbb{Z}_+$ is the set of nonnegative integers, $C^\infty(\mathbb{R}^n)$ is the set of smooth functions from $\mathbb{R}^n$ to $\mathbb{R}$, and $$ \|f\|_{\alpha,\beta}=\sup_{x\in\mathbf{R}^n} \left |x^\alpha D^\beta f(x) \right |. $$

It is said in this note by Hunter we have the following characterization of tempered distributions:

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By definition, $\varphi_n\to\varphi$ in $S$ if and only if $\|\varphi_n-\varphi\|_{\alpha,\beta}\to 0$ for every pair $(\alpha,\beta)$.

Here is my question

How does such $d$ in the statement above possibly exist and why would it imply the continuity of the functional $T$ on $S$?

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This is a general fact about seminormed spaces. You have a family of seminorms $\|\cdot\|_{\alpha,\beta}$. The sets $U_{\alpha,\beta,r} = \{\phi : \|\phi\|_{\alpha,\beta}<r\}$ form a sub-basis of neighborhoods of zero, meaning that their finite intersections yield an basis of such neighborhoods. This is precisely the topology for which the convergence of sequences is described as "$\|\phi_n-\phi\|_{\alpha,\beta}\to 0$ for all $\alpha,\beta$".

Suppose $T$ is a continuous linear functional. Then the set $\{\phi : |T(\phi)|\le 1\}$ is an open set containing the origin. Therefore, it contains a basis neighborhood: a set of the form $\bigcap_{k=1}^n U_{\alpha_k,\beta_k,r_k}$. By homogeneity, it follows that $$ |T(\phi)| \le \max_{k=1,\dots, n} r_k^{-1} \|\phi\|_{\alpha_k,\beta_k} \tag1$$ The existence of $d$ follows from (1): take the maximum of $|\alpha_k|$, $|\beta_k|$.

Conversely,(1) implies the continuity of $T$: this follows directly from the definition.

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  • $\begingroup$ Thanks for your answer. The family of seminorms $\|\|_{\alpha,\beta}$ is infinite. Would you elaborate how you choose the $\alpha_k,\beta_k,r_k$? $\endgroup$ – Jack May 29 '16 at 20:48
  • $\begingroup$ There is no explicit choice. They exist because of how seminorms induce topology. $\endgroup$ – user147263 May 29 '16 at 21:56
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A more general theorem in Folland's Real Analysis can answer this question:

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