I was reading sec 6.3 form Rudin Functional analysis enter image description here

enter image description here

I do not understand why highlighted set belong to $\beta$?

$D(\Omega)$ is test function space on open set

Please help me

Any Help will be appreciated

  • $\begingroup$ Your question is unclear. Are you tried to say "I do not understand highlighted set , which belongs to $\beta$", or "I do not understand why the highlighted set belongs to $\beta$". In either case, it would help if you wrote out the definition of $\mathscr D(\Omega)$ for us. $\endgroup$ – Omnomnomnom Jan 16 at 13:51
  • $\begingroup$ By your previous question, it seems that $\mathscr D(\Omega)$ is the set of all $C^\infty$ functions over $\Omega$ with compact support. $\endgroup$ – Omnomnomnom Jan 16 at 13:54
  • $\begingroup$ Yes Sir.That is defination Thanks a lot $\endgroup$ – MathLover Jan 16 at 13:54

The set $S = \{\phi \in \mathscr{D}(\Omega): |\phi(x_n)| < c_n\; n \in \mathbb{N}\}$ is obviously convex and balanced.

Now take arbitrary compact $K \subset \Omega$ and consider $S \bigcap \mathscr{D}_K$. Since sequence $x_n$ does not have a limit point in $\Omega$ then $K \bigcap \{x_n:\; n \in \mathbb{N}\}$ is finite because $K$ is compact. Let $\{x_{n_1},\dots,x_{n_m}\} = K \bigcap \{x_n:\; n \in \mathbb{N}\}$. Then $S \bigcap \mathscr{D}_K = \{\phi \in \mathscr{D}_K: |\phi(x_{n_k})| < c_{n_k}\; k = 1,\dots,m\}$. This set is obviously open in $\mathscr{D}_K$ since it is a finite intersection of open sets $\bigcap\limits_{k = 1}^{m} \{\phi \in \mathscr{D}_K: |\phi(x_{n_k})| < c_{n_k}\} = S \bigcap \mathscr{D}_K \in \tau_K$.

By definition 6.3 (b) $S$ belongs to $\beta$.

  • $\begingroup$ Thanks a lot Sir $\endgroup$ – MathLover Jan 18 at 4:15

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