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Let $G$ be an open subset of $\mathbb R ^n$ and $D(G)$ denotes the set of smooth functions with compact support in $G$.

Consider following families of seminorms,

  1. For $f\in D(G)$

$||f||_N = \sup \{ | D^{\alpha}f(x)|: x\in G,|\alpha| \leq N \}$ for $N\in \mathbb N$.

  1. Let $K_n$ be a nested sequence of compact sets which exhaust $G$ then for $f \in D(G)$ define

$\nu_N (f) = \sup \{ | D^{\alpha}f(x)|: x\in K_N,|\alpha| \leq N \}$ for $N\in\mathbb N$.

where $\alpha$ is a multi-index.

Are topologies induced on $D(G)$ by above two families same?

My guess is NO but I am unable to prove it.

EDIT- I am interested in this question because on $C^\infty(G)$ we give topology induced by family 2. In chapter 6 of Rudin’s Functional Analysis, it says family 1 doesn’t give good topology on $D(G)$ (it’s not complete). But Rudin didn’t talk about family 2 even though it’s subspace topology on $D(G)$ as a subset of $C^\infty(G)$.

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  • $\begingroup$ In the definition of $\nu_N$, you wrote $x\in K_N$. Did you mean $x\in K_m$? Since $N$ is already used as the size of the maximal multi index? $\endgroup$ Dec 26 '18 at 7:33
  • $\begingroup$ It’s $x\in K_N$. In fact it’s a topology which we give on $C^\infty (G)$. $\endgroup$
    – Mayuresh L
    Dec 26 '18 at 8:07
  • $\begingroup$ Can you define the natural basis of neighborhoods of $0$ in both topology ? (the translates of those neighborhoods are a basis of the topology). Will a neighborhood in the 1st topology be open in the 2nd topology, or is it containing some open set of the 2nd topology ? $\endgroup$
    – reuns
    Dec 27 '18 at 6:13
  • $\begingroup$ @reuns I tried that but couldn’t figure out! $\endgroup$
    – Mayuresh L
    Dec 27 '18 at 6:39
  • $\begingroup$ Let $\|f\|_{K_n,n}= \sup_{x \in K_n, |\alpha| \le n}|D^\alpha f(x)|$ and $U_1(r,N) = \{ f \in C^\infty_c(G), \forall n \le N, \|f\|_{G,n}<r_n \}$. Is it open in the 1st topology ? Does it contain an open set in the 2nd topology ? (let $T(C^\infty_c(G), \|.\|_{K_n,n})$ be the set of open sets of the topological vector space $C^\infty_c(G), \|.\|_{K_n,n}$, then $T_2 = \bigcap_n T(C^\infty_c(G), \|.\|_{K_n,n})$) $\endgroup$
    – reuns
    Dec 27 '18 at 12:19
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With the topology from 2, for functions in $C^\infty(\Omega)$, $\lim_n f_n = f$ can be thought as of $D^\alpha f_n \rightarrow D^\alpha f$ uniformly on any compact subsets, and for each $|\alpha| < \infty$.

One can construct $f_n\in \mathcal{D}(\Omega)$ and yet the limit $f\in C^\infty(\Omega)$ does not have compact support.

2 gives a weaker topology than 1 on $\mathcal{D}(\Omega)$, since 1 is "converge uniformly" and 2 is "converge locally uniformly". For example, let $\phi$ have support in $[0,1]$, define $$\psi_m(x) = \phi(x) + \phi(x-1) + \cdots + \phi(x-m)$$ this sequence is Cauchy in 2, but not Cauchy in 1. This also shows that $\mathcal{D}(\Omega)$ with the topology induced by 2 is not complete. (Similar to Rudin's example on page 151.)

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  • $\begingroup$ So we are not interested in topology 2 because it also is not complate. But how to justify the line “2 gives a weaker topology than 1 on $D(\Omega)$”? $\endgroup$
    – Mayuresh L
    Dec 28 '18 at 15:37
  • $\begingroup$ Two families of semi-norms generate the same topology if and only if any semi-norm from one family can be bounded by some semi-norm of the other family. In this case $ \nu_N \leq C\|\cdot \|_N$ is clear, and I don't see a way to have the bound the other way around. $\endgroup$
    – Xiao
    Dec 28 '18 at 16:32
  • $\begingroup$ Maybe the more precise statement is given two families of semi-norms $\mathcal P$ and $\mathcal Q$, $\tau_P$ is finer, i.e. $\tau_Q \subset \tau_p$ if for each $q\in \mathcal {Q}$, there exist $n$ semi-norms $p_1, \cdots, p_n \in \mathcal {P}$ and a constant $C$ such that $$q\leq C \max \{ p_1, \cdots p_n\}.$$ $\endgroup$
    – Xiao
    Dec 28 '18 at 16:43
  • $\begingroup$ Could you give reference for statement in your last comment?. Thanks a lot for your help. $\endgroup$
    – Mayuresh L
    Dec 28 '18 at 17:08
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    $\begingroup$ @MayureshL It should be easy to see with the metric explicitly. For $k,l\geq M$, $|\psi_k -\psi_l|$ will vanish on $[0, M]$. This means $\nu_N (|\psi_k -\psi_j|)$ will all be zero for $N=1, 2,\cdots, M'$. $\endgroup$
    – Xiao
    Dec 31 '18 at 17:58

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