# Are topologies induced by following families of seminorms same?

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 $$C^\infty(G)$$.

• 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? Dec 26 '18 at 7:33
• It’s $x\in K_N$. In fact it’s a topology which we give on $C^\infty (G)$. Dec 26 '18 at 8:07
• 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 ? Dec 27 '18 at 6:13
• @reuns I tried that but couldn’t figure out! Dec 27 '18 at 6:39
• 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 $T_2 = \bigcap_n T(C^\infty_c(G), \|.\|_{K_n,n})$) Dec 27 '18 at 12:19

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.)

• 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 $D(\Omega)$”? Dec 28 '18 at 15:37
• 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.
– Xiao
Dec 28 '18 at 16:32
• 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 $$q\leq C \max \{ p_1, \cdots p_n\}.$$
– Xiao
Dec 28 '18 at 16:43
• Could you give reference for statement in your last comment?. Thanks a lot for your help. Dec 28 '18 at 17:08
• @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 $N=1, 2,\cdots, M'$.
– Xiao
Dec 31 '18 at 17:58