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This is probably a trivial question,

Referring to this question. If we define with $\mathcal{D}_K$ the space of all functions in $C^{\infty}(\mathbb{R}^N)$ whose support is a subset of $K$ (which is a compact set in $\mathbb{R}^N$. We know that $\mathcal{D}_K$ is a closed subspace of $C^{\infty}(\Omega)$ where $\Omega$ is an open set in $\mathbb{R}^N$. A topology on such space can be defined by the family of seminorms

$$ p_N(f) = \max \left\{|D^{\alpha}f(x)| : x \in K_N , |\alpha| \leq N \right\} $$

where $K_N$ is a sequence of compact sets such that $K_N \subset \text{int }K_{N+1}$ and $\bigcup K_N = \Omega$.

I'm trying to work out the details for a specific exercise, and I've found a solution here, in this solution however the seminorms used to define the topology is given by

$$ q_\alpha(f) = \sup_{x \in K} |D^\alpha f(x)| $$

What I wonder now is whether the two families of seminorms define the same topology on $\mathcal{D}_K$, if they do it means I can use one rather than the other, whatever the benefits are.

the thing is I don't have a clue how to prove this, I know somehow I need to show that every open generated by $\left\{ p_N \right\}$ can be generated by $\left\{ q_\alpha \right\}$ and viceversa.

There's an observation, given my hypothesis on $\left\{ K_N \right\}$ there's an index, say $l$ such that

$$ K \subset K_l $$

and would mean that for any $k \geq l$ we have

$$ p_k(f) = \max_{|\alpha| \leq N} \left\{ q_i(f) \right\} $$

however I don't think this would suffice to show the two topologies are the same, can anyone give me any insight?

This also reminds me of this exercise from Rudin's Functional Analysis

Problem 8: a) Suppose $\mathcal{P}$ is a separating family of seminorms on a vector space $X$. Let $\mathcal{Q}$ be the smallest family of seminorms on $X$ that contains $\mathcal{P}$ and is closed under max. [This means: if $p_1, p_2 \in \mathcal{Q}$ and $p = \max(p_1,p_2)$ then $p \in \mathcal{Q}$]. If the construction of Theorem 1.37 is applied to $\mathcal{P}$ and $\mathcal{Q}$ show that the two resulting topologies coincide. The main difference is that $\mathcal{Q}$ leads directly to a base, rather than a subbase. b) Suppose $\mathcal{Q}$ as in part (a) and $\Lambda$ is a linear functional on $X$. Show that $\Lambda$ is continuous if and only if there exists a $p \in \mathcal{Q}$ such that $|\Lambda x | \leq M p(x)$ for all $x \in X$ and some constant $M < \infty$.

Especially part (a) of the problem seems to be related to my question.

Thanks

Update

I believe that the answer to my question is given in section 6.2, where the space $\mathcal{D}(\Omega)$ is described, I'll write down the relevant bit, and I'll highlight what confuses me.

Consider a non empty open set $\Omega \in \mathbb{R}^N$. For each compact $K \in \Omega$, the Frechet space $\mathcal{D}_K$ was described in section 1.46. The union of the spaces $\mathcal{D}_K$, as $K$ ranges over all compact subsets of $\Omega$, is the test function space $\mathcal{D}(\Omega)$. It is clear that $\mathcal{D}(\Omega)$ is a vector space, with respect to the usual definition of addition and scalar multiplication of complex functions. Explicitly, $\phi \in \mathcal{D}(\Omega)$ if and only if $\phi \in C^{\infty}(\Omega)$ and the support of $\phi$ is a compact subset of $\Omega$. Let us introduce the norms $$ \left\lVert \phi \right\rVert_N = \max \left\{ |D^{\alpha}\phi(x) | : x \in \Omega, |\alpha| \leq N \right\}, \;\;\;\; (1) $$ for $\phi \in \mathcal{D}(\Omega)$ and $N = 0, 1, 2...$. The restrictions of these norms to any fixed $\mathcal{D}_K \subset \mathcal{D}(\Omega)$ induce the same topology on $\mathcal{D}_K$ as do the seminorms $p_N$. To see this, note that to each $K$ corresponds an integer $N_0$ such that $K \subset K_N$ for all $N \geq N_0$. For these $N$, $\left\lVert \phi \right\rVert_N = p_N(\phi)$ if $\phi \in \mathcal{D}_K$. Since $$ \left\lVert \phi \right\rVert_N \leq \left\lVert \phi \right\rVert_{N+1} \leq \;\; \text{and} \;\; p_N(\phi) \leq p_{N+1}(\phi), \;\;\; (2) $$ the topologies induced by either sequence of seminorms are unchanged if we let $N$ start at $N_0$ rather than $1$.

Here is my only question, I cannot figure why the ordering matters, and not understanding this doesn't allow me to understand the equivalence of the topologies.

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  • $\begingroup$ Replace the $q$-seminorms by their max-s up to $N$, this makes them ordered, call them $Q_N$. According to Rudin's exercise, the topology remains the same. The $p$-seminorms are already ordered. So your identity shows that every $p$-seminorm is bounded by a $Q$-seminorm and vice versa. Hence they define the same topology. $\endgroup$ – Conifold Dec 18 '19 at 11:47
  • $\begingroup$ Am I right when I say I would need to show the open sets generated are the same? (Regardless of Rudin exercise). $\endgroup$ – user8469759 Dec 18 '19 at 11:49
  • $\begingroup$ It suffices to do that just for bases or subbases. $\endgroup$ – Conifold Dec 18 '19 at 11:56
  • $\begingroup$ can you elaborate? a bit more maybe in a full answer? $\endgroup$ – user8469759 Dec 18 '19 at 11:59
  • $\begingroup$ I also don't follow your order argument, why is the ordering important? $\endgroup$ – user8469759 Dec 18 '19 at 12:11

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