# Equivalence of topologies generated by two seminorms and section 6.2 Rudin's Functional analysis

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.

• 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. – Conifold Dec 18 '19 at 11:47
• Am I right when I say I would need to show the open sets generated are the same? (Regardless of Rudin exercise). – user8469759 Dec 18 '19 at 11:49
• It suffices to do that just for bases or subbases. – Conifold Dec 18 '19 at 11:56
• can you elaborate? a bit more maybe in a full answer? – user8469759 Dec 18 '19 at 11:59
• I also don't follow your order argument, why is the ordering important? – user8469759 Dec 18 '19 at 12:11