Equivalent family of norms on the Schwartz space

Question

We consider the following two families of seminorms on the Schwartz space and want to show that they induce the same topology.

First family: $(\| \cdot \|_N) $ for $N \in \mathbb{N}$ defined by $$ \| f \|_N=\sup_{\vert \alpha \vert \leq N} \sup_{x\in \mathbb{R}^n} (1+\vert x \vert ^2)^N \vert (\partial^\alpha f)(x) \vert. $$

Second family: $(\| \cdot \|_{(K,\alpha)})$ for $K \in \mathbb{N}$ and $\alpha$ a multi-index defined by $$ \| f \|_{(K,\alpha)}=\sup_{x \in \mathbb{R}^n} (1+\vert x \vert)^K \vert (\partial^\alpha f)(x)\vert. $$

I am not very familiar with Frechet spaces but from my topological knowledge I suspect that it is sufficient to show $$ \| f \|_N \leq C \sum_{i=1}^l \| f \|_{(K_i,\alpha_i)} $$ for some for each $N \in \mathbb{N}$, $C>0$ and $l,K_i\in \mathbb{N}$ and some multi-indices $\alpha_i$ where $C,l,K_i,\alpha_i$ may depend on $N$ and

$$ \| f \|_{(K,\alpha)} \leq C \sum_{i=1}^l \| f \|_{N_i} $$

for each $K \in \mathbb{N}$ and each multi-index $\alpha$ for some $C>0$, some $l \in \mathbb{N}$ and $N_i \in \mathbb{N}$ where $C,l,N_i$ may depend on $K$ and $\alpha$.

Is this approach correct or do I need to change something?

Answer 1

It seems that the problem has already been solved but it happened that I have typed down the answer for myself so let's share.

A slightly more general statement can be found in Reed and Simon, vol. 1, p.126 (proof left as an exercise...) and I will stick to their framework and just proove one crucial ingredient.

Let $X$ be a vector space with a topology induced by a family $(\rho_{\alpha})_{\alpha \in A}$ of semi-norms. Let $p: X \mapsto [0,+\infty )$ be another semi-norm. It is continuous at $0$ iff $$\exists\, C >0,\ (\alpha_1,\ \cdots, \alpha_n) \in A^n\, \big/\ \forall\, x\in X,\quad p(x) \leq C \big( \rho_{\alpha_1}(x)+ \cdots + \rho_{\alpha_n}(x)\big)$$ A variant of this could also be $$\exists\, C >0,\ (\alpha_1,\ \cdots, \alpha_n) \in A^n\, \big/\ \forall\, x\in X,\quad p(x) \leq C \sup_{1\leq i \leq n} \rho_{\alpha_i}(x)$$

$(\Rightarrow):$ Let's write continuity of $p$ at $0$ with $\epsilon = C > 0$: there exists a neighborhood V of $0$ such that $p(V) \subset [0,M [$ (open or closed, not important, just a neighborhood of $p(0)=0$ in $\mathbb{R}_+$). Now since the topology of $X$ is generated by $(\rho_{\alpha})_{\alpha \in A},\ V$ contains an open subset/can be chosen of the form $$ V:= \big\lbrace y \in X,\ \rho_{\alpha_i}(y-0) =\rho_{\alpha_i}(y) < \delta,\ \forall\, i\in [\![1,n]\!]\big\rbrace $$ (implicitly, "there exists $\delta >0,\ n\in N$ and a subset $\{\alpha_1,\ \cdots, \alpha_n\}\subset A$" such that ...).

Now for an arbitrary $x\in X$,

• either every $\rho_{\alpha_i}(x)=0$ so that $x\in V$ and $p(x) < C$ (then by homogeneity of $p$, one must have $p(x)=0$),
• or $\rho_{\alpha_1}(x)+ \cdots + \rho_{\alpha_n}(x) > 0,\ $ resp. $ \sup_{1\leq i \leq n} \rho_{\alpha_i}(x) > 0$ and $$ \frac{x}{\rho_{\alpha_1}(x)+ \cdots + \rho_{\alpha_n}(x)} \in V,\quad \frac{x}{\sup_{1\leq i \leq n} \rho_{\alpha_i}(x) } \in V $$ (again, homogeneity of the $\rho_{\alpha_i}$) so that $$ p\bigg( \frac{x}{\rho_{\alpha_1}(x)+ \cdots + \rho_{\alpha_n}(x)}\bigg) = \frac{p(x)}{\rho_{\alpha_1}(x)+ \cdots + \rho_{\alpha_n}(x)} < C,\quad \frac{p(x)}{\sup_{1\leq i \leq n} \rho_{\alpha_i}(x) } < C $$

$(\Leftarrow):$ If the inequality holds, then for any $\epsilon > 0$, consider the following neighborhood of $0$: $$ V := \Big\lbrace y\in X,\ \rho_{\alpha_i}(\mathbf{y}) < \frac{\epsilon}{n\, C},\ \forall\, i\in [\![1,n]\!] \Big\rbrace $$ (or for the variant $$ V := \Big\lbrace y\in X,\ \rho_{\alpha_i}(\mathbf{y}) < \frac{\epsilon}{1+ C},\ \forall\, i\in [\![1,n]\!] \Big\rbrace $$ )

Then $p(V) \subset [0,\epsilon[$.

Remarks: This little proposition justifies that one can choose a directed family of semi-norms and it will play a role in the proof that the Schwartz space is Fréchet (metrizable complete). Here we do not talk about completeness.