Finding a set of "graded" seminorms on the Schwartz space

For any fixed $$N \in \mathbb{N}$$, let $$\mathcal{S}(\mathbb{R}^N)$$ be the Schwartz space.

Then, it is well-known that $$\mathcal{S}(\mathbb{R}^N)$$ is a Fréchet space with the seminorms: $$$$\lVert f \rVert_{n, \alpha} := \sup_{ x \in \mathbb{R}^N } (1 + \lvert x \rvert^n) \lvert \partial^\alpha f \rvert$$$$ where $$n$$ is any non-negative integer and $$\alpha$$ is any multi-index on $$(x_1, \cdots, x_N)$$.

Now, I wonder if it is possible to find a collection of seminorms $$\lVert \cdot \rVert_m$$ for nonnegative intergers $$m$$ giving the same topology on $$\mathcal{S}(\mathbb{R}^N)$$ as $$\lVert \cdot \rVert_{n, \alpha}$$'s but further satisfing the "graded" property $$$$\lVert \cdot \rVert_0 \leq \lVert \cdot \rVert_1 \leq \lVert \cdot \rVert_2 \leq \cdots.$$$$

A possible candidate I have come up with is the following: $$$$\lVert f \rVert_m := \sum_{ n,\lvert \alpha \rvert \leq m} \lVert f \rVert_{n, \alpha}$$$$

However, I am not fully sure if these seminorms indeed give rise to the same Fréchet topology as $$\lVert \cdot \rVert_{n, \alpha}$$'s.

Could anyone please clarify for me?

• This should be true. Let $d_1$ be the metric induced by the first set of seminorms and $d_2$ the metric induced by the second set of seminorms. Then consider the map $\iota:(\mathcal{S}(\mathbb{R}^N), d_2) \rightarrow (\mathcal{S}(\mathbb{R}^N), d_1),\iota(f)=f.$ This map is clearly a bijection. If you can show that it is also continuous, then it is a homeomorphism by the open mapping theorem for Fréchet spaces. Commented Feb 21 at 6:16
• Is there a reason you are not accepting my answer? Commented Feb 27 at 0:03
• @SeverinSchraven Oops sorry. I forgot about it.. Commented Feb 27 at 3:01

This is indeed true. Pick an enummeration $$\varphi: \mathbb{N}\rightarrow \mathbb{N} \times\mathbb{N}^N$$ and define the metrics

$$d_1(f,g) = \sum_{n\in \mathbb{N}} 2^{-n} \frac{\Vert f-g \Vert_{\varphi(n)}}{1+\Vert f-g \Vert_{\varphi(n)}}$$

and

$$d_2(f,g)= \sum_{n\in \mathbb{N}} 2^{-n} \frac{\Vert f-g \Vert_n}{1+\Vert f-g\Vert_n}.$$

The topologies are the same if we can show that the map $$\iota: (\mathcal{S}(\mathbb{R}^N), d_2) \rightarrow (\mathcal{S}(\mathbb{R}^N), d_1), \iota(f)=f$$ is a homeomorphism. Clearly this map is a bijection. Next we check that it is also continuous.

Pick $$(f_n)_{n\in \mathbb{N}}\subseteq \mathcal{S}(\mathbb{R}^N)$$ converging to $$f\in (\mathcal{S}(\mathbb{R}^N), d_2)$$, we now want to show that $$\lim_{n\rightarrow \infty} d_1(f_n, f) =0.$$

Pick $$\varepsilon>0$$ and $$M\in \mathbb{N}$$ such that $$\sum_{n\geq M+1} 2^{-n}<\varepsilon/2$$. Then we have $$d_1(f_n, f) \leq \varepsilon/2+ \sum_{k=1}^M \frac{\Vert f_n -f \Vert_{\varphi(k)}}{1+\Vert f_n -f \Vert_{\varphi(k)}}.$$

Thus, all we need to show is that $$\Vert \cdot \Vert_{\varphi(k)} : (\mathcal{S}(\mathbb{R}^N, d_2) \rightarrow \mathbb{R}$$ is continuous, which is easy to see. Namely, $$d_2(f_n, f) \rightarrow 0$$ implies $$\Vert f_n -f \Vert_{m,\alpha}\rightarrow 0$$ for all $$(m,\alpha)\in \mathbb{N}\times \mathbb{N}^N$$, which implies that $$\Vert f_n -f \Vert_k \rightarrow 0$$ (as this only a finite sum of the $$\Vert f_n -f \Vert_{m,\alpha}$$.

To conclude we can either use the open mapping theorem for Fréchet spaces, or we can just check by hand that the inverse is continuous too. To check it by hand, we just need to verify that $$\Vert \cdot \Vert_{k} : (\mathcal{S}(\mathbb{R}^N, d_1) \rightarrow \mathbb{R}$$ is continuous (which is again easy to see).

• I am very sorry for delay in accepting it. Perhaps, could you help me with the following question as well: mathoverflow.net/questions/466023/… Commented Feb 27 at 17:47
• @Keith No worries. I was just wondering whether my answer was incomprehensible. Your linked question looks quite fun, I'll give it a go once I have a bit more time. Commented Feb 27 at 18:23