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

Question

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: \begin{equation} \lVert f \rVert_{n, \alpha} := \sup_{ x \in \mathbb{R}^N } (1 + \lvert x \rvert^n) \lvert \partial^\alpha f \rvert \end{equation} 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 \begin{equation} \lVert \cdot \rVert_0 \leq \lVert \cdot \rVert_1 \leq \lVert \cdot \rVert_2 \leq \cdots. \end{equation}

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

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?

Answer 1

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).