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I do have a question about the Sobolev spaces of infinite order. Let me first define them:

Let $H^s(\mathbb{R}^n)$ denote the Sobolev space of order $s \in \mathbb{R}$. We can naturally identify $H^{-s}(\mathbb{R}^n)$ with the dual space of $H^s(\mathbb{R}^n)$. Furthermore, for all $s > t$ we do have continuous embeddings $\mathcal{S}(\mathbb{R}^n) \to H^s(\mathbb{R}^n) \to H^t(\mathbb{R}^n) \to \mathcal{S}^\prime(\mathbb{R}^n)$, where $\mathcal{S}(\mathbb{R}^n)$ is the Schwartz space and $\mathcal{S}^\prime(\mathbb{R}^n)$ its dual space (the space of tempered distributions).

We define the Sobolev spaces of infinite order as $H^{\infty}(\mathbb{R}^n) := \bigcap_s H^s(\mathbb{R}^n)$ and $H^{-\infty}(\mathbb{R}^n) := \bigcup_s H^s(\mathbb{R}^n)$. We endow $H^{\infty}(\mathbb{R}^n)$ with the Frechet topology and $H^{-\infty}(\mathbb{R}^n)$ with the weak topology that it inherits as the dual of $H^{\infty}(\mathbb{R}^n)$.

From above we know that we have embeddings $\mathcal{S}(\mathbb{R}^n) \to H^{\infty}(\mathbb{R}^n) \to H^{-\infty}(\mathbb{R}^n) \to \mathcal{S}^\prime(\mathbb{R}^n)$. I think this embeddings are continuous?

Now my main question: Do we have $\mathcal{S}(\mathbb{R}^n) = H^{\infty}(\mathbb{R}^n)$ and $H^{-\infty}(\mathbb{R}^n) = \mathcal{S}^\prime(\mathbb{R}^n)$?

Thanks in advance.

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No. Consider $u(x) = (1 + x^2)^{-1}$. Then $u \in H^{\infty}(\mathbb{R})$ according to your definition, but $u \notin \mathcal{S}(\mathbb{R})$.

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