Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.