We denote by $\mathcal{S}(\mathbb{R})$ the space of smooth and rapidly decreasing functions. We define on $\mathcal{S}(\mathbb{R})$ the family of semi-norms $$\lVert \varphi \lVert_{n,m} = \lVert (1+|\cdot|^m) \varphi^{(n)} \lVert_\infty.$$ This family of semi-norms defines a topology $\tau$ on $\mathcal{S}(\mathbb{R})$. The space $(\mathcal{S}(\mathbb{R}),\tau)$ is then well-known to be nuclear.

If $\mathcal{M}(\mathbb{R})$ is the space of measurable functions in $\mathbb{R}$, I define the space $\mathcal{R}(\mathbb{R})$ of rapidly decreasing functions (without smoothness condition) by $$\mathcal{R}(\mathbb{R}) = \{ \varphi \in \mathcal{M}(\mathbb{R}), \ \lVert \varphi \lVert_{0,m} < \infty \}.$$ Moreover, the family of (semi-)norms $(\lVert \cdot \lVert_{0,m})_{m\in \mathbb{N}}$ defines a topology $\tau'$ on $\mathcal{R}(\mathbb{R})$.

**Question: Is $(\mathcal{R}(\mathbb{R}),\tau')$ a nuclear space?**

NB. The proofs I found for the nuclearity of $\mathcal{S}(\mathbb{R})$ are essentially using the facts that the space $s$ of sequences with quick decay is nuclear and $\mathcal{S}(\mathbb{R})$ is homeomorphic to a subspace of $s$ (Treves: Topological vector spaces, III-51).

Several authors are claiming that the nuclear structure is strongly related with the condition of infinite smoothness. The space $\mathcal{D}(\mathbb{R})$ of compactly supported and smooth function (with the inductive topology coming from the spaces $\mathcal{D}([-n,n])$), and the space $\mathcal{C}^\infty(\mathbb{R})$ (projective limit of the spaces $\mathcal{D}([-n,n])$) are other famous examples of nuclear structure that go in that sense. This suggests that the space $\mathcal{R}(\mathbb{R})$ may be not nuclear or at least that the usual technics are not available.