# Do tempered distributions form a topological subspace of the space of distributions?

I'm learning about distributions and tempered distributions. From what I understand, by "enlarging" the space of test functions $\mathcal{D}$ to the Schwarz space $\mathcal{S}$ and correspondingly "shrinking" the space of distributions $\mathcal{D}'$ (continuous linear functionals on $\mathcal{D}$) to $\mathcal{S}'$, we get a space which is closed under Fourier transforms. It's clear that $\mathcal{D}$ is a linear subspace of $\mathcal{S}$, and $\mathcal{S}'$ is a linear subspace of $\mathcal{D}'$ (correct me if I'm wrong).

Question: Is it true that $\mathcal{D}$ is a topological subspace of $\mathcal{S}$? And is $\mathcal{S}'$ a topological subspace of $\mathcal{D}'$ if both have the weak-$*$ topology? Additionally, are the inclusion maps $\mathcal{D} \hookrightarrow \mathcal{S}$ and $\mathcal{S}' \hookrightarrow \mathcal{D}'$ continuous?

I'm using the definitions of this Wikipedia page which I believe are fairly standard. I'm not well-versed in the theory of topological vector spaces (yet) so I don't quite understand the definitions of the topologies on $\mathcal{D}$ and $\mathcal{S}$ in terms of semi-norms. However, I'm interested in knowing the answer to this.

My work: My guess is that the answer is no since it seems to me that you can find a sequence $(\varphi_k)$ of functions in $\mathcal{D}$ (which are then also in $\mathcal{S}$) which converge e.g. to the $0$ function in the topology of $\mathcal{S}$ but not in the topology of $\mathcal{D}$. This is because in order to converge in $\mathcal{D}$, there must exist a compact set $K$ containing the support of each $\varphi_k$ and the sequence must converge to $0$ on $K$. However, to converge to $0$ in $\mathcal{S}$ we require only that $$\lim_{k \to \infty} \sup \{|x^{\alpha} D^{\beta} \varphi_k| \;/\; x \in \mathbb{R^n} \} = 0$$ for every $\alpha$, $\beta$. I don't think this implies that there exists a compact set $K$ containing the support of each $\varphi$. For example, letting $$\gamma(x) = \begin{cases} e^{-(1-|x|^2)^{-1}} & \text{if } |x| < 1\\ 0 & \text{if } |x| \geq 1 \end{cases}$$ and $\varphi_k(x) = e^{-k} \gamma(\frac{x}{k})$, we get a sequence of smooth functions with compact support which converges in $\mathcal{S}$ but not in $\mathcal{D}$ because the support gets increasingly larger. Does this imply that the subspace topology of $\mathcal{D}$ on the set $\mathcal{S}$ is finer than the usual topology of $\mathcal{S}$?

Intuitively, I think that a positive/negative answer to the question for $\mathcal{D}$ and $\mathcal{S}$ should imply the same answer for the dual spaces but I could be wrong.

You have answered almost everything by yourself. Let $\tau$ be the usual topology on $\mathcal{D}$ and $\sigma$ the subspace topology of $\mathcal{S}$ on $C_c^\infty$. You are asking for $\tau=\sigma$.
You constructed a sequence in $\mathcal{D}$ converging with respect to $\sigma$ but not with respect to $\tau$. This shows $\tau\nsubseteq\sigma$.
To see $\sigma\subseteq\tau$, you need to show the continuity of the inclusion map $\iota\colon\mathcal{D}\to\mathcal{S}$. But as $\tau$ is the final topology of the $\mathcal{D}_K$, where $K$ runs through the compact sets, $\iota$ is continuous iff each $\iota_K\colon\mathcal{D}_K\to\mathcal{S}$ is continuous. And this is immediate to see.
For the dual space, take a non-tempered distribution $T$ and $\psi_n$ a function identical $1$ on the ball with radius $n$ with support in the ball with radius $n+1$. Define $S_n=\psi_nT$. Then each $S_n$ is tempered (as it has compact support), $(S_n)_n\to T$ in $\mathcal{D}'$, but as the limit is non-tempered, it cannot converge in the weak-star topology of $\mathcal{S}'$.