Consider the left regular representation of $\mathbb{R}$ in $\mathrm{L}^2(\mathbb{R})$. Let us denote by $\mathrm{L}^2(\mathbb{R})^\infty$ the algebraic subspace of smooth vectors, or equivalently, the Gårding subspace (the theorem of Dixmier-Malliavin assumed to be at our disposal). I want to prove


$$\mathrm{L}^2(\mathbb{R})^\infty =
 \{f\in\mathcal{C}^\infty(\mathbb{R}) \mid
 f^{(n)}\in\mathrm{L}^2(\mathbb{R})\:\:\text{for all}\:\:n\geq 0\}.$$


Actually, I proved the inclusion $\subseteq$ by making use of an exercise in Knapp's *Representation Theory of Semisimple Groups* (Problem 5 on p. 57). My questions are: Is the above equality really true? If so, could you please give a reference, or an argument for the proof of $\supseteq$?



On a characterization of the Gårding subspace of the left regular representation of reals
Yes, your desired equality is true: regarding the left regular representation as
$$
\operatorname{Ind}\_{\{0\}}^{\mathbf R}1,
$$
it becomes a special case of the characterization of smooth vectors in induced representations by N. S. Poulsen, [*On $C^\infty$-vectors and intertwining bilinear forms for representations of Lie groups*](http://dx.doi.org/10.1016/0022-1236%2872%2990016-X). J. Functional Analysis **9** (1972), 87–120, Theorem 5.1.


**Edit** to clear up your extra question: For $G=\mathbf R$, we have $\mathfrak g=\mathbf R$ and $\exp$ is just the identity $\mathbf R\to\mathbf R$. So Poulsen's second displayed formula on p. 113 says
\begin{align}
(Xf)(x)
&=\Bigl.\frac d{dt}f(\exp(-tX)\cdot x)\Bigr|\_{t=0}\\
&=\Bigl.\frac d{dt}f(x-tX)\Bigr|\_{t=0}= - Xf'(x).
\end{align}
Fixing $X=-1$ (basis of $\mathfrak g$) we get $X^\alpha f=f^{(\alpha)}$ and so Poulsen's first displayed formula on p. 114,
$\mathbf D\_\infty(U\_2)=\{f\in C^\infty(G)\mid X^\alpha f\in L^2(G) \text{ for all } \alpha\}$, is exactly your desired equality. 


