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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 $\supseteq$?

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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. 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.

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  • $\begingroup$ Thank you for the reference, but are you sure that Poulsen's results imply my equality? In his notation (see p. 113), we have $(Xf)(x)=-Xxf'(x)$ for $X\in\mathrm{Lie}(\mathbb{R})$, so $X$ is not simply the derivative operator; consequently, the right-hand side of the first displayed equality on his p. 114 is not my right-hand side above. $\endgroup$ Aug 12, 2013 at 14:06
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    $\begingroup$ @MuratGüngör: Yes, I am sure. (Your formula for $(Xf)(x)$ has a stray $x$; compare my edited answer.) $\endgroup$ Aug 13, 2013 at 14:16
  • $\begingroup$ You are right. My computation above is definitely incorrect. Thanks for the clarification. $\endgroup$ Aug 13, 2013 at 14:50
  • $\begingroup$ By the way, I realized that the inclusion $\supseteq$ in my question (more generally, the inclusion $\supseteq$ in the first displayed equality on p. 114 of Poulsen's) follows easily from Lemma 3.13 of Knapp's as well. $\endgroup$ Aug 13, 2013 at 14:54

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