Do we have $C^\infty \cap \mathcal{O}_C' = \mathcal{S}$ and/or $C^\infty \cap \mathcal{S}' = \mathcal{O}_M$?

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We define the following traditional function spaces from distribution theory.

  • $\mathcal{S}$ the space of rapidly decreasing smooth functions.
  • $\mathcal{S}'$ the space of tempered distributions, dual of $\mathcal{S}$.
  • $\mathcal{O}_M$ the space of slowly increasing smooth functions (i.e., the function and all its derivatives are bounded by a polynomial).
  • $\mathcal{O}_C'$, the space of rapidly decreasing distributions, i.e. distributions $u \in \mathcal{S}'$ such that $u * f \in \mathcal{S}$ for every $f\in \mathcal{S}$. (We have that $\mathcal{F}\ \mathcal{O}_C' = \mathcal{O}_M$ with $\mathcal{F}$ the Fourier transform on $\mathcal{S}'$.)

Now are my questions. Do we have the following equalities between sets: $$C^\infty \cap \mathcal{O}_C' = \mathcal{S},$$ and $$C^\infty \cap \mathcal{S}' = \mathcal{O}_M?$$

Thanks for your attention.