Thank you for the clarification. Is there a way to see that $\mathcal{A}'=\mathcal{A}$ without using the Fourier transform? In fact one has that if I apply the Fourier transform to $\mathcal{A}$ this space becomes $L^{\infty}(\mathbb{R})$. How can one use this?

@YCor Sorry of course I mean $B(L^2(\mathbb{R}))$.