$\{f \in L^1(\mathbb R^n) \cap L^2(\mathbb R^n): \mathcal Ff \in L^2(\mathbb R^n)\}$ dense in $L^1(\mathbb R^n) \cap L^2(\mathbb R^n)$?

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This belongs to the Plancherel Theorem: For $f,g \in L^1(\mathbb R^n) \cap L^2(\mathbb R^n)$ and $ h(x) := f \ast g(-\cdot)(x)$, we have $h(0) = \langle f ,g\rangle$ and $\hat h(\xi) = (2\pi)^{-n/2} \hat f(\xi) \overline{\hat g(\xi)}$.

If $\hat f, \hat g \in L^2(\mathbb R^n) \Rightarrow \hat h \in L^1(\mathbb R^n)$. Using that, it is easy to show: $\langle f, g\rangle = \langle \hat f, \hat g \rangle$.

To generalize this for all $f,g \in L^1(\mathbb R^n) \cap L^2(\mathbb R^n)$, I need density arguments but I don't why this holds.

EDIT: It is not allowed to use that $\mathcal F$ is bijective on $L^2(\mathbb R^n)$

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The space of $C^\infty$ functions with compact support is contained in $\{f \in L^1(\mathbb R^n) \cap L^2(\mathbb R^n): \mathcal Ff \in L^2(\mathbb R^n)\}$ and dense in $L^1(\mathbb R^n) \cap L^2(\mathbb R^n)$.