I want to ask a detail in the book "Classical Fourier Analysis" by Loukas Grafakos
Proof of proposition2.4.7 in page 141 in Classical Fourier Analysis,3rd: A detail about $\langle u,\Psi\rangle=\langle\hat{u},\hat{\Psi}\rangle$. In the proof, why does the author use $\langle u,\Psi\rangle=\langle\hat{u},\hat{\Psi}\rangle$? Does it right? Why not $\langle u,\Psi\rangle=\langle\hat{u},\check{\Psi}\rangle$?
Notation:$u$ represents a tempered distribution and in $\mathcal{C}^{\infty}(\mathbb{R}^{n}\setminus\lbrace0\rbrace)$,$\Psi$ is a nonnegtive, radial, smooth, and nonzero function on $\mathbb{R}^{n}$ support in the annulus $1<|x|<2$.
Is Plancherel's theorem true for tempered distribution? I have answered this question under the above post and now for readers' convenience, I copy the answer here:
Yes. For any $f\in\mathcal{S}'(\mathbb{R}^{n})$, there exists a sequence $f_{n}\in \mathcal{C}^{\infty}_{0}(\mathbb{R}^{n})$ such that $f_{n}\to f$ under the topology of $\mathcal{S}'(\mathbb{R}^{n})$. According to Plancherel's theorem, we have $\langle f_{n},g\rangle=\langle\hat{{f}_{n}},\hat{g}\rangle$. Then by tending $n$ to $\infty$ we have $\langle f,g\rangle=\langle\hat{f},\hat{g}\rangle$.
Notation: if $f_{n}\to f$ under the topology of $\mathcal{S}'(\mathbb{R}^{n})$, we have $\hat{f_{n}}\to\hat{f}$ under the topology of $\mathcal{S}'(\mathbb{R}^{n})$.