There is a technical detail I am trying to understand in the proof that the Fourier transform is a bijection on the Schwartz space.
Theorem. $\mathcal{F} \in \mathcal{L}\left(\mathcal{S}\left(\mathbb{R}^n\right), \mathcal{S}\left(\mathbb{R}^n\right)\right)$ is bijective, and for $\varphi \in \mathcal{S}\left(\mathbb{R}^n\right)$ we have the inversion formula $$ \left(\mathcal{F}^{-1} \varphi\right)(x)=(2 \pi)^{-n / 2} \int_{\mathbb{R}^n} e^{i \xi \cdot x} \varphi(\xi) \mathrm{d} \xi=\hat{\varphi}(-x), \quad x \in \mathbb{R}^n . $$ PROOF. Fix $\varphi \in \mathcal{S}\left(\mathbb{R}^n\right)$ and $x \in \mathbb{R}^n$. I et us first motivate the subsequent computations and emphasize, that, since $$ \left[(y, \xi) \mapsto e^{i \xi \cdot x} e^{-i \xi \cdot y} \varphi(y)\right] \notin L_1\left(\mathbb{R}^n \times \mathbb{R}^n\right), $$ we cannot apply Fubini's theorem directly to interchange the integration on the right-hand side of $$ \int_{\mathbb{R}^n} e^{i \xi \cdot x} \hat{\varphi}(\xi) \mathrm{d} \xi=(2 \pi)^{-n / 2} \int_{\mathbb{R}^n} e^{i \xi \cdot x} \int_{\mathbb{R}^n} \varphi(y) e^{-i \xi \cdot y} \mathrm{~d} y \mathrm{~d} \xi . $$
I am trying to understand why the statement:
$ \left[(y, \xi) \mapsto e^{i \xi \cdot x} e^{-i \xi \cdot y} \varphi(y)\right] \notin L_1\left(\mathbb{R}^n \times \mathbb{R}^n\right), $
is true, that $ \left[(y, \xi) \mapsto e^{i \xi \cdot x} e^{-i \xi \cdot y} \varphi(y)\right]$ is not integrable.
To be integrable in $L^1(\Bbb R^n\times \Bbb R^n)$, you need a function that decays in both the $y$ and $\xi$ variables, but $$ |e^{i\xi\cdot x}\,e^{-i\xi\cdot y}\,\varphi(y)| = |\varphi(y)| $$ is a constant function of the variable $\xi$, that is for any $y$, the function $\xi\mapsto |\varphi(y)|$ is constant, and so certainly not integrable. More precisely, if you assume by contradiction that $(y,\xi)\mapsto |\varphi(y)|$ is integrable in $L^1(\Bbb R^n\times \Bbb R^n)$, then you get from the Fubini theorem $$ \int_{\Bbb R^n\times \Bbb R^n} |\varphi(y)|\,\mathrm d y\,\mathrm d \xi = \int_{\Bbb R^n} |\varphi(y)|\left(\int_{\Bbb R^n} 1\,\mathrm d \xi\right)\mathrm d y = \infty $$ since $\int_{\Bbb R^n} 1\,\mathrm d \xi = \infty$, and this contradicts the fact that you assumed $(y,\xi)\mapsto |\varphi(y)|$ to be integrable.