Fourier transform agrees with l one and plancherel transforms

Let $f\in L^1$ and $\phi \in \mathcal{S}$. Schwartz decay makes $\phi \in L^1$, so $\int \text{\hspace{0.17em}⁣}\int \mathrm{\mid }f(x)\phi (\xi )\mathrm{\mid }dxd\xi <\mathrm{\infty }$. Absolute Fubini is therefore applicable. [F2, F3] $⟨\mathcal{F}{u}_f,\phi ⟩=\int f(x)\text{\hspace{0.17em}⁣}(\int \phi (\xi )e^{-2\pi ix\cdot \xi }d\xi )dx=\int \widehat{f}(\xi )\phi (\xi )d\xi \mathrm{.}$ Since $\widehat{f}$ is bounded, [F1] makes the last functional tempered. This proves the $L^1$ assertion.

Let $f\in L^2$ and choose $f_j\in \mathcal{S}$ with $f_j\to f$ in $L^2$. For a fixed $\phi \in \mathcal{S}$, also $\mathcal{F}\phi \in L^2$, and Hölder gives the first convergence below. [F4, F6] $\int (f_j-f)\mathcal{F}\phi ⟶0.$ Plancherel gives ${\mathcal{F}}_2{f}_j\to {\mathcal{F}}_{2}f$ in $L^2$, so a second Hölder estimate gives $\int ({\mathcal{F}}_2{f}_j-{\mathcal{F}}_{2}f)\phi \to 0$.

Each $f_j$ belongs to $L^1\cap L^2$, and the two agreement results give the displayed identity. [F5, step 1.1] $\int f_j\mathcal{F}\phi =\int ({\mathcal{F}}_2{f}_j)\phi \mathrm{.}$ Passing to the two limits from step 1.2 yields $⟨\mathcal{F}{u}_f,\phi ⟩=⟨u_{{\mathcal{F}}_{2}f},\phi ⟩$. Since this holds for every Schwartz test, the $L^2$ assertion follows.