**Proof plan.** The [distributional](/page/Distribution) [Fourier transform](/page/Fourier%20Transform) $\widehat{T_f}$ is defined by $\widehat{T_f}(\varphi) = T_f(\hat{\varphi}) = \int f(x)\hat{\varphi}(x)\,d\mathcal{L}^n(x)$. By expanding $\hat{\varphi}(x) = \int \varphi(\xi)e^{-ix\cdot\xi}\,d\mathcal{L}^n(\xi)$ and applying Fubini's theorem to swap the order of integration (justified by the [integrability](/page/Integral) condition on $f$ and the Schwartz decay of $\varphi$), the double integral reorganises into $\int \hat{f}(\xi)\varphi(\xi)\,d\mathcal{L}^n(\xi) = T_{\hat{f}}(\varphi)$. **Step 1 (Setup).** Let $\varphi \in \mathcal{S}(\mathbb{R}^n)$. By the definition of the distributional Fourier transform ([Fourier Transform as Automorphism of Tempered Distributions](/theorems/230)): \begin{align*} \widehat{T_f}(\varphi) &= T_f(\hat{\varphi}) = \int_{\mathbb{R}^n} f(x)\,\hat{\varphi}(x)\,d\mathcal{L}^n(x). \end{align*} By the definition of the Fourier transform on $\mathcal{S}(\mathbb{R}^n)$: \begin{align*} \hat{\varphi}(x) &= \int_{\mathbb{R}^n}\varphi(\xi)\,e^{-ix\cdot\xi}\,d\mathcal{L}^n(\xi). \end{align*} Substituting: \begin{align*} \widehat{T_f}(\varphi) &= \int_{\mathbb{R}^n} f(x)\left(\int_{\mathbb{R}^n}\varphi(\xi)\,e^{-ix\cdot\xi}\,d\mathcal{L}^n(\xi)\right)d\mathcal{L}^n(x). \end{align*} **Step 2 (Application of Fubini's theorem).** [claim:Fubini Integrability Condition] The function $(x,\xi) \mapsto f(x)\,\varphi(\xi)\,e^{-ix\cdot\xi}$ is in $L^1(\mathbb{R}^n \times \mathbb{R}^n, \mathcal{L}^{2n})$. [/claim] [proof] Since $|e^{-ix\cdot\xi}| = 1$ for all $x, \xi \in \mathbb{R}^n$: \begin{align*} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|f(x)\,\varphi(\xi)\,e^{-ix\cdot\xi}|\,d\mathcal{L}^n(\xi)\,d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n}|f(x)|\,d\mathcal{L}^n(x)\cdot\int_{\mathbb{R}^n}|\varphi(\xi)|\,d\mathcal{L}^n(\xi) \\ &= \|f\|_{L^1(\mathbb{R}^n)}\cdot\|\varphi\|_{L^1(\mathbb{R}^n)}. \end{align*} The first factor is finite because $f \in L^1(\mathbb{R}^n)$. The second is finite because $\varphi \in \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$ (Schwartz [functions](/page/Function) are integrable, since rapid decay gives $|\varphi(\xi)| \leq C(1+|\xi|)^{-(n+1)}$, which is integrable). Therefore the product is finite and Fubini's theorem applies. [/proof] By Fubini's theorem, we may swap the order of integration: \begin{align*} \widehat{T_f}(\varphi) &= \int_{\mathbb{R}^n}\varphi(\xi)\left(\int_{\mathbb{R}^n}f(x)\,e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)\right)d\mathcal{L}^n(\xi) = \int_{\mathbb{R}^n}\hat{f}(\xi)\,\varphi(\xi)\,d\mathcal{L}^n(\xi) = T_{\hat{f}}(\varphi). \end{align*} Since $\varphi \in \mathcal{S}(\mathbb{R}^n)$ was arbitrary, $\widehat{T_f} = T_{\hat{f}}$ as elements of $\mathcal{S}'(\mathbb{R}^n)$.