The strategy is to regularise the integral $\int \hat{f}(\xi)\, d\mathcal{L}^n(\xi)$ using a Gaussian convergence factor, evaluate both sides using the known [Fourier transform](/page/Fourier%20Transform) of a Gaussian, and take the regularisation parameter to zero. **Step 1 (Gaussian Fourier transform).** For $a > 0$, define $\chi_a(\xi) = e^{-a|\xi|^2}$. The Fourier transform of a Gaussian is itself a Gaussian: \begin{align*} \hat{\chi}_a(x) = (2a)^{-n/2} e^{-|x|^2/(4a)}. \end{align*} This is verified by factoring into one-dimensional [integrals](/page/Integral): it suffices to check the case $n = 1$, $a = 1/2$, where $g(x) = e^{-x^2/2}$ satisfies the ODE $\hat{g}'(\xi) = -\xi\hat{g}(\xi)$ (obtained by writing $\hat{g}'(\xi) = \int (-ix) e^{-ix\xi} e^{-x^2/2}\, d\mathcal{L}^1(x)$ and integrating by parts) with initial condition $\hat{g}(0) = (2\pi)^{-1/2}\int e^{-x^2/2}\, d\mathcal{L}^1(x) = 1$, giving $\hat{g}(\xi) = e^{-\xi^2/2}$. **Step 2 (Regularised identity at $x = 0$).** For $f \in \mathcal{S}(\mathbb{R}^n)$ and $\varepsilon > 0$, Fubini's theorem gives \begin{align*} \int_{\mathbb{R}^n} \hat{f}(\xi)\, \chi_\varepsilon(\xi)\, d\mathcal{L}^n(\xi) = \int_{\mathbb{R}^n} f(x)\, \hat{\chi}_\varepsilon(x)\, d\mathcal{L}^n(x). \end{align*} The left side converges to $\int \hat{f}(\xi)\, d\mathcal{L}^n(\xi)$ as $\varepsilon \to 0$ by dominated convergence, since $|\hat{f}(\xi) \chi_\varepsilon(\xi)| \le |\hat{f}(\xi)| \in L^1(\mathbb{R}^n)$ (because $f \in \mathcal{S}$ implies $\hat{f} \in \mathcal{S} \subset L^1$). For the right side, substituting $y = x/\sqrt{2\varepsilon}$ gives \begin{align*} \int_{\mathbb{R}^n} f(x)\, (2\varepsilon)^{-n/2} e^{-|x|^2/(4\varepsilon)}\, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n} f(\sqrt{2\varepsilon}\, y)\, e^{-|y|^2/2}\, d\mathcal{L}^n(y). \end{align*} Since $|f(\sqrt{2\varepsilon}\, y) e^{-|y|^2/2}| \le \|f\|_{L^\infty} e^{-|y|^2/2} \in L^1(\mathbb{R}^n)$ and $f(\sqrt{2\varepsilon}\, y) \to f(0)$ pointwise as $\varepsilon \to 0$ (by [continuity](/page/Continuity) of $f$), dominated convergence gives \begin{align*} \lim_{\varepsilon \to 0} \int_{\mathbb{R}^n} f(\sqrt{2\varepsilon}\, y)\, e^{-|y|^2/2}\, d\mathcal{L}^n(y) = f(0) \int_{\mathbb{R}^n} e^{-|y|^2/2}\, d\mathcal{L}^n(y) = (2\pi)^{n/2} f(0). \end{align*} Combining: $\int \hat{f}(\xi)\, d\mathcal{L}^n(\xi) = (2\pi)^{n/2} f(0)$, which is the inversion formula at $x = 0$. **Step 3 (General $x$).** For arbitrary $x \in \mathbb{R}^n$, define $h(y) = f(x + y)$. Then $h \in \mathcal{S}(\mathbb{R}^n)$ and $\hat{h}(\xi) = e^{ix \cdot \xi} \hat{f}(\xi)$ (by the translation property of the Fourier transform). Applying Step 2 to $h$: \begin{align*} (2\pi)^{n/2} f(x) = (2\pi)^{n/2} h(0) = \int_{\mathbb{R}^n} \hat{h}(\xi)\, d\mathcal{L}^n(\xi) = \int_{\mathbb{R}^n} e^{ix \cdot \xi} \hat{f}(\xi)\, d\mathcal{L}^n(\xi), \end{align*} which gives $f(x) = (2\pi)^{-n/2} \int e^{ix \cdot \xi} \hat{f}(\xi)\, d\mathcal{L}^n(\xi)$ as claimed. **Step 4 (Bijectivity).** The formula $\mathcal{F}^{-1}(g)(x) = (2\pi)^{-n/2} \int e^{ix \cdot \xi} g(\xi)\, d\mathcal{L}^n(\xi)$ defines a map $\mathcal{S} \to \mathcal{S}$ (by the [automorphism theorem](/theorems/228), since $\mathcal{F}^{-1}$ differs from $\mathcal{F}$ only by replacing $\xi$ with $-\xi$ and conjugating, both of which preserve $\mathcal{S}$). Steps 2–3 show $\mathcal{F}^{-1} \circ \mathcal{F} = \mathrm{Id}$, and an analogous argument gives $\mathcal{F} \circ \mathcal{F}^{-1} = \mathrm{Id}$. Thus $\mathcal{F}$ is a bijection on $\mathcal{S}(\mathbb{R}^n)$ with continuous inverse (continuity follows from the automorphism property applied to $\mathcal{F}^{-1}$).