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. Throughout we use the symmetric normalisation $\hat{f}(\xi) = (2\pi)^{-n/2}\int f(x)\,e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)$. **Step 1 (Gaussian Fourier transform).** For $a > 0$, define $\chi_a(\xi) = e^{-a|\xi|^2}$. The symmetric Fourier transform of $\chi_a$ is: \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 $\hat{g}(\xi) = (2\pi)^{-1/2}\int e^{-x^2/2}\,e^{-ix\xi}\,d\mathcal{L}^1(x)$. The ODE $\hat{g}'(\xi) = -\xi\hat{g}(\xi)$ (obtained by writing $\hat{g}'(\xi) = (2\pi)^{-1/2}\int(-ix)e^{-ix\xi}e^{-x^2/2}\,d\mathcal{L}^1(x)$ and integrating by parts) with initial condition $\hat{g}(0) = (2\pi)^{-1/2}\int e^{-x^2/2}\,d\mathcal{L}^1(x) = (2\pi)^{-1/2}\sqrt{2\pi} = 1$ gives $\hat{g}(\xi) = e^{-\xi^2/2}$. In particular, the Gaussian $e^{-x^2/2}$ is a fixed point of $\mathcal{F}$ under the symmetric normalisation. The general case follows by rescaling. **Step 2 (Regularised identity at $x = 0$).** For $f \in \mathcal{S}(\mathbb{R}^n)$ and $\varepsilon > 0$, the Parseval-type identity (a consequence of 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 $\hat{f} \in \mathcal{S} \subset L^1$ by the [automorphism theorem](/theorems/228)). For the right side, using $\hat{\chi}_\varepsilon(x) = (2\varepsilon)^{-n/2}\, e^{-|x|^2/(4\varepsilon)}$ and substituting $y = x/(2\sqrt{\varepsilon})$: \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(2\sqrt{\varepsilon}\,y)\, \frac{(2\sqrt{\varepsilon})^n}{(2\varepsilon)^{n/2}}\, e^{-|y|^2}\, d\mathcal{L}^n(y). \end{align*} Since $(2\sqrt{\varepsilon})^n/(2\varepsilon)^{n/2} = 2^{n/2}\varepsilon^{n/2}/\varepsilon^{n/2}\cdot 2^{n/2} \cdot 1 = 2^{n/2}$ (explicitly: $(2\sqrt{\varepsilon})^n = 2^n \varepsilon^{n/2}$ and $(2\varepsilon)^{n/2} = 2^{n/2}\varepsilon^{n/2}$, so the ratio is $2^{n/2}$), the integral equals \begin{align*} 2^{n/2}\int_{\mathbb{R}^n} f(2\sqrt{\varepsilon}\,y)\, e^{-|y|^2}\, d\mathcal{L}^n(y). \end{align*} As $\varepsilon \to 0$, the integrand converges pointwise to $f(0)\,e^{-|y|^2}$, and is dominated by $\|f\|_{L^\infty}\, e^{-|y|^2} \in L^1$. By dominated convergence and $\int e^{-|y|^2}\, d\mathcal{L}^n(y) = \pi^{n/2}$: \begin{align*} \lim_{\varepsilon \to 0} \int f(x)\, \hat{\chi}_\varepsilon(x)\, d\mathcal{L}^n(x) = 2^{n/2}\, \pi^{n/2}\, f(0) = (2\pi)^{n/2}\, f(0). \end{align*} Combining: $\int \hat{f}(\xi)\, d\mathcal{L}^n(\xi) = (2\pi)^{n/2}\, f(0)$. **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, which holds with the same form under any normalisation). 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*} giving $f(x) = \frac{1}{(2\pi)^{n/2}} \int e^{ix \cdot \xi} \hat{f}(\xi)\, d\mathcal{L}^n(\xi)$. **Step 4 (Bijectivity).** The map $\mathcal{F}^{-1}(g)(x) = \frac{1}{(2\pi)^{n/2}} \int e^{ix \cdot \xi} g(\xi)\, d\mathcal{L}^n(\xi)$ sends $\mathcal{S}$ to $\mathcal{S}$ (it satisfies analogous exchange identities with reversed signs), and Steps 2–3 show $\mathcal{F}^{-1} \circ \mathcal{F} = \mathrm{Id}$. A symmetric argument gives $\mathcal{F} \circ \mathcal{F}^{-1} = \mathrm{Id}$, so $\mathcal{F}$ is a bijection with [continuous](/page/Continuity) inverse.