[proofplan] We establish the $L^2$ isometry on the dense subspace $L^1 \cap L^2$ using Gaussian regularisation: smooth $f$ by convolving with $g_t$, apply the [Fourier Inversion Formula](/theorems/528) and [Convolution Theorem](/theorems/527) to compute $\|f_t\|_2^2$ in terms of $\hat{f}$, and take $t \to 0$ using the [Monotone Convergence Theorem](/theorems/509). Extension to all of $L^2$ follows from completeness. Throughout we use the symmetric normalisation $\hat{f}(\xi) = (2\pi)^{-d/2}\int f(x)\,e^{-ix\cdot\xi}\,d\mathcal{L}^d(x)$. [/proofplan] [step:Establish the Parseval identity for Gaussian-smoothed functions] [claim:Parseval For Regularised Functions] For $f \in L^1 \cap L^2$ and $t > 0$, with $f_t = f * g_t$, \begin{align*} \int_{\mathbb{R}^d} |f_t(x)|^2\,d\mathcal{L}^d(x) = \int_{\mathbb{R}^d} |\hat{f}(u)|^2\,e^{-t|u|^2}\,d\mathcal{L}^d(u). \end{align*} [/claim] [proof] Under the symmetric normalisation, the [Convolution Theorem](/theorems/527) gives $\widehat{f * g}(\xi) = (2\pi)^{d/2}\,\hat{f}(\xi)\,\hat{g}(\xi)$. With the Gaussian $g_t(x) = (2\pi t)^{-d/2}\,e^{-|x|^2/(2t)}$ (which satisfies $\hat{g}_t(u) = (2\pi)^{-d/2}\,e^{-t|u|^2/2}$ ... actually, let us work directly). The key identity is that for $f_t = f * g_t$, we have $\hat{f}_t(u) = (2\pi)^{d/2}\,\hat{f}(u)\,\hat{g}_t(u)$. Instead, we use the Parseval-type identity directly. Define $h(x) = \overline{f_t(-x)}$, so $\hat{h}(u) = \overline{\hat{f}_t(u)}$. Then $(f_t * h)(0) = \|f_t\|_2^2$. By the [Fourier Inversion Formula](/theorems/528) at $x = 0$: \begin{align*} \|f_t\|_2^2 = (f_t * h)(0) = (2\pi)^{d/2}\int_{\mathbb{R}^d} \hat{f}_t(u)\,\hat{h}(u)\,d\mathcal{L}^d(u) \cdot (2\pi)^{-d/2} = \int_{\mathbb{R}^d} |\hat{f}_t(u)|^2\,d\mathcal{L}^d(u). \end{align*} The last equality uses $\hat{h}(u) = \overline{\hat{f}_t(u)}$. More precisely, by the inversion formula $(f_t * h)(0) = (2\pi)^{-d/2}\int \widehat{f_t * h}(u)\,d\mathcal{L}^d(u) = (2\pi)^{-d/2}\int (2\pi)^{d/2}\hat{f}_t(u)\hat{h}(u)\,d\mathcal{L}^d(u) = \int |\hat{f}_t(u)|^2\,d\mathcal{L}^d(u)$. Since under symmetric normalisation the Fourier transform of a Gaussian $\chi_\varepsilon(\xi) = e^{-\varepsilon|\xi|^2}$ gives $\hat{\chi}_\varepsilon(x) = (2\varepsilon)^{-d/2}e^{-|x|^2/(4\varepsilon)}$, we can choose the Gaussian smoothing so that $\hat{f}_t(u) = \hat{f}(u)\,e^{-t|u|^2/2}$ (adjusting parameters). Then \begin{align*} \|f_t\|_2^2 = \int_{\mathbb{R}^d} |\hat{f}(u)|^2\,e^{-t|u|^2}\,d\mathcal{L}^d(u). \end{align*} [/proof] [/step] [step:Take $t \to 0$ to obtain the Plancherel identity] [claim:Limit Gives Plancherel] $\|f_t\|_2^2 \to \|f\|_2^2$ and $\int |\hat{f}|^2 e^{-t|u|^2}\,d\mathcal{L}^d \to \|\hat{f}\|_2^2$ as $t \to 0$. [/claim] [proof] For the left side: $\|f_t\|_2 \leq \|f\|_2$ (convolution with a probability measure is a contraction on $L^2$ by Jensen) and $f_t \to f$ in $L^1$, so a subsequence $f_{t_k} \to f$ a.e. By [Fatou's lemma](/theorems/510), $\|f\|_2^2 \leq \liminf \|f_{t_k}\|_2^2 \leq \|f\|_2^2$, giving $\|f_t\|_2^2 \to \|f\|_2^2$. For the right side: $|\hat{f}(u)|^2 e^{-t|u|^2}$ increases monotonically to $|\hat{f}(u)|^2$ as $t \downarrow 0$. By the [Monotone Convergence Theorem](/theorems/509), \begin{align*} \int_{\mathbb{R}^d} |\hat{f}(u)|^2\,e^{-t|u|^2}\,d\mathcal{L}^d(u) \uparrow \|\hat{f}\|_2^2. \end{align*} Combining: $\|f\|_2^2 = \|\hat{f}\|_2^2$, i.e., $\|\hat{f}\|_2 = \|f\|_2$. [/proof] [/step] [step:Extend to all of $L^2$ by completeness] The map $f \mapsto \hat{f}$ is an isometry from the dense subspace $L^1 \cap L^2$ into $L^2$. Since $L^2$ is complete, this isometry extends uniquely to a bounded linear operator $\mathcal{F}: L^2(\mathbb{R}^d) \to L^2(\mathbb{R}^d)$ satisfying $\|\mathcal{F}(f)\|_2 = \|f\|_2$ for all $f \in L^2$. Surjectivity follows from the density of $\mathcal{F}(L^1 \cap L^2)$ in $L^2$ (using the inversion formula) and the closed range property of isometries. Hence $\mathcal{F}$ is a unitary operator. [/step]