[proofplan] We extend the Fourier transform to a unitary operator on $L^2(\mathbb{R}^n)$. First, we establish the Plancherel identity $\|\hat{f}\|_{L^2} = \|f\|_{L^2}$ on the Schwartz space by computing $\|f\|_{L^2}^2 = (f * g)(0)$ (where $g(x) = \overline{f(-x)}$) in two ways: directly and via the Fourier inversion theorem. Then density of $\mathcal{S}$ in $L^2$ and the isometry property allow a unique continuous extension to all of $L^2$. Surjectivity follows from the density of $\mathcal{S}$ in $L^2$ and closedness of the range of an isometry. Throughout we use the symmetric normalisation $\hat{f}(\xi) = (2\pi)^{-n/2}\int f(x)\,e^{-ix\cdot\xi}\,d\mathcal{L}^n(x)$. [/proofplan] [step:Establish the Plancherel identity on the Schwartz space] Let $f \in \mathcal{S}(\mathbb{R}^n)$. Define $g(x) = \overline{f(-x)}$. Then $g \in \mathcal{S}(\mathbb{R}^n)$ and $\hat{g}(\xi) = \overline{\hat{f}(\xi)}$ (by substituting $y = -x$ in the definition and using the conjugate-symmetry of the exponential). The convolution $f * g$ evaluated at $0$ gives \begin{align*} (f * g)(0) &= \int_{\mathbb{R}^n} f(-y) \, \overline{f(-y)} \, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} |f(y)|^2 \, d\mathcal{L}^n(y) = \|f\|_{L^2}^2. \end{align*} On the other hand, under the symmetric normalisation the convolution theorem gives $\widehat{f * g}(\xi) = (2\pi)^{n/2}\,\hat{f}(\xi)\,\hat{g}(\xi)$. Applying the [Fourier Inversion Theorem](/theorems/246) at $x = 0$: \begin{align*} (f * g)(0) &= (2\pi)^{-n/2} \int_{\mathbb{R}^n} \widehat{f * g}(\xi) \, d\mathcal{L}^n(\xi) = (2\pi)^{-n/2} \int_{\mathbb{R}^n} (2\pi)^{n/2}\,\hat{f}(\xi)\,\overline{\hat{f}(\xi)} \, d\mathcal{L}^n(\xi) = \|\hat{f}\|_{L^2}^2. \end{align*} Combining: $\|\hat{f}\|_{L^2}^2 = \|f\|_{L^2}^2$ for all $f \in \mathcal{S}(\mathbb{R}^n)$. [/step] [step:Extend $\mathcal{F}$ to $L^2$ by density] The [Schwartz space](/page/Schwartz%20Space) $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^2(\mathbb{R}^n)$. For any $f \in L^2(\mathbb{R}^n)$, choose $\{f_m\} \subset \mathcal{S}(\mathbb{R}^n)$ with $\|f - f_m\|_{L^2} \to 0$. By the isometry from the previous step, \begin{align*} \|\hat{f}_m - \hat{f}_k\|_{L^2} &= \|f_m - f_k\|_{L^2} \to 0 \quad \text{as } m, k \to \infty. \end{align*} The sequence $\{\hat{f}_m\}$ is Cauchy in $L^2(\mathbb{R}^n)$, which is complete, so it converges to some $F \in L^2(\mathbb{R}^n)$. Define $\hat{f} := F$. This is independent of the approximating sequence: if $\{f_m'\}$ also satisfies $f_m' \to f$ in $L^2$, then $\|\hat{f}_m - \hat{f}_m'\|_{L^2} = \|f_m - f_m'\|_{L^2} \to 0$. [/step] [step:Verify the extension is isometric] By continuity of the $L^2$ norm: \begin{align*} \|\hat{f}\|_{L^2} &= \lim_{m \to \infty} \|\hat{f}_m\|_{L^2} = \lim_{m \to \infty} \|f_m\|_{L^2} = \|f\|_{L^2}. \end{align*} [/step] [step:Prove surjectivity to conclude $\mathcal{F}$ is unitary] The operator $\mathcal{F}$ is an isometry of $L^2(\mathbb{R}^n)$ into itself. Its range is closed (isometries always have closed range: if $\mathcal{F}f_m \to g$, then $\{f_m\}$ is Cauchy by the isometry property, so $f_m \to f$ and $\mathcal{F}f = g$). On $\mathcal{S}(\mathbb{R}^n)$, the [Fourier Inversion Theorem](/theorems/246) shows that $\mathcal{F}$ is surjective onto $\mathcal{S}(\mathbb{R}^n)$: for every $g \in \mathcal{S}$, the function $f = \mathcal{F}^{-1}g \in \mathcal{S}$ satisfies $\mathcal{F}f = g$. Since $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^2(\mathbb{R}^n)$ and the range of $\mathcal{F}$ contains $\mathcal{S}(\mathbb{R}^n)$ and is closed, $\mathcal{F}$ is surjective onto $L^2(\mathbb{R}^n)$. Hence $\mathcal{F}$ is a unitary isomorphism. [/step]