[proofplan] We derive the Parseval identity from the Plancherel theorem by polarisation. The Plancherel identity $\|\hat{f}\|_{L^2} = \|f\|_{L^2}$ (under the symmetric normalisation) controls norms; we recover the inner product by applying the polarisation identity to both the spatial and frequency sides, then using linearity of $\mathcal{F}$ to match corresponding terms. [/proofplan] [step:Recall the Plancherel identity] By the [Plancherel Theorem](/theorems/247), for every $f \in L^2(\mathbb{R}^n; \mathbb{C})$: \begin{align*} \|\hat{f}\|_{L^2}^2 &= \|f\|_{L^2}^2. \end{align*} [/step] [step:Apply the polarisation identity to both sides] The inner product on any [Hilbert space](/page/Hilbert%20Space) can be recovered from the norm via the polarisation identity. For $f, g \in L^2(\mathbb{R}^n; \mathbb{C})$: \begin{align*} \int_{\mathbb{R}^n} f(x) \, \overline{g(x)} \, d\mathcal{L}^n(x) &= \frac{1}{4}\left(\|f + g\|_{L^2}^2 - \|f - g\|_{L^2}^2 + i\|f + ig\|_{L^2}^2 - i\|f - ig\|_{L^2}^2\right). \end{align*} Apply the Plancherel identity to each of the four norms. Since $\mathcal{F}$ is linear, $\widehat{f + g} = \hat{f} + \hat{g}$, $\widehat{f - g} = \hat{f} - \hat{g}$, $\widehat{f + ig} = \hat{f} + i\hat{g}$, and $\widehat{f - ig} = \hat{f} - i\hat{g}$. Therefore: \begin{align*} \|f + g\|_{L^2}^2 &= \|\hat{f} + \hat{g}\|_{L^2}^2, \\ \|f - g\|_{L^2}^2 &= \|\hat{f} - \hat{g}\|_{L^2}^2, \\ \|f + ig\|_{L^2}^2 &= \|\hat{f} + i\hat{g}\|_{L^2}^2, \\ \|f - ig\|_{L^2}^2 &= \|\hat{f} - i\hat{g}\|_{L^2}^2. \end{align*} [/step] [step:Reassemble to obtain the Parseval identity] Substituting into the polarisation identity: \begin{align*} \int_{\mathbb{R}^n} f(x) \, \overline{g(x)} \, d\mathcal{L}^n(x) &= \frac{1}{4}\left(\|\hat{f} + \hat{g}\|_{L^2}^2 - \|\hat{f} - \hat{g}\|_{L^2}^2 + i\|\hat{f} + i\hat{g}\|_{L^2}^2 - i\|\hat{f} - i\hat{g}\|_{L^2}^2\right). \end{align*} The expression on the right is the polarisation identity applied to $\hat{f}$ and $\hat{g}$, which equals $\int_{\mathbb{R}^n} \hat{f}(\xi) \, \overline{\hat{g}(\xi)} \, d\mathcal{L}^n(\xi)$. Therefore \begin{align*} \int_{\mathbb{R}^n} f(x) \, \overline{g(x)} \, d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n} \hat{f}(\xi) \, \overline{\hat{g}(\xi)} \, d\mathcal{L}^n(\xi). \end{align*} [/step]