The Plancherel theorem for locally compact abelian groups

Given: A locally compact Hausdorff abelian group $G$ with Haar measure $m_G$, dual $\widehat{G}$ with the compatible dual Haar measure, and the isometric extension $\mathcal{F}$ of the Fourier transform.

The Fourier transform restricts to a linear isometry ${\mathcal{F}}_0$ on the dense subspace $L^1\cap L^2(G)\subseteq L^2(G)$ and has a unique linear isometric extension $\mathcal{F}:L^2(G)\to L^2(\widehat{G})$ with $\mathrm{\parallel }\mathcal{F}f{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. (Plancherel isometric extension on LCA groups, The space $L^p(\mu )$ as the quotient by null functions)

The range of $\mathcal{F}$ is dense in $L^2(\widehat{G})$. (The Plancherel transform range is dense in L^2 of the dual)

$L^2(G,m_G)$ and $L^2(\widehat{G},m_{\widehat{G}})$ are complete; a linear isometry from a complete space has closed range, and a closed dense subspace is the whole space. (Riesz-Fischer completeness of $L^p$ for $1\le p\le \mathrm{\infty }$)

If a linear map between inner product spaces satisfies $\mathrm{\parallel }Tu\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }$ for all $u$, then $⟨Tu,Tv⟩=⟨u,v⟩$ for all $u,v$: the inner product is recovered from the norm by the polarisation formula. (Jordan–von Neumann: a norm is induced by an inner product exactly when it satisfies the parallelogram law)