Theorem 1 (Plancherel)

The restriction of the Fourier transform to the subspace A(G) of $L^2(G)$ extends in a unique way to an isometry Φ of $L^2(G)$ onto $L^2(\widehat{G})$.

Moreover, if $f\in L^1(G)\cap L^2(G)$, its Fourier transform belongs to $L^2(\widehat{G})$ and coincides in $L^2(\widehat{G})$ with $\Phi (f)$.

By formula (22), the restriction of $\mathscr{F}$ to A(G) is an isometry of the subspace A(G) of $L^2(G)$ onto the subspace $\widehat{A}(G)$ of $L^2(\widehat{G})$. Since A(G) is dense in $L^2(G)$ (cor. 1 of II, p. 212), the Fourier transform extends uniquely to an isometry Φ of $L^2(G)$ onto a closed subspace of $L^2(\widehat{G})$. But since its image contains $\widehat{A}(G)$, which is dense in $L^2(\widehat{G})$ (lemma 5), the mapping Φ is surjective.

Let now $f\in L^1(G)\cap L^2(G)$; let us show that its Fourier transform belongs to $L^2(\widehat{G})$. By prop. 8, (iv) of II, p. 211, and the fact that A(G) is an ideal of $L^1(G)$, there exists a filter basis $\mathfrak{B}$ on A(G) which converges to $f$ both in $L^1(G)$ and in $L^2(G)$. We then have

$\Phi (f) =$ lim$_{g,\mathfrak{B}}\Phi (g) =$ lim$_{g,\mathfrak{B}}\mathscr{F}(g)$ in $L^2(\widehat{G})$ and $\mathscr{F}(f) =$ lim$_{g,\mathfrak{B}}\mathscr{F}(g)$ in $\mathscr{C}_0(\widehat{G})$. There therefore exists a

sequence $(g_n)$ in A(G) such that $\mathscr{F}(g_n)$ converges to $\Phi (f)$ in $L^2(\widehat{G})$ and to $\mathscr{F}(f)$ in $\mathscr{C}_0(\widehat{G})$. By INT, IV, §3, No.$^o4$, th. 3 and cor. 1, we have $\mathscr{F}(f) = \Phi (f)$, and in particular $\mathscr{F}(f)\in L^2(\widehat{G})$. This completes the proving of the theorem.

We still denote by $\mathscr{F}$ the isometry of $L^2(G)$ onto $L^2(\widehat{G})$ defined in theorem 1, and we call it the Fourier transform in $L^2(G)$. Analogously, the inverse Fourier transform admits a unique isometric extension to $L^2(G)$, still called the inverse Fourier transform and denoted by $\overline{\mathscr{F}}$.