Plancherel isometric extension on LCA groups

(Pointwise inversion and isometry on $C_c$.) For $h\in E$, set $H(x):={\int }_{\widehat{G}}\widehat{h}(\gamma )\gamma (x)d{m}_{\widehat{G}}(\gamma )$, which is absolutely defined since $\widehat{h}\in L^1$ by [F2]. We first show $H$ is continuous without using sequential convergence: given $x_0\in G$ and $\epsilon >0$, choose compact $C\subseteq \widehat{G}$ with ${\int }_{\widehat{G}\setminus C}\mathrm{\mid }\widehat{h}\mathrm{\mid }d{m}_{\widehat{G}}<\epsilon \mathrm{/}4$ by [F7]. Joint continuity of evaluation and compactness of $C$ give a neighbourhood $U$ of $x_0$ such that $\mathrm{\mid }\gamma (x)-\gamma (x_0)\mathrm{\mid }<\epsilon \mathrm{/}(2(1+\mathrm{\parallel }\widehat{h}{\mathrm{\parallel }}_1))$ for every $x\in U$ and $\gamma \in C$: take product neighbourhoods at each $(x_0,\gamma )$ and a finite subcover of $C$. Since characters have modulus one, for $x\in U$ we obtain $\mathrm{\mid }H(x)-H(x_0)\mathrm{\mid }\le \frac{\epsilon \mathrm{\parallel }\widehat{h}{\mathrm{\parallel }}_1}{2(1+\mathrm{\parallel }\widehat{h}{\mathrm{\parallel }}_1)}+2{\int }_{\widehat{G}\setminus C}\mathrm{\mid }\widehat{h}\mathrm{\mid }d{m}_{\widehat{G}}<\epsilon \mathrm{.}$ Thus $H$ is continuous. By [F2], $h=H$ almost everywhere; both functions are continuous, so they agree everywhere, since a nonzero continuous difference would stay nonzero on a nonempty open set of positive Haar measure [F7]. In particular $h(0)=H(0)={\int }_{\widehat{G}}\widehat{h}d{m}_{\widehat{G}}$. Parseval [F3] now gives $\mathrm{\parallel }\widehat{h}{\mathrm{\parallel }}_2=\mathrm{\parallel }h{\mathrm{\parallel }}_2$. For $g\in C_c(G;\mathbb{C})$, take $h:=g\ast \tilde{g}\in E$: by [F1] and the pointwise identity just proved, $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=h(0)={\int }_{\widehat{G}}\widehat{h}d{m}_{\widehat{G}}={\int }_{\widehat{G}}\mathrm{\mid }\widehat{g}{\mathrm{\mid }}^{2}d{m}_{\widehat{G}}=\mathrm{\parallel }\widehat{g}{\mathrm{\parallel }}_2^2,$ where the last equality uses $\widehat{h}=\mathrm{\mid }\widehat{g}{\mathrm{\mid }}^2$ from [F1]; hence the Fourier transform is isometric on $C_c(G;\mathbb{C})$, and it is linear by [F5].

(Simultaneous density of $C_c$ in $L^1\cap L^2$.) Let $f\in L^1(G,m_G)\cap L^2(G,m_G)$ and $\epsilon >0$. Inner regularity of the finite Radon measures $\mathrm{\mid }f\mathrm{\mid }d{m}_G$ and $\mathrm{\mid }f{\mathrm{\mid }}^{2}d{m}_G$ [F4] gives a compact $K$ with ${\int }_{G\setminus K}\mathrm{\mid }f\mathrm{\mid }<\epsilon \mathrm{/}2$ and ${\int }_{G\setminus K}\mathrm{\mid }f{\mathrm{\mid }}^2<{\epsilon }^2\mathrm{/}4$; set $f_1:=f{\mathbf{1}}_K$. Convolving with an approximate identity $u_U$ gives $\phi :=f_1\ast u_U\in C_c(G;\mathbb{C})$ with $\mathrm{\parallel }\phi -f_1{\mathrm{\parallel }}_1<\epsilon \mathrm{/}2$ and $\mathrm{\parallel }\phi -f_1{\mathrm{\parallel }}_2<\epsilon \mathrm{/}2$ for small identity neighbourhoods $U$, by norm continuity of translations and the approximate-identity limits in $L^1$ and $L^2$ [F4]. Here $f_1\ast u_U$ is continuous since its differences are bounded by $\mathrm{\parallel }{f}_1{\mathrm{\parallel }}_1{\sup }_t\mathrm{\mid }{u}_U(t+z)-u_U(t)\mathrm{\mid }$, which tends to zero by uniform continuity of $u_U\in C_c$; it vanishes outside the compact $K+\mathrm{supp}{u}_U$. Hence $\mathrm{\parallel }f-\phi {\mathrm{\parallel }}_1<\epsilon$ and $\mathrm{\parallel }f-\phi {\mathrm{\parallel }}_2<\epsilon$.

(The extension.) Step 1.1 makes the transform a linear isometry $T:C_c(G;\mathbb{C})\to L^2(\widehat{G},m_{\widehat{G}})$; $C_c(G;\mathbb{C})$ is dense in $L^2(G,m_G)$ and $L^2(\widehat{G},m_{\widehat{G}})$ is complete [F4, F5], so by [F6] $T$ has a unique linear isometric extension $\mathcal{F}:L^2(G,m_G)\to L^2(\widehat{G},m_{\widehat{G}})$ with $\mathrm{\parallel }\mathcal{F}f{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$ for all $f$.

(${\mathcal{F}}_0$ is the restriction of $\mathcal{F}$.) Let $f\in L^1\cap L^2$ and choose ${\phi }_n\in C_c(G;\mathbb{C})$ with $\mathrm{\parallel }{\phi }_n-f{\mathrm{\parallel }}_1\to 0$ and $\mathrm{\parallel }{\phi }_n-f{\mathrm{\parallel }}_2\to 0$ by step 1.2. Then $\mathrm{\parallel }\widehat{{\phi }_n}-\widehat{f}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{\phi }_n-f{\mathrm{\parallel }}_1\to 0$ by [F5], so $\widehat{{\phi }_n}\to \widehat{f}$ pointwise everywhere; on the other hand $\widehat{{\phi }_n}=\mathcal{F}{\phi }_n\to \mathcal{F}f$ in $L^2(\widehat{G},m_{\widehat{G}})$ by step 2.1, so a subsequence of $(\widehat{{\phi }_n})$ converges to $\mathcal{F}f$ almost everywhere [F5]. Hence $\widehat{f}=\mathcal{F}f$ $m_{\widehat{G}}$-almost everywhere; in particular $\widehat{f}\in L^2(\widehat{G},m_{\widehat{G}})$ and, by step 2.1, $\mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathcal{F}f{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Thus the pointwise transform on $L^1\cap L^2$ is the restriction of $\mathcal{F}$ to that subspace, and ${\mathcal{F}}_0$ is a linear isometry.

(Density and uniqueness.) $C_c(G;\mathbb{C})\subseteq L^1\cap L^2\subseteq L^2(G,m_G)$ and $C_c(G;\mathbb{C})$ is dense in $L^2$ [F4], so $L^1\cap L^2$ is dense in $L^2$. If ${\mathcal{F}}^{\mathrm{\prime }}$ is another linear isometric extension of ${\mathcal{F}}_0$, then $\mathcal{F}-{\mathcal{F}}^{\mathrm{\prime }}$ is a bounded linear map vanishing on the dense subspace $L^1\cap L^2$, hence ${\mathcal{F}}^{\mathrm{\prime }}=\mathcal{F}$; the extension is unique.

Steps 2.1 and 3.1 exhibit the linear isometry ${\mathcal{F}}_0$ on the dense subspace $L^1\cap L^2$ and its unique linear isometric extension $\mathcal{F}$; no surjectivity of $\mathcal{F}$ is claimed, and no use of the biduality identification is made.