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Use additive notation and fix Haar measure $m$ on $G$. A continuous unitary character is a continuous homomorphism $\chi :G\to \mathbb{T}$. With the conventionthe Fubini theorem and the character identity imply $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Also $|\hat{f}(\chi )|\le \Vert f{\Vert }_1$, and choosing $f=\chi a$ for a nonzero nonnegative $a\in C_c(G)$ shows that evaluation at $\chi$ is a nonzero functional. Thus each character gives a character of an algebra on the L1 convolution algebra. Here a multiplicative linear functional means a nonzero multiplicative linear functional; the zero functional must be excluded for the claimed bijection.

Conversely, let $\phi$ be such a functional. It is automatically bounded with $|\phi (f)|\le \Vert f{\Vert }_1$: extend it to the unitization of an algebra by $\tilde{\phi }(f+z1)=\phi (f)+z$. If $|z|>\Vert f{\Vert }_1$, the Neumann series makes $z1-f$ invertible, so its image under the unital homomorphism cannot be zero. Hence $\phi (f)$ cannot lie outside the norm disc.

Choose $a\in L^1(G)$ with $\phi (a)\ne 0$, and let $T_{x}a(t)=a(t-x)$. The Abelian translation identities $(T_{x}a)\ast b=a\ast (T_{x}b)$ giveTherefore, defining $c(x)=\phi (T_{x}a)/\phi (a)$, one has $\phi (T_{x}b)=c(x)\phi (b)$ for every $b$. Applying this to two successive translations gives $c(x+y)=c(x)c(y)$ and $c(0)=1$. By translation continuity in Lp on a locally compact group, $c$ is continuous. It is uniformly bounded by $\Vert a{\Vert }_1/|\phi (a)|$; applying this bound to $c(nx)=c(x{)}^n$ for positive and negative integers forces $|c(x)|=1$. Thus $\chi (x)=\overline{c(x)}$ is a continuous unitary character.

The Bochner integral identity $a\ast b={\int }_{G}b(y)T_{y}adm(y)$ holds in $L^1$, since $\Vert T_{y}a{\Vert }_1=\Vert a{\Vert }_1$. Applying the bounded functional yieldsCancel $\phi (a)\ne 0$ to obtain the required correspondenceThe recovery formula $\overline{\chi (x)}=\phi (T_{x}a)/\phi (a)$ proves uniqueness. These functionals have norm exactly one, since $b=\chi a$ with $a\ge 0$ and $\int a=1$ has $\Vert b{\Vert }_1=\phi (b)=1$.

For the topology, identify the Pontryagin dual group with these functionals and give it the Gelfand topology, namely pointwise convergence of $\hat{f}(\chi )$ for every $f\in L^1$. If a net ${\chi }_{\lambda }$ converges to $\gamma$ uniformly on every compact subset of $G$, then for $f\in C_c(G)$,Density of $C_c$ in $L^1$ and the uniform functional norm bound one extend this convergence to every $f\in L^1$.

Conversely, suppose convergence in the Gelfand topology and choose $a$ with $\hat{a}(\gamma )\ne 0$. For compact $K\subseteq G$, the set $\{T_{x}a:x\in K\}$ is compact in $L^1$ by translation continuity in Lp on a locally compact group. Evaluation convergence is uniform on this compact set: a finite $\delta$-net reduces its supremum to finitely many convergent evaluations plus $2\delta$, using the common functional norm one. ThusNow $\widehat{T_{x}a}(\chi )=\overline{\chi (x)}\hat{a}(\chi )$, and $\hat{a}({\chi }_{\lambda })\to \hat{a}(\gamma )\ne 0$. Dividing this identity proves ${\sup }_{x\in K}|{\chi }_{\lambda }(x)-\gamma (x)|\to 0$. Hence the Gelfand topology is exactly the compact-open topology, with the requested neighbourhood basisFor compact $K$, the continuous difference attains its maximum, so this strict pointwise inequality on $K$ is equivalent to a strict uniform bound. Finite intersections contain another such set by taking the union of the compact sets and the minimum of the tolerances.

Fix Haar measure $m_G$ and use $\hat{f}(\chi )={\int }_{G}f(x)\overline{\chi (x)}d{m}_G(x)$. A positive-definite function $p$ satisfies ${\sum }_{i,j}{z}_i\overline{z_j}p(x_i-x_j)\ge 0$ for every finite choice of points and coefficients. The Bochner theorem states that a continuous positive-definite function on a locally compact Hausdorff Abelian group has a unique representationwhere ${\mu }_p$ is a Radon measure. Conversely, any finite positive Radon measure on the Pontryagin dual group has a continuous positive-definite inverse transform. The sign convention here is the one compatible with the displayed Fourier transform.

First note that the dual is a locally compact Hausdorff topological group. Its group operations are continuous in the compact-open topology proved in question 3. For local compactness, adjoin the zero functional to the character space of an algebra of $L^1(G)$. All its elements lie in the closed unit ball of $L^1(G{)}^{\ast }$, and the equations $\phi (a\ast b)=\phi (a)\phi (b)$ define a closed subset in the weak-star topology. The Banach-Alaoglu theorem makes that subset compact Hausdorff. Removing the zero functional gives an open, hence locally compact, subspace. This provides the setting for constructing the measure on the dual, without assuming a Fourier inversion theorem.

Let $u^{\ast }(x)=\overline{u(-x)}$ and let $\mathcal{P}$ be the finite sums of convolution squares $u\ast u^{\ast }$, $u\in C_c(G)$. Each such square is continuous with compact support, and is positive-definite becauseIts transform is $|\hat{u}{|}^2$, so every $p\in \mathcal{P}$ has $\hat{p}\ge 0$. Also $p\ast q\in \mathcal{P}$: convolution of two squares is $(u\ast v)\ast (u\ast v{)}^{\ast }$, and finite sums distribute. Insert the Bochner representation of $q$ into $p\ast q$ and use the Fubini theorem to obtainThe representing measure on the right is finite and positive. Uniqueness in the Bochner theorem and commutativity now give the Bochner consistency identity for convolution squares

For every compact $K\subseteq \hat{G}$ there is $p\in \mathcal{P}$ with $\hat{p}>0$ throughout $K$. Indeed, at any ${\chi }_0$ choose $u={\chi }_{0}a$, where $a\ge 0$ is a nonzero member of $C_c(G)$. Then $\hat{u}({\chi }_0)=\int a>0$, and continuity supplies a neighbourhood where it stays nonzero. A finite subcover of $K$ and the sum of the corresponding squares give such a $p$.

For $\psi \in C_c(\hat{G})$, choose $p$ positive in transform on its support and defineThe quotient is set to zero away from the support; its denominator is bounded away from zero there. If $q$ is another choice, the consistency identity, multiplied by $\psi /(\hat{p}\hat{q})$ on that compact support, shows that the integral is unchanged. Choosing one $p$ for a union of supports proves linearity; positivity follows from positivity of ${\mu }_p$. On each fixed compact support it is bounded by ${\mu }_p(K)/{\min }_K\hat{p}$ times the uniform norm. Moreover, for every $p\in \mathcal{P}$ and $\psi \in C_c(\hat{G})$, choosing $q$ positive on the support of $\psi$ givesA nonzero square has ${\mu }_p(\hat{G})=p(0)=\Vert u{\Vert }_2^2>0$; a suitable nonnegative compactly supported $\psi$ consequently makes this last expression positive. Thus $J$ is nonzero.

To prove translation invariance, fix $\eta \in \hat{G}$ and modulate $p$ by $p_{\eta }(x)=\eta (x)p(x)$. This again belongs to $\mathcal{P}$, since modulation takes $u\ast u^{\ast }$ to $(\eta u)\ast (\eta u{)}^{\ast }$. Direct calculation and Bochner uniqueness giveHence, for ${\psi }_{\eta }(\chi )=\psi ({\eta }^{-1}\chi )$, using $p_{\eta }$ as the denominator function and changing variables gives $J({\psi }_{\eta })=J(\psi )$. The Riesz representation on compactly supported continuous functions now produces a nonzero translation-invariant positive Radon measure $\nu$ on the dual. It is finite on compact sets, so it is Haar measure. This is the compatible dual Haar measure, denoted $m_{\hat{G}}=\nu$.

The identity for $J(\psi \hat{p})$ says, by uniqueness of Radon measures,Thus $\hat{p}\in L^1(\hat{G},\nu )$ and the Bochner representation immediately provesIt also proves the formula for all finite complex linear combinations of these functions. The measure scale is fixed by this identity: at $x=0$, a nonzero square gives $\int |\hat{u}{|}^{2}d\nu =\Vert u{\Vert }_2^2$.

A wider inversion class is $f\in L^1(G)$ with $\hat{f}\in L^1(\hat{G},\nu )$. To prove this extension, choose nonnegative $a_U\in C_c(G)$ with integral one and support in a small identity neighbourhood $U$, and put $p_U=a_U\ast a_U^{\ast }$. These functions form an approximate identity: they are nonnegative, have integral one and support in $U-U$. Also $0\le {\hat{p}}_U=|{\hat{a}}_U{|}^2\le 1$, and ${\hat{p}}_U\to 1$ uniformly on compact subsets of the dual. For this uniform convergence use equicontinuity of compact families of characters: evaluation is jointly continuous, as follows from the translation quotient in question 3, and a finite cover of a compact family gives uniform control near the identity.

Convolving the already established inversion formula for $p_U$ with $f$, the Fubini theorem givesThis is justified by $\Vert f{\Vert }_1{\mu }_{p_U}(\hat{G})<\infty$. Compact-set uniform convergence and an integrable-tail split implyExplicitly, first choose a compact set outside which $\int |\hat{f}|$ is small, use the bound $|{\hat{p}}_U-1|\le 2$ on that tail, and use uniform convergence on the compact remainder. This argument works for neighbourhood-indexed nets, without assuming metrizability. Consequently $f\ast p_U$ converges uniformly to the continuous inverse integral. On the other hand, translation continuity in Lp on a locally compact group gives $f\ast p_U\to f$ in $L^1$; testing these two limits against $C_c(G)$ shows they agree almost everywhere. At every continuity point of the chosen representative of $f$, the shrinking-support approximate identity also converges pointwise to $f(x)$. We obtain Fourier inversion on a locally compact abelian group:valid everywhere for continuous $f\in L^1(G)$ with integrable transform, and in general as equality with a continuous representative almost everywhere. This class includes all convolution squares above, and their linear span.

Finally, for arbitrary $h\in C_c(G)$, apply the inversion formula at zero to $h\ast h^{\ast }$:Thus the integral Fourier transform is linear and norm-preserving on $C_c(G)$. That space is dense in $L^2(G)$, so for $f\in L^2(G)$ choose $h_n\in C_c(G)$ with $h_n\to f$ in $L^2$ and define $\mathcal{F}f$ as the $L^2(\hat{G},\nu )$ limit of ${\hat{h}}_n$. The norm identity makes this limit exist and independent of the approximating sequence, and preserves linearity. This proves the Plancherel theorem for locally compact abelian groups in the requested form:On $L^1\cap L^2$, choose the approximating sequence to converge in both norms; uniform convergence of the integral transforms and their $L^2$ convergence show that this extension agrees almost everywhere with the original transform. For general $L^2$ functions the extension is defined by norm limits, not by presuming that the pointwise integral exists.