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Not research but advertising this question from mse in case someone wants to answer.

I'm struggling with some bookkeeping associated with the Pontryagin duality theorem. I'm thinking about the first equation of the second last line in the proof of $4.32$ in Folland's harmonic analysis, which I feel requires some explanation along the lines of my question below.

Notation: Let $(G, d_G)$ be a locally compact abelian group with fixed choice of Haar measure. Let $\hat{G}$ be the set of all characters on $G$ equipped with the compact-open topology (or, equivalently, the weak$^*$-topology inherited from $L^{\infty}(G, d_G)$). Ponrtyagin duality says that the map $$\Phi: G\to \hat{\hat{G}};\ \ x\mapsto (\cdot)(x)$$ is a homeomorphism as well as an isomorphism of groups. Further, write $M(G), M(\hat{G}), M(\hat{\hat{G}})$ to be the set of regular complex Borel measures on $G, \hat{G}, \hat{\hat{G}}$ respectively. If $\mu \in M(\hat{G})$ let $$T_\mu(x) := \int \xi(x)d\mu(\xi).$$ $T$ gives an injective norm decreasing map from $M(\hat{G})$ into the bounded continuous functions on $G$ (In fact, onto the complex span of the set of functions of positive type on $G$). Thus the image will contain all convolutions of functions having compact support. We have our test function spaces, $\mathcal{B}(G):= \{T_\mu : \mu \in M(\hat{G})\}$ and its intersection with $L^1(G, d_G)$, $\mathcal{B}^1(G)$.

The theorem Fourier Inversion I ($4.21$ in the book) says that, there is a unique choice of Haar measure $d_{\hat{G}}$ on $\hat{G}$ such that: \begin{equation} \label{1} \tag{1} \text{if $\ f\in \mathcal{B}^1(G)\ $ with $f = T_{\mu_f}$, then $\ \hat{f} \in L^1(\hat{G})\ $ and $\ \mu_{f}= \hat{f}(\xi)d_{\hat{G}}(\xi)\ $ on the dual.} \end{equation} This, in particular, implies $\ f(x) =\int \xi(x)\hat{f}(\xi)d_{\hat{G}}(\xi)$. Note, the fourier transform on $L^1(G, d_G)$ is defined by $\hat{f}(\xi) = \int \overline{\xi(x)}f(x)d_G(x).$ The same Fourier inversion I shows that we have a uniquely defined Haar measure $d_{\hat{\hat{G}}}$ on $\hat{\hat{G}}$ satisfying the corresponding equations in \eqref{1}.

Setup: In the proof of Fourier inversion II ($4.32$ in the book), there is a point where the author takes an $f\in L^1(G, d_G)$ with $\hat{f} \in L^1(\hat{G}, d_{\hat{G}})$ and arrives at the conclusion that $\hat{f}$ must belong to $\mathcal{B}^1(\hat{G})$ with the corresponding measure $d\mu_{\hat{f}}(\eta) = f\circ \Phi^{-1}(\eta^{-1})d(\Phi_*d_G)(\eta)\ \left(= \Phi_*(f(x^{-1})d_G(x))\right)$ on $\hat{\hat{G}}$. This and Fourier inversion I applied to $\hat{G}$ then leads the the equality of measures \begin{equation} \label{2} \tag{2} \hat{\hat{f}}(\eta)d_{\hat{\hat{G}}}(\eta) = c f\circ \Phi^{-1}(\eta^{-1})d_{\hat{\hat{G}}}(\eta) \end{equation} where the constant $c$ is defined by the relation $\Phi_*(d_G) = cd_{\hat{\hat{G}}}$. The above equation \eqref{2} leads to the equality $$ \hat{\hat{f}}(\Phi(x)) = cf(x^{-1}) $$ rather than what is stated in the text. This leads to:

Question: On pushing forward, do we have the equality $\Phi_*(d_G) = d_{\hat{\hat{G}}}$?

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If I understand author correctly - For $f \in L^1(G)$ with $\hat{f}\in L^1(\hat{G})$, you get $$ f(x) = c^{-1}\hat{\hat{f}}(\Phi(x^{-1})) $$ where $\phi_*(d_G) = cd_{\hat{\hat{G}}}$.

On the other hand, if $h \in \mathcal{B}^1(G)$, from 'inversion I' we know $\hat{h}\in L^1(\hat{G})$ and that, $$ h(x) = \int \xi(x)\hat{h}(\xi)d_{\hat{G}}(\xi) =: \hat{\hat{h}}(\Phi(x^{-1})). $$ This shows $c$ must be equal to $1$.

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