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I have some problems to follow the proof of the anti commutativity property of the convolution and involution operations defined using a Haar measure as presented in Pedersen's book Analysis Now, chapter 6 section 6, theorem 6.6.21. To explain the problem I need to recall the definition he makes:

Given $G$ locally compact Hausdorff group, let $C_c(G)$ be the family of continuous functions from $G$ to $\mathbb{C}$ with compact support. A Radon integral $\int: C_c(G)\to\mathbb{C}$ defines an Haar measure if it is a linear continuous map which maps positive functions on positive real numbers and is left invariant (i.e$\int f_y=\int f$ for all $y\in G$ where $f_y(x)=f(y^{-1}\cdot x)$).

The modulus function $\Delta :G\to\mathbb{R}$ is the unique group homomorphism defined by the request that $\Delta(x)\int f(yx)d y=\int f(y)dy$ for all $x\in G$ and $f\in C_c(G)$.

The convolution of $f$ and $g$ is defined by:

$f\times g(x)=\int f(y)g_y(x)dy$.

The involution is defined by:

$f^*(x)=\overline{f(x^{-1})}\cdot \Delta(x^{-1})$.

Theorem 6.6.21 states (among other identities which are less problematic to me):

$(g^*\times f^*)=(f\times g)^*$.

The proof of this identity is as follows:

  1. $(g^*\times f^*)(x)=$

  2. $=\int g^*(y)f^*(y^{-1}x)dy=$

  3. $=\int \overline{f(x^{-1}y)}\Delta(x^{-1}y)\overline{g(y^{-1})}\Delta(y^{-1})dy=$

  4. $=\int \overline{f(x^{-1}y)}\overline{g(y^{-1})}dy\Delta(x^{-1})=$

  5. $=\int \overline{f(y)}\overline{g(y^{-1}x^{-1})}dy\Delta(x^{-1})=$

  6. $=\overline{\int f(y)g(y^{-1}x^{-1})dy}\Delta(x^{-1})=$

  7. $=(f\times g)^*(x)$.

I can't really understand how is it justified the passage from (4) to (5). Can someone help me? Thanks.

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The passage from (4) to (5) is justified because of the left invariance of the Radon integral. For fixed $x \in G$ set $$h(y) = \overline{f(x^{-1}y)}\overline{g(y^{-1})}, \quad y \in G.$$ Then (4) is just $\int h(y) \text{d}y \Delta(x^{-1})$ and (5) is $\int h_{x^{-1}}(y) \text{d}y \Delta(x^{-1})$.

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