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:
-
$(g^*\times f^*)(x)=$
-
$=\int g^*(y)f^*(y^{-1}x)dy=$
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$=\int \overline{f(x^{-1}y)}\Delta(x^{-1}y)\overline{g(y^{-1})}\Delta(y^{-1})dy=$
-
$=\int \overline{f(x^{-1}y)}\overline{g(y^{-1})}dy\Delta(x^{-1})=$
-
$=\int \overline{f(y)}\overline{g(y^{-1}x^{-1})}dy\Delta(x^{-1})=$
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$=\overline{\int f(y)g(y^{-1}x^{-1})dy}\Delta(x^{-1})=$
-
$=(f\times g)^*(x)$.
I can't really understand how is it justified the passage from (4) to (5).
Can someone help me? Thanks.
Best Answer
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 $\int h_{x^{-1}}(y) \text{d}y \Delta(x^{-1})$.