Let $\gamma \colon (\mathbb{R}_+,+) \to (\mathbb{C}^\times,\cdot) $ a continous Homomorphism such that $\|\gamma\|_\infty \le 1$. I want to prove that for $ f,g \in L^1(\mathbb{R}_+) $: $$ \int_{\mathbb{R}_+} (f \ast g ) (x) \gamma(x) \mathrm{d}x = \int_{0}^{\infty} g(y) \gamma(y) \mathrm{d}x \int_{0}^{\infty} f(x) \gamma(x) \mathrm{d}y $$.
My try:
\begin{align} \int_{\mathbb{R}_+} ( f \ast g ) (x) \gamma(x) \mathrm{d}x &= \int_{0}^{\infty} \gamma(x) \int_{0}^{x} f(x-y)g(y) \mathrm{d}y \mathrm{d}x \\ &= \int_{0}^{\infty} \gamma(y) \int_{0}^{x} f(x-y)g(y) \gamma(x-y) \mathrm{d}y \mathrm{d}x \\ &= \int_{0}^{\infty} \int_{0}^{\infty } f(x-y) g(y) \gamma(x-y) \gamma(y) 1_{ [ 0,x ] } (y) \mathrm{d}y \mathrm{d}x \\ &= \int_{0}^{\infty} \int_{0}^{\infty } f(x-y) g(y) \gamma(x-y) \gamma(y) 1_{ [ 0,x ] } (y) \mathrm{d}x \mathrm{d}y \\ &= \int_{0}^{\infty} \int_{0}^{\infty } f(x-y) g(y) \gamma(x-y) \gamma(y) \mathrm{d}x \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \int_{0}^{\infty } f(x-y) \gamma(x-y) \mathrm{d}x \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \left[ \int_{0}^{\infty} f(x-y) \gamma(x-y) \mathrm{d}x \right] \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \left[ \int_{0}^{\infty} f(x) \gamma(x) \mathrm{d}x \right] \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \mathrm{d}x \int_{0}^{\infty} f(x) \gamma(x) \mathrm{d}y \end{align}
Explanation where needed:
Line 2: $\gamma(x-y)\gamma(y) = \gamma(x)$ if $x-y \ge 0$
Line 3: Transform integral limits into characteristic function
Line 4: Fubini
Line 5: The Integral over this function results into $1_{[0,\infty)}(x)$. This is a step where i do not know any rigor way to prove this.
Line 8: Extending $f $ and $\gamma$ by zero on $(-\infty,0)$ and due to the translation invariance of the lebesgue measure this holds.
I now know how to better prove the step in line 5:
It is $1_{[0,x]}(y)$ with $0 \le y \le \infty$ and $x \ge 0$. This is equivalent to $1_{[0,\infty)}(x-y)$ and $0 \le y < \infty$. So the step 5 should look like in the following calculation:
\begin{align} \int_{\mathbb{R}_+} ( f \ast g ) (x) \gamma(x) \mathrm{d}x &= \int_{0}^{\infty} \gamma(x) \int_{0}^{x} f(x-y)g(y) \mathrm{d}y \mathrm{d}x \\ &= \int_{0}^{\infty} \gamma(y) \int_{0}^{x} f(x-y)g(y) \gamma(x-y) \mathrm{d}y \mathrm{d}x \\ &= \int_{0}^{\infty} \int_{0}^{\infty } f(x-y) g(y) \gamma(x-y) \gamma(y) 1_{ [ 0,x ] } (y) \mathrm{d}y \mathrm{d}x \\ &= \int_{0}^{\infty} \int_{0}^{\infty } f(x-y) g(y) \gamma(x-y) \gamma(y) 1_{ [ 0,x ] } (y) \mathrm{d}x \mathrm{d}y \\ &= \int_{0}^{\infty} \int_{0}^{\infty } f(x-y) g(y) \gamma(x-y) \gamma(y) 1_{[0,\infty)}(x-y) \mathrm{d}x \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \int_{0}^{\infty } f(x-y) \gamma(x-y) 1_{[0,\infty)}(x-y) \mathrm{d}x \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \left[ \int_{0}^{\infty} f(x-y) \gamma(x-y) \mathrm{d}x \right] \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \left[ \int_{0}^{\infty} f(x) \gamma(x) \mathrm{d}x \right] \mathrm{d}y \\ &= \int_{0}^{\infty} g(y) \gamma(y) \mathrm{d}x \int_{0}^{\infty} f(x) \gamma(x) \mathrm{d}y \end{align}
Where in line 7 i can remove $1_{[0,\infty)}(x-y)$ due to the fact that it already holds $f(x-y) = 0 $ for $x-y<0$.