Comparing $X+Y$ and $X-Y$ for independent random variables with values in an abelian locally compact group

Question

Let $G$ be an abelian locally (separable?) compact group with Haar measure $\mu$. Inspired by the interesting proof of A sum of two binomial random variables : Let $X$ and $Y$ be $G$-valued independent random variables with $\mu$-density $f_X$ resp. $f_Y$. Let $f_{X \pm Y}$ be the $\mu$-density of $X \pm Y$. Is it always true that $$\int f_{X-Y}^2 d\mu = \int f_{X+Y}^2 d\mu \ ?$$ This holds true for $G = \mathbb{Z}$ (here both integrals are finite) .

Answer 1

The answer is yes. Indeed, let $dx:=\mu(dx)$ for brevity. We have \begin{align} I_{X,Y}&:=\iiint f_X(x'+y'-y)f_Y(y)f_Y(y')f_X(x')dx'\,dy\,dy' \\ &=\iint f_{X+Y}(x'+y')f_Y(y')f_X(x')dx'\,dy' \\ &=\iint f_{X+Y}(t)f_Y(t-x')f_X(x')dt\,dx' \\ &=\int f_{X+Y}(t)^2dt. \end{align} Also, \begin{align} I_{X,Y}&=\iiint f_X(x'-u'+u)f_Y(-u)f_Y(-u')f_X(x')dx'\,du\,du' \\ &=\iiint f_X(x'+u-u')f_{-Y}(u)f_{-Y}(u')f_X(x')dx'\,du\,du' \\ &=\iiint f_X(x'+u'-u)f_{-Y}(u)f_{-Y}(u')f_X(x')dx'\,du\,du' \\ &=I_{X,-Y} =\int f_{X-Y}(t)^2dt, \end{align} by what was shown in the previous display.

Thus, $$\int f_{X+Y}(t)^2dt=\int f_{X-Y}(t)^2dt,$$ as desired.