I'm trying to show $(L^1(\mathbb{R}^n), \ast)$ is a Banach algebra and I'm ending up with some variables in switched positions.
Let $f, g \in L^1(\mathbb{R}^n)$ and define the convolution of $f$ with $g$ by $$ (f \ast g)(x) = \int_{\mathbb{R}^n} f(y)g(x-y)dy. $$ Now $(L^1(\mathbb{R}^n), \ast)$ is a Banach algebra. I am trying to show this by showing that for $a,b,c \in (L^1(\mathbb{R}^n), \ast)$ we have the associative property $(a\ast b) \ast c = a \ast (b \ast c)$.
First we have \begin{align} ((a\ast b) \ast c) (x) & = \int_{\mathbb{R}^n} (a\ast b)(y)c(x-y)dy \\ & = \int_{\mathbb{R}^n} \int_{\mathbb{R}^n}a(z)\mathbf{b(y-z)}dz \ c(x-y)dy, \end{align} and I have bolded the term that I will have an issue with later.
And then \begin{align} (a \ast (b \ast c)) (x) & = \int_{\mathbb{R}^n} a(y)(b\ast c)(x-y)dy \\ & = \int_{\mathbb{R}^n} a(y)\int_{\mathbb{R}^n}b(z)c(x-y-z)dzdy \\ \end{align} Making the change of variables $\phi = y-z$ we have $z = y - \phi, \ d\phi = - dz$ and then \begin{align} (a \ast (b \ast c)) (x) & = -\int_{\mathbb{R}^n} a(y)\int_{\mathbb{R}^n}b(y-\phi)c(x-\phi)d\phi dy \\ & = -\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} a(y)b(y-\phi)dy \ c(x-\phi)d\phi \\ \end{align} Now replacing $y$ with $z$ and $\phi$ with $y$ I have $$ (a \ast (b \ast c)) (x) = -\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} a(z)\mathbf{b(z-y)}dy \ c(x-y)dy, $$ and I have bolded the term corresponding to the one I bolded earlier. The function $b$ has the variables switched in the last expression compared to the earlier one, and there is also a minus sign in the final expression. I don't know whether it's possible to 'bring the minus sign into b' in the final expression? Have I made a mistake somewhere?
You have $c (x-y-z) = c (x- (y+z)) $ in one of the integrals and then substitute $\phi = y-z $ to get $c (x-\phi) $. This is incorrect, you should have used $\phi = y+z) $.
Also, you do not really have $d\phi = -dz $, since the change of variables formula for the Lebesgue integral uses the absolute value of the jacobian. Of course, this problem also vanishes if you use the correct substitution from above.