If $f,g\in L_1$ we define the product $\Box$ as $$f\Box g(x)=f(x)\int_{-\infty}^xg(y)dy + g(x)\int_{-\infty}^x f(y)dy $$
The first question is if this product has a special name.
So to show that it has a associative property I need to show: $[(f\Box g)\Box h](x) = [f\Box(g\Box h)](x)$. My idea was quite simple, just expand both sides and then look at what I get
$$\begin{align}[(f\Box g)\Box h](x) &=f(x)\int_{-\infty}^xg(y)dy\int_{-\infty}^xh(y)dy + g(x)\int_{-\infty}^xf(y)dy\int_{-\infty}^xh(y)dy \\ &+ h(x)\int_{-\infty}^xf(y)dy\int_{-\infty}^yg(z)dzdy+h(x)\int_{-\infty}^xg(y)\int_{-\infty}^yf(z)dzdy \end{align}$$ and $$\begin{align}[f\Box(g\Box h)](x) &= f(x)\int_{-\infty}^xg(y)\int_{-\infty}^yh(z)dzdy+f(x)\int_{-\infty}^xh(y)\int_{-\infty}^yg(z)dzdy\\ &+g(x)\int_{-\infty}^xh(y)dy\int_{-\infty}^xf(y)dy+h(x)\int_{-\infty}^xg(y)dy\int_{-\infty}^xf(y)dy \end{align}$$
The problem is that I don't see why these two things should be the same. Hope someone can help a bit.
First, you have to change the order of intergration:
$$ \int\limits_{-\infty}^x dz \; f(z) \int\limits_{-\infty}^z dy \; g(y) = \int\limits_{-\infty}^x dy \; g(y) \int\limits_y^x dz \; f(z). $$
Now we notice that two last terms of $(f \square g) \square h$ sum up to $$ h(x) \int\limits_{-\infty}^x dy \; g(y) \left[ \int\limits_y^x dz \; f(z) + \int\limits_{-\infty}^y dz \; f(z) \right] = h(x) \left[ \int\limits_{-\infty}^x dy \; g(y) \right] \; \left[ \int\limits_{-\infty}^x dz \; f(z) \right]. $$
And whole expression becomes symmetric
$$ (f \square g) \square h \, (x) = f(x) \left[ \int\limits_{-\infty}^x dy \; g(y) \right] \; \left[ \int\limits_{-\infty}^x dz \; h(z) \right] + \\ + g(x) \left[ \int\limits_{-\infty}^x dy \; h(y) \right] \; \left[ \int\limits_{-\infty}^x dz \; f(z) \right]+ h(x) \left[ \int\limits_{-\infty}^x dy \; g(y) \right] \; \left[ \int\limits_{-\infty}^x dz \; f(z) \right] $$