Can convolution fails to be commutative and distributive in the sense of (generalized) Riemann integral?

Question

For two (say continuous) function $f,g\in C(\mathbb R^n)$, let us define \begin{equation} f\ast g(x):=\lim_{R\to+\infty}\int_{B^n(0,R)}f(x-y)g(y)dy, \end{equation} whenever the integral converges.

It is partially clear that under such definition the convolution can be non-associative (e.g. this post). Indeed by taking Fourier transform convolutions become multiplications, where the multiplications of distributions can be non-associative without certain restrictions.

My question is, can it fails to be commutative and distributive as well? More precisely,

1. Can we find $f,g\in C(\mathbb R^n)$ such that $f\ast g$ and $g\ast f$ both converge, but not equal?

2. Can we find $f,g,h\in C(\mathbb R^n)$ such that $f\ast h$, $g\ast h$ and $(f+g)\ast h$ all converge, but $f\ast h+g\ast h\neq(f+g)\ast h$?

Note that both questions don't have analogy on the products of distributions.

For a bonus part from Proposition 8.6 in Folland's Real Analysis, set $\tau_hf(x)=f(x-h)$,

3. Can we find $f,g\in C(\mathbb R^n)$ and $h\in\mathbb R^n$ such that $f\ast g$ and $(\tau_hf)\ast g$ are both converge, but $\tau_h(f\ast g)\neq(\tau_hf)\ast g$? Can we do the similar thing to $f\ast(\tau_hg)$ as well?

Answer 1

This convolution is distributive, since for all $x\in\mathbb{R}^n$ and $R>0$, we have: $$ \int_{B^n(0,R)} (f+g)(x-y)h(y)dy = \int_{B^n(0,R)} f(x-y)h(y)dy + \int_{B^n(0,R)} g(x-y)h(y)dy, $$ and so taking the limit $R\to\infty$, we get $(f+g)*h = f*g + g*h$ by definition. Similarly, it is distributive on the other side, with $f*(g+h) = f*g + f*h$.

In the same vein, we also have: $$ \tau_h(f*g)(x) = f*g(x-h) = \lim_{R\to\infty} \int_{B^n(0,R)} f(x-h-y)g(y)dy = \lim_{R\to\infty} \int_{B^n(0,R)} \tau_h f(x-y)g(y)dy = (\tau_h f)*g(x). $$

However, it is NOT commutative. For simplicity's sake, $f$ will be discontinuous but that won't change the arguments. We'll also work in dimension $n=1$.

Take $f(x)=sgn(x)$ the sign function, and $g\equiv 1$ constant. We have for $R>|x|$: $$ \int_{-R}^R f(x-y)g(y) = \int_{-R}^x 1dy + \int_x^R (-1)dy = x+R - (R-x) = 2x, $$ and so $f*g(x)=2x$. However: $$ \int_{-R}^R g(x-y)f(y)dy = \int_{-R}^R sgn(y)dy = 0, $$ and so $f*g\neq g*f$. We can extend the arguments to continuous functions by just taking an approximation of $sgn(x)$, and to higher dimensions $n$ by taking $f(x)\approx sgn(x_1)\cdot\boldsymbol{1}_{|x_i|<1, \ i\geq2}$ instead. This also means there are $f,g$ such that $\tau_h(f*g)\neq f*(\tau_h g)$.