The convolution operator is defined as
$${\displaystyle (f*g)(t)\triangleq \ \int _{-\infty }^{\infty }f(\tau )g(t-\tau )\,d\tau .} $$
where it shares a relationship with the Laplace transform such that:
$${\mathcal{L}\left\{ {f * g} \right\} = F\left( s \right)G\left( s \right)\hspace{0.25in}\hspace{0.25in}{\mathcal{L}^{\, - 1}}\left\{ {F\left( s \right)G\left( s \right)} \right\} = \left( {f * g} \right)\left( t \right) }$$
However, there are many more types of integral transforms than just the laplace transform. For example, the Mellin transform:
$${\displaystyle \left\{{\mathcal {M}}f\right\}(s)=\varphi (s)=\int _{0}^{\infty }x^{s-1}f(x)\,dx.}$$
which has an 'equivalent' convolutional operator of: $${\displaystyle (f*g)(x)= \ \int _{0 }^{\infty }f(y)g\left( \frac{x}{y} \right) \frac{dy}{y}\,} $$
Keeping these two examples in mind, is there a a more general convolution operation that has the property:
$$\left( {f * g} \right)\left( x \right) = {\mathcal{T}^{\, - 1}}\left\{ { \mathcal{T}\left\{ {f} \right\}\mathcal{T}\left\{ {g} \right\} } \right\}$$
where the transform T is an integral transform of the form:
$${\displaystyle (Tf)(u)=\int \limits _{t_{1}}^{t_{2}}f(t)\,K(t,u)\,dt}$$
Thanks!
References: https://en.wikipedia.org/wiki/Integral_transform https://www.johndcook.com/blog/2017/12/03/transforms-and-convolutions/