Fourier convolution is defined as follows.
(1) $\quad\left(f(x)\ *_\mathcal{F}\ g(x)\right)(y)=\int\limits_{-\infty}^\infty f(x)\ g(y-x)\ dx$
The Hilbert transform of $f(x)$ defined in (2) below is based on Fourier convolution where $g(x)=\frac{1}{x}$. Note the inverse Hilbert transform is defined in (3) below is the negative of the Hilbert transform defined in (2) below $\left(\mathcal{H}_y^{-1}[f(y)](x)=-\mathcal{H}_y[f(y)](x)\right)$.
(2) $\quad\hat{f}(y)=\mathcal{H}_x[f(x)](y)=\int\limits_{-\infty}^\infty f(x)\ \frac{1}{y-x}\ dx$
(3) $\quad f(x)=\mathcal{H}_y^{-1}[f(y)](x)=-\int\limits_{-\infty}^\infty f(y)\ \frac{1}{x-y}\ dy$
Mellin convolution can be defined in a couple of ways as follows.
(4) $\quad\left(f(x)\ *_{\mathcal{M}_1}\ g(x)\right)(y)=\int\limits_0^\infty f(x)\ g(\frac{y}{x})\ \frac{dx}{x}$
(5) $\quad\left(f(x)\ *_{\mathcal{M}_2}\ g(x)\right)(y)=\int\limits_0^\infty f(x)\ g(y\ x)\ dx$
Question (1): Are there any notable transforms based on either of the Mellin convolutions defined above?
Question (2): Is there a function $g(x)$ for which either of the Mellin convolutions defined above is its own inverse or the negative of its own inverse as illustrated in (6) below analogous to the inverse Hilbert transform defined in (3) above?
(6) $\quad \hat{f}(y)=(f(x)\ *_\mathcal{M}\ g(x))(y)\implies f(x)=\pm\left(\hat{f}(y)\ *_\mathcal{M}\ g(y)\right)(x)$
Question (3): Is question (2) above misguided? Should I really be looking for a function $g(x)$ that satisfies something like the following as the Mellin convolutions analogous to the Fourier convolutions illustrated formulas (2) and (3) above?
(7) $\quad \hat{f}(y)=\left(f(x)\ *_\mathcal{M}\ g(x)\right)(y)\implies f(x)=\left(\hat{f}(y)\ *_\mathcal{M}\ g(\frac{1}{y})\right)(x)$
