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Simply put, are Mellin convolution and usual convolution same thing?

1) Mellin convolution:

$$\int_0^\infty f(x)g\Big(\frac{x}{y}\Big) \frac{dy}{y}$$

2) convolution:

$$\int_{-\infty}^\infty f(x-y)g(y) dy$$

would be nice if you provide some insight on their differences.

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In view of Pontryagin duality the classical Fourier transform and Mellin transform are both examples of the abstract Fourier transform on locally compact abelian (LCA) group and each is connected to its corresponding convolution that is the multiplication on each banach algebra.

For classical Fourier transform the LCA group is the real numbers with addition as group operation and the (up to scaling) Haar measure is the Lebesgue measure dx.

For Mellin the LCA group is positive real numbers with multiplication as group operation and Haar measure dx/x. This LCA group is isomorphic to (R,+) by the log map.

https://en.m.wikipedia.org/wiki/Pontryagin_duality

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Assume the following definitions of the Fourier transform, inverse transform, and related convolution.

(1) $\quad F(s)=\mathcal{F}_x[f(x)](s)=\int\limits_{-\infty}^{\infty}f(x)\,e^{-i\,s\,x}\,dx$

(2) $\quad f(x)=\mathcal{F}_s^{-1}[F(s)](x)=\frac{1}{2\,\pi}\int_{-\infty}^{\infty}F(s)\,e^{i\,s\,x}\,ds$

(3) $\quad f*_{\mathcal{F_x}}g\,(y)=\int\limits_{-\infty}^{\infty}f(x)\,g(y-x)\,dx\,,\quad y\in\mathbb{R}$

Assume the following definitions of the Mellin transform, inverse transform, and related convolution.

(4) $\quad F(s)=\mathcal{M}_x[f(x)](s)=\int\limits_0^{\infty}f(x)\,x^{s-1}\,dx$

(5) $\quad f(x)=\mathcal{M}_s^{-1}[F(s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{c-i \infty}^{c+i\,\infty}F(s)\,x^{-s}\,ds$

(6) $\quad f*_{\mathcal{M_x}}g\,(y)=\int\limits_0^{\infty}f(x)\,g\left(\frac{y}{x}\right)\frac{dx}{x}\,,\quad y>0$

The Fourier transform of the Fourier convolution $f(x)*_{\mathcal{F}}g(x)$ is the product of the Fourier transforms of $f(x)$ and $g(x)$ as illustrated in (7) below.

(7) $\quad\mathcal{F}_x[f(x)*_{\mathcal{F}}g(x)](s)=\mathcal{F}_x[f(x)](s)\,\mathcal{F}_x[g(x)](s)$

Likewise, The Mellin transform of the Mellin convolution $f(x)*_{\mathcal{M}}g(x)$ is the product of the Mellin transforms of $f(x)$ and $g(x)$ as illustrated in (8) below.

(8) $\quad\mathcal{M}_x[f(x)*_{\mathcal{M}}g(x)](s)=\mathcal{M}_x[f(x)](s)\,\mathcal{M}_x[g(x)](s)$

The relationships between the Fourier and Mellin transforms are as follows. Note in (9) and (10) below the Mellin transforms are evaluated at $-i\,s$.

(9) $\quad\mathcal{F}_u\left[f\left(e^u\right)\right](s)=\mathcal{M}_x[f(x)](-i\,s)$

(10) $\quad\mathcal{M}_u\left[f(\log(u))\right](-i\,s)=\mathcal{F}_x[f(x)](s)$

The derivation of the Mellin transform from the Fourier transform in (9) above can be verified with the variable substitution $x=e^u$ in the Fourier transform integral in (11) below. Since $x=e^u$, $e^{-i\,s\,u}=x^{-i\,s}$, $du=\frac{dx}{e^u}=\frac{dx}{x}$, the lower integration limit becomes $e^{-\infty}=0$, and the upper integration limit becomes $e^{\infty}=\infty$.

(11) $\quad\mathcal{F}_u\left[f\left(e^u\right)\right](s)=\int\limits_{-\infty}^{\infty}f\left(e^u\right)\,e^{-i\,s\,u}\,du=\int\limits_0^{\infty}f(x)\,x^{-i\,s-1}\,dx=\mathcal{M}_x[f(x)](-i\,s)$

The derivation of the Fourier transform from the Mellin transform in (10) above can be verified with the variable substitution $x=\log(u)$ in the Mellin transform integral in (12) below. Since $x=\log(u)$, $u=e^x$, $dx=\frac{du}{u}$, the lower integration limit becomes $\log(0)=-\infty$, and the upper integration limit becomes $\log(\infty)=\infty$.

(12) $\quad\mathcal{M}_u\left[f(\log(u))\right](-i\,s)=\int_0^{\infty}f(\log(u))\,u^{-i\,s-1}\,du=\int_{-\infty}^{\infty}f(x)\,e^{-i\,s\,x}\,dx=\mathcal{F}_x[f(x)](s)$

Assuming $F(u)=f(e^u)$ and $G(u)=g(e^u)$, the Mellin convolution can be derived from the Fourier convolution as follows where the Fourier convolution is evaluated with the variable substitution $x=e^u$. Since $x=e^u$, $e^{\log(y)-u}=\frac{e^{\log(y)}}{e^u}=\frac{y}{x}$, $du=\frac{dx}{e^u}=\frac{dx}{x}$, the lower integration limit becomes $e^{-\infty}=0$, and the upper integration limit becomes $e^{\infty}=\infty$.

(13) $\quad F*_{\mathcal{F_u}}G\,(log(y))=\int\limits_{-\infty}^{\infty}f\left(e^u\right)g\left(e^{\log(y)-u}\right) \,du=\int\limits_0^{\infty}f(x)\,g\left(\frac{y}{x}\right)\frac{dx}{x}=f*_{\mathcal{M_x}}g\,(y)$

Assuming $F(u)=f(log(u))$ and $G(u)=g(log(u))$, the Fourier convolution can be derived from the Mellin convolution as follows where the Mellin convolution is evaluated with the variable substitution $x=\log(u)$. Since $x=\log(u)$, $\log\left(\frac{e^y}{u}\right)=\log\left(e^y\right)-\log(u)=y-x$, $dx=\frac{du}{u}$, the lower integration limit becomes $\log(0)=-\infty$, and the upper integration limit becomes $\log(\infty)=\infty$.

(14) $\quad F*_{\mathcal{M_u}}G\,(e^y)=\int_0^{\infty}f(\log(u))g\left(\log\left(\frac{e^y}{u}\right)\right)\frac{du}{u}=\int_{-\infty }^{\infty }f(x)\,g(y-x)\,dx=f*_{\mathcal{F_x}}g\,(y)$

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  • $\begingroup$ This is a mess. Mellin convolution $f \star g(x) = \int_0^\infty f(y) g(\frac{x}y)\frac{dy}{y}$ ($x > 0$), Fourier convolution $F \ast G(u) = \int_{-\infty}^\infty F(v)G(u-v)dv$ ($u \in \mathbb{R}$). Let $F(u) = f(e^{u}), G(u) = g(e^{u})$ and use the change of variable $y = e^v, dy = e^v dv$ you'll obtain $f \star g(x) = \int_0^\infty f(y) g(\frac{x}{y})\frac{dy}{y} = \int_{-\infty}^\infty f(e^{v}) g(\frac{x}{e^v})\frac{e^v dv}{e^v} = \int_{-\infty}^\infty F(v) G(\log x - v)= F \ast G(\log x)$. $\endgroup$ – reuns Dec 29 '17 at 20:28
  • $\begingroup$ In (14) above my goal was to derive $f\,*\,g\,(y)$, not $F\,*\,G\,(\log\,y)$. Personally I find your derivation more obscure than mine because you substitute function definitions as well as variables. But your derivation has the advantage that both the starting point and ending point meet the definition of a convolution, whereas the left sides of (13) and (14) above do not. $\endgroup$ – Steven Clark Dec 30 '17 at 0:53
  • $\begingroup$ ?? The answer to the question takes 2 lines and there is no alternative to what I wrote (up to renaming and $x= e^u$). Why do you start with the Fourier and Mellin transform ? And you didn't realize yet the variable is real in the Fourier transform and complex (not purely imaginary) in the Mellin and (bilateral) Laplace transform ? $\endgroup$ – reuns Dec 30 '17 at 1:15
  • $\begingroup$ I edited my answer to make things more clear. Thanks for your insight. $\endgroup$ – Steven Clark Dec 30 '17 at 15:55
  • $\begingroup$ I started with the Fourier and Mellin transforms in an attempt to show the relationships between Fourier and Mellin convolutions are analogous to the relationships between Fourier and Mellin transforms, but the results are not quite what I expected, so perhaps this didn't really add much insight. $\endgroup$ – Steven Clark Dec 30 '17 at 18:08

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