Questions on Commutativity of Mellin and Fourier Convolutions

Lately I've been investigating the following Mellin convolutions:

(1) $\quad f\,*_{\mathcal{M}_1}\,g=\int_0^\infty\frac{f(x)}{x}\,g\left(\frac{y}{x}\right)\,dx$

(2) $\quad f\,*_{\mathcal{M}_2}\,g=\int_0^\infty f(x)\,g(y\,x)\,dx$

I'm wondering about the commutativity of the two Mellin convolutions above as illustrated in (3) and (4) below.

(3) $\quad\int_0^\infty\frac{f(x)}{x}\,g\left(\frac{y}{x}\right)\,dx\stackrel{?}{=}\int_0^\infty\frac{g(x)}{x}\,f\left(\frac{y}{x}\right)\,dx$

(4) $\quad\int_0^\infty f(x)\,g(y\,x)\,dx\stackrel{?}{=}\int_0^\infty g(x)\,f(y\,x)\,dx$

I haven't been able to find much information on either Mellin convolution above, but I've found a bit more information on the first one than the second one. I've read the Mellin convolution (1) above is commutative, but I haven't found this to be true in general. A case of particular interest to me that seems to violate commutativity is illustrated in (5) and (6) below. Note (5) and (6) below are equivalent for $y>0$, but (5) below is valid for $y$ in general whereas (6) below is only valid for $y>0$.

(5) $\quad\delta(x-1)\,*_{\mathcal{M}_1}\,g(x)=\int_0^\infty\frac{\delta(x-1)}{x}g\left(\frac{y}{x}\right)\,dx=g(y)$

(6) $\quad g(x)\,*_{\mathcal{M}_1}\,\delta(x-1)=\int_0^\infty\frac{g(x)}{x}\delta\left(\frac{y}{x}-1\right)\,dx=g(y)\,,\quad y>0$

I've noticed the Mellin convolution (2) above also seems to violate commutativity for a case of particular interest to me illustrated in (7) and (8) below.

(7) $\quad\delta(x-1)\,*_{\mathcal{M}_2}\,g(x)=\int_0^\infty\delta(x-1)\,g(y\,x)\,dx=g(y)$

(8) $\quad g(x)\,*_{\mathcal{M}_2}\,\delta(x-1)=\int_0^\infty g(x)\,\delta(y\,x-1)\,dx=\frac{f\left(\frac{1}{y}\right)\,\theta \left(\frac{1}{y}\right)}{\left|y\right|}\,,\quad y\in\mathbb{R}$

Question 1: What are the conditions under which the Mellin convolution (1) above is commutative?

Question 2: What are the conditions (if any) under which the Mellin convolution (2) above is commutative?

Fourier convolution is defined as follows.

(9) $\quad f\,*\,g=\int_{-\infty}^\infty f(x)\,g(y-x)\,dx$

I'm also wondering about the commutativity of the Fourier convolution (9) above as illustrated in (10) below.

(10) $\quad\int_{-\infty}^\infty f(x)\,g(y-x)\,dx\stackrel{?}{=}\int_{-\infty}^\infty g(x)\,f(y-x)\,dx$

I've read the Fourier convolution (9) above is commutative, but I haven't found this to be true in general. I've noticed the Fourier convolution (9) above seems to violate commutativity for a case of particular interest to me illustrated in (11) and (12) below. Note (11) below is valid for $y$ in general whereas (12) below is only valid for $y\in\mathbb{R}$.

(11) $\quad\delta(x)\,*\,g(x)=\int_{-\infty}^\infty\delta(x)\,g(y-x)\,dx=g(y)$

(12) $\quad g(x)\,*\,\delta(x)=\int_{-\infty}^\infty g(x)\,\delta(y-x)\,dx=g(y)\,,\quad y\in\mathbb{R}$

Question 3: What are the conditions under which the Fourier convolution (9) above is commutative?

Perhaps the problem is with Mathematica versus convolution theory, but to illustrate a more concrete example with respect to the problems I'm having with Mellin convolution (1) above I used Mathematica to derive formulas (13) and (14) below where $J_0$ is a Bessel function of the first kind of order zero. Note these two formulas only evaluate the same for $\Re(y)>0$. Mathematica provided both of these results as unconditional implying they're both valid for all $y$.

(13) $\quad \int_0^{\infty}\frac{\cos(2\,\pi\,x)}{x}\,\sin\left(2\,\pi\,\frac{y}{x}\right)\,dx=\frac{\pi\,y}{2\,\sqrt{y^2}}\,J_0\left(\frac{4\pi}{\sqrt[4]{\frac{1}{y^2}}}\right)$

(14) $\quad\int_0^{\infty }\frac{\sin(2\,\pi\,x)}{x}\,\cos\left(2\,\pi\,\frac{y}{x}\right)\,dx=\frac{\pi}{2}\,J_0\left(\frac{4\,\pi}{\sqrt[4]{\frac{1}{y^2}}}\right)$

• The Mellin convolution is just the usual convolution after a change of variable $\int_0^\infty f(x) g(t/x) \frac{dx}{x} = \int_{-\infty}^\infty f(e^u) g(e^{\log t - u})du = \int_{-\infty}^\infty F(u) G(\log t - u)du = F \ast G(\log t)$ where $F(u) = f(e^u), G(u) = g(e^u)$. Jul 4 '17 at 22:34
• Of course the convolution (of functions and distributions, when it is well-defined) is commutative and distributive. And the pairing and is compatible with the change of variable, ie. $\int_{-\infty}^\infty f(x) \delta(ax-b)dx=\int_{-\infty}^\infty f(y/a) \delta(y-b)d(y/a) = \frac{f(b/a)}{a}$ Jul 4 '17 at 22:43
• The answer is in my comments and Donald's answer. Mathematica didn't simplify $y/\sqrt{y^2}$. So what ? Jul 5 '17 at 1:37
• @user1952009 Please see the example I added following Question (3) in the main body above. Jul 5 '17 at 1:40
• Come on stop playing with mathematica and read the books. In complex analysis we consider analytic continuation of functions. Jul 5 '17 at 1:45

Equation $(3)$ is easily shown by the substitution $x=y/ t$ so $dx =\frac{-y}{t^2} dt$ \begin{eqnarray*} f \star_{M_1} g \;\;(y) &=&\int_{0}^{\infty} \frac{f(x)}{x} g\left(\frac{x}{y}\right) dx = \int_{\infty}^0 \frac{1}{\left(\frac{y}{t}\right)} f\left(\frac{y}{t}\right) g\left(\frac{y}{\left(\frac{y}{t}\right)}\right) \frac{-y}{t^2} dt \\ &=&\int_{0}^{\infty} \frac{g(t)}{t} f\left(\frac{t}{y}\right) dt =g \star_{M_1} f \; \;(y). \end{eqnarray*}

Equation $(4)$ does not workout, try the substitution $x=ty$. We get \begin{eqnarray*} y \int_{0}^{\infty} f(ty)g(y^2 t) dt \end{eqnarray*} So commutativity of $\star_{m_2}$ would require $y g( y^2t) =g(t)$.

The commutativity of Fourier convolution is easily shown by the substitution $t=y-x$ \begin{eqnarray*} f \star g (y)&=& \int_{-\infty}^{\infty} f(x)g(y-x) dx = -\int_{\infty}^{-\infty} f(y-t)g(t) dt \\ &=& \int_{-\infty}^{\infty} g(t)f(y-t) dt = g \star f (y). \end{eqnarray*}

• In the proof of equation (3): shouldn't $g\left(\frac{y}{\left(\frac{y}{x}\right)}\right)$ be $g\left(\frac{y}{\left(\frac{y}{t}\right)}\right)$? Jul 5 '17 at 13:29
• In the proof of Fourier convolution commutativity: shouldn't $g(t)dx$ be $g(t)dt$? Jul 5 '17 at 13:34
• In the proof of equation (3): shouldn't $\frac{-y}{t^2}$ be $\frac{y}{t^2}$ since a minus sign was added preceding the integral? Jul 5 '17 at 15:47
• Yes all three were typos ... I think we have them all now ... Thank you for pointing them out. $\ddot \smile$ Jul 5 '17 at 19:19
• I'm still a bit confused with respect to the commutativity of Fourier convolution. Mathematica tells me Fourier convolution involving the $\delta(x)$ function is only commutative for $y\in \Re$ (see (11) and (12) above). I've also read Fourier convolution is a commutative algebra without identity. It seems to me if Fourier convolutions involving the $\delta(x)$ function are commutative then the $\delta(x)$ function would be the identity function with respect to Fourier convolutions. Jul 25 '17 at 15:22