# Questions on Mellin and Fourier Convolutions and Evaluation with Generalized Functions (Distributions)

This question is a follow-on and expansion of the following question.

Questions on Commutativity of Mellin and Fourier Convolutions

Lately I've been investigating the following Mellin and Fourier convolutions:

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

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

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

I've read and been told Mellin convolution $*_{\mathcal{M}_1}$ defined in (1) above is commutative, but I haven't found this to be true for a case of particular interest to me illustrated in (4) and (5) below. Note (4) and (5) below are only equivalent for $y\in\mathbb{R}$ and $y>0$.

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

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

Question 1: Is it true in general that Mellin convolution $*_{\mathcal{M}_1}$ defined in (1) above is only commutative for $y\in\mathbb{R}$ and $y>0$, or is this a result of convolving with a generalized function (distribution) versus a normal function?

I've also read and been told Fourier convolution defined in (3) above is commutative, but again I haven't found this to be true for a case of particular interest to me illustrated in (6) and (7) below. Note (6) and (7) below are only equivalent for $y\in\mathbb{R}$.

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

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

Question 2: Is it true in general that Fourier convolution defined in (3) above is only commutative for $y\in\mathbb{R}$, or is this a result of convolving with a generalized function (distribution) versus a normal function?

I've also been investigating the following convolution which I consider to be a Mellin convolution.

(8) $\quad f\,*_{\mathcal{M}_3}\,g=\int_0^\infty f(x)\,g(y+1-x)\,dx$

I believe Mellin convolution $*_{\mathcal{M}_3}$ defined in (8) above can be shown to be commutative via integration by substitution, but again I haven't found this to be true for a case of particular interest to me illustrated in (9) and (10) below. Note (9) and (10) below are only equivalent for $y\in\mathbb{R}$ and $y>0$.

(9) $\quad\delta(x-1)\,*_{\mathcal{M}_3}\,g(x)=\int_0^{\infty }\delta (x-1)\,g(y+1-x)\,dx=g(y)$

(10) $\quad g(x)\,*_{\mathcal{M}_3}\,\delta(x-1)=\int_0^\infty g(x)\,\delta(y-x)\,dx=g(y)\,\theta(y)\,,\quad y\in\mathbb{R}$

Question 3: Is it true in general that Mellin convolution $*_{\mathcal{M}_3}$ defined in (8) above is only commutative for $y\in\mathbb{R}$ and $y>0$, or is this a result of convolving with a generalized function (distribution) versus a normal function?

Relationships (11) and (12) below illustrate an analogy between evaluation of the Fourier convolution defined in (3) above with successive derivatives of $\delta(x)$ and evaluation of the Mellin convolution $*_{\mathcal{M}_3}$ defined in (8) above with successive derivatives of $\delta(x-1)$.

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

(12) $\quad g^{(n)}(y)=\delta^{(n)}(x-1)\,*_{\mathcal{M}_3}\,g(x)=\int_0^\infty\delta^{(n)}(x-1)\,g(y+1-x)\,dx$

Relationship (13) below illustrates Mellin convolution $*_{\mathcal{M}_2}$ defined in (2) above exhibits a slightly more complicated pattern with respect to evaluation with successive derivatives of $\delta(x-1)$.

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

Relationships (14) to (17) below illustrate Mellin convolution $*_{\mathcal{M}_1}$ defined in (1) above exhibits a considerably more complicated pattern with respect to evaluation with successive derivatives of $\delta(x-1)$.

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

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

(16) $\quad\delta''(x-1)\,*_{\mathcal{M}_1}\,g(x)=\int_0^\infty\frac{\delta''(x-1)}{x}g\left(\frac{y}{x}\right)\,dx=2\,g(y)+y\left(4\,g'(y)+y\,g''(y)\right)$

(17) $\quad\delta^{(3)}(x-1)\,*_{\mathcal{M}_1}\,g(x)=\int_0^\infty\frac{\delta^{(3)}(x-1)}{x}g\left(\frac{y}{x}\right)\,dx=6\,g(y)+y\left(18\,g'(y)+y\left(9\,g''(y)+y\,g^{(3)}(y)\right)\right)$

The relationship between Fourier convolution and the Fourier transform is illustrated in (18) below where $k$ is a constant that depends on the specific normalization of the Fourier transform. The relationship between Mellin convolution $*_{\mathcal{M}_1}$ and the Mellin transform is analogous as illustrated in (19) below.

(18) $\quad\mathcal{F}\{f\,*\,g(x)\}=k\cdot\mathcal{F}\{f\}\cdot\mathcal{F}\{g\}$

(19) $\quad\mathcal{M}\{f\,*_{\mathcal{M}_1}\,g\}=\mathcal{M}\{f\}\cdot\mathcal{M}\{g\}$

Mellin convolution $*_{\mathcal{M}_1}$ seems to be the most widely known Mellin convolution. For example, the Wolfram Language implements Mellin convolution $*_{\mathcal{M}_1}$ as the MellinConvolve function. I've found a tidbit of information on Mellin convolution $*_{\mathcal{M}_2}$ at Mellin Transform Methods which indicates the following relationship between Mellin convolution $*_{\mathcal{M}_2}$ and the Mellin transform.

(20) $\quad\mathcal{M}\{f\,*_{\mathcal{M}_2}\,g\}(z)=\mathcal{M}\{f\}(1-z)\cdot\mathcal{M}\{g\}(z)$

I don't believe relationship (20) above is correct. I believe the correct relationship between Mellin convolution $*_{\mathcal{M}_2}$ and the Mellin transform is illustrated in (21) below.

(21) $\quad\mathcal{M}\{f\,*_{\mathcal{M}_2}\,g\}(z)=\mathcal{M}\{f\}(1+z)\cdot\mathcal{M}\{g\}(z)$

Question (4): What is the correct relationship between Mellin convolution $*_{\mathcal{M}_2}$ and the Mellin transform?

I haven't been able to find any information at all on Mellin convolution $*_{\mathcal{M}_3}$ but I believe Mellin convolution $*_{\mathcal{M}_3}$ is related to the Mellin transform as illustrated in (22) below which is analogous to relationship (21) above.

(22) $\quad\mathcal{M}\{f\,*_{\mathcal{M}_3}\,g\}(z)=\mathcal{M}\{f\}(1+z)\cdot\mathcal{M}\{g\}(z)$

Question (5): Is the relationship between Mellin convolution $*_{\mathcal{M}_3}$ and the Mellin transform illustrated in (22) above correct? If not, what is this relationship?

Question (6): Can anyone point me to more information regarding Mellin convolutions $*_{\mathcal{M}_2}$ and $*_{\mathcal{M}_3}$ defined in (2) and (8) above respectively?

To clarify further, I'm trying to understand the definition of "commutative" with respect to the convolutions defined in (1) and (3) above. I've seen unconditional claims these two convolutions are commutative in several places, but Mathematica seems to indicate the convolution defined in (1) above is only commutative for $y\in\mathbb{R}_{>0}$ and the convolution defined in (3) above is only commutative for $y\in\mathbb{R}$ (at least when evaluated with $f(x)=\delta(x-1)$ and $f(x)=\delta(x)$ respectively). I'm trying to understand if this is true in general or only when convolving with generalized functions (distributions), and also why this is true for either or both cases. I'll note that convolutions (1) and (3) above can both be used to derive functions which converge for $y\in\mathbb{C}$ or a subset thereof (typically $\Re(y)<0$ or $\Re(y)>0$), but even if one only considers $y\in\mathbb{R}$ convolution (1) above only seems to be commutative for a subset of $y\in\mathbb{R}$ (i.e. $y>0$).

• Read a course on the Fourier transform of functions and distributions, the Mellin transform and convolution being the same as the Fourier transform and convolution up to a change of variable. Yes of course for the Mellin convolution we only look at $y > 0$. How do you define $\delta(y-1)$ for $y$ complex ? Netherless, if $g$ is analytic then analytic continuation might provide an extension of $\int_0^\infty f(x) g(y/x) \frac{dx}{x}$ to $y$ complex. – reuns Jul 30 '17 at 22:33

A more-general viewpoint may help clarify what would be a reasonable thing to do, or not. For any (implicitly, locally compact, Hausdorff, probably countably based) topological group $G$ (let's say unimodular, for simplicity, that is, that a left Haar measure is a right Haar measure), and for $\varphi,\psi\in C^o_c(G)$ (compactly supported continuous functions) there is the convolution $$(\varphi*\psi)(g) \;=\; \int_G \varphi(gh^{-1})\,\psi(h)\;dh$$ We do not have to depend on an epiphany to acquire this formula: this is the provably unique "product" on $C^o_c(G)$ such that $\varphi\cdot (\psi\cdot v)=(\varphi*\psi)\cdot v$ for the integrated action on any reasonable (=quasi-complete, locally convex) representation space of $G$.

When $G$ is abelian, by a change of variables in the integral it follows that convolution is commutative. Done.

(Notably, in looking at Mellin transforms and/or convolutions related to these, we are looking at the multiplicative group $(0,\infty)$, and the appropriate integral is $\int_0^\infty f(xy^{-1})\,g(y)\;{dy\over y}$, rather than just $dy$ without the division by $y$.)

To give a robust sense to convolutions involving distributions (on Lie groups, say), we have to be careful, for more than one reason. For example, on the group $\mathbb R$, there's the classic chestnut $$1*(\delta'*H)\;=\; 1*\delta \;=\; 1 \;\not=\; 0 \;=\; 0 * H \;=\; (1*\delta')*H$$ showing non-associativity if we are too grandiose in our claims/definitions.

The main reason is that, as noted above, the genuine sense of "convolution" is a manifestation of associativity of integral operators on representation spaces. How should distributions act on representation spaces?

Compactly-supported distributions on Lie groups are demonstrably derivatives of $C^o_c$ functions. Thus, again requiring associativity of actions, compactly-supported distributions can act on smooth vectors in representations, and thus have a well-characterized convolution extending that of $C^o_c$ by being compatible with differentiation (by the Lie algebra).

The issue about extending distributions or functions to some sort of complexification of a given real Lie group is a bit of a red herring, but not entirely unreasonable. Cf. Paley-Wiener theorems, and more recent work of various people in repn theory.

The only thing you need to know about distributions is that $\delta(t)$ is defined by $$\int_{-\infty}^\infty \delta(t) \phi(t) dt \quad\overset{def}=\quad \lim_{\epsilon \to 0^+} \int_{-\epsilon}^\epsilon \frac{1_{|t| < \epsilon}}{2 \epsilon} \phi(t) dt$$ If $\phi$ is continuous at $t=0$ then the result of this limit is $\phi(0)$.

This simple definition, when we replace $\frac{1_{|t| < \epsilon}}{2 \epsilon}$ by a smoothed version $\psi_\epsilon(t)= \frac{\psi(t/\epsilon)}{ \epsilon}$ (with $\psi > 0$ smooth, compactly supported and $\int_{-\infty}^\infty \psi(t)dt=1$) allows us to define the derivative $\delta'$

$$\int_{-\infty}^\infty \delta'(t) \phi(t) dt \quad\overset{def}=\quad \lim_{\epsilon \to 0^+} \int_{-\epsilon}^\epsilon \psi_\epsilon'(t) \phi(t) dt$$ (check this is consistent with the integration by parts)

as well as the Fourier transform, the convolution and everything we need.

To ensure rigorously those limits exist (and don't depend on the chosen sequence $\psi_\epsilon$) we need some sort of continuity : this is what the test function topology and the induced distribution topology are about.

• @StevenClark My answer (and my comment) explain why your question is a mess. What do you want to know/prove ? Every distribution $T(x)$ is the limit (in the sense of distributions) of a sequence of $C^\infty_c$ functions, namely $T_n(x) = (T \ast n\varphi(n x)) \varphi(x/n)$, which means $\int_{-\infty}^\infty T(x) \phi(x)dx= \lim_{n \to \infty} \int_{-\infty}^\infty T_n(x) \phi(x)dx$ whenever $\phi \in C^\infty_c$. That's indeed the definition of distributions. – reuns Aug 2 '17 at 16:11
• Perhaps my questions were not well formulated, so I added a paragraph to the end of the main body above in an attempt to clarify my questions. In simpler terms, I'm looking for a precise mathematical definition of "commutative" with respect to convolutions (1) and (3) above. – Steven Clark Aug 2 '17 at 19:30
• @StevenClark Very funny. You want some help for understanding why $\int_{-\infty}^\infty f(x) g(y-x)dx = \int_{-\infty}^\infty f(y-x_2) g(x_2)dx_2$ when $y \in \mathbb{R}$ with the change of variable $x = y-x_2$ ? – reuns Aug 2 '17 at 19:55
• No. I'm primarily interested in independent evaluation of the two integrals without a change of variables. I'm also more interested in Mellin convolution than Fourier convolution. What is the definition of commutative with respect to Mellin convolution (1)? Also, I'm interested in definition of the conditions under which the range of commutativity can be extended. For example, if Mellin convolution (1) is defined to be commutative for $y>0$ (or $y\ge 0$?), are there conditions under which the commutativity can be extended to $y\in\mathbb{R}$? – Steven Clark Aug 3 '17 at 16:42
• Also, what are the conditions for which Mellin convolution (1) and Fourier convolution (3) can be extended from the reals to the complexes or some subset of the complexes? Would this require both f and g to be analytic? – Steven Clark Aug 3 '17 at 16:43