# Convolution notations and some miscellanous convolution questions

Convolution between distributions over the real number line, as it is mentioned in the optics course in my uni and also in wolfram is defined as:

$$f(t) * g(t)=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau$$

For convolutions, there's a nice theorem called Convolution Theorem which states for any distributions $f$ and $g$:

$$\mathcal{F}(f(t)g(t))=\mathcal{F}(f(t)) *\mathcal{F}(g(t))$$

In this link and also this, it was found that for a 2 dimensional convolution, it is expressed as follows:

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

So I suspect in general, an n dimensional convolution will be written as:

$$f(a_1,\dots,a_n) * g(a_1,\dots,a_n)=\int_{\mathbb{R}^n}f(a_1',\dots,a_n')g(a_1-a_1',\dots,a_n-a_n')d^na'$$

But what if I want to carry out some sort of "partial convolution", i.e.

$$?=\int_{-\infty}^{\infty}f(x',y')g(x-x')dx'\hspace{15mm}[1]$$

1. Is there a notation in terms of $*$ for $[1]$, or generalisation of it? Or is it whenever we wrote the convolution symbol $*$, we always imply it is convolution for all the independent variables present?

For a concrete example of $[1]$, suppose I have a laser beam with Gaussian profile $\exp(-x^2)$ which passes through a rectangular slit modeled by a product of boxcar functions $\Pi_{-a,a}(x)\Pi_{-b,b}(y)$. The far field diffraction pattern produced then landed on another filter which is $\delta(y)$.

Therefore the final diffraction pattern $\phi$ after being Fourier transformed again by a lens placed behind the 2nd filter (In our optics course we use the convention such that the prefactor in the Fourier transform is $2\pi i$ thus the normalisation factor is 1), would been given as:

$$\phi(x,y)=\int_{-\infty}^{\infty}\mathcal{F}\left[\int_{-\infty}^{\infty}e^{-x^2}\Pi_{-a,a}(x-x')\Pi_{-b,b}(y)dx'\right]\delta(v')dv'$$

So question 1 was asking whether we can express the above expression in terms of $*$ thus we can apply Convolution Theorem and common Fourier transform pairs to simplify it.

For question 2, we know that if some 2 variable distributions $k$ and $l$ are separable, then there is a relation between convolution and multiplication

$$k(x)*l(y)=k(x)l(y)$$

But I am having some trouble try to convince myself by proving it for the following case:

$$e^{-x^2}*\delta(y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x'^2}\delta(y-x'-y')dx'dy'$$ $$=\int_{-\infty}^{\infty}e^{-x'^2}\delta(x')dx'=e^{-0^2}=1$$ $$\color{red}{\neq e^{-x^2}\delta(y)}$$

1. What kind of rigorous treatment on distributions that was missing in the above physical usage (possibly abuse of notation) that can help rectify the problem highlighted in red?

1. Partial convolutions do exists, there are two known ways to notate them. for example suppose we have the following partial covolution $\int_{-\infty}^{\infty}f(x')g(x-x',y)dx'$, it can be notated as:
• $\bigl(f*g(\cdot,y)\bigr)(x)$
• $f *_1 g$, where the subscript indicate the argument you are convolving with