Convolution of scaled variable

Question

A rather simple question...

Is the following true?

$$f * f(\frac{x}{a}) = \int_\mathbb{R} f(u)f(\frac{x}{a}-u)du$$

Or is it

$$f * f(\frac{x}{a}) = \int_\mathbb{R} f(\frac{u}{a})f(\frac{x}{a}-u)du$$

or something else...?

The answer is rather obvious, will answer it myself in a little while. — Benjamin Lindqvist, Nov 05 '15 at 15:08

score 0 · Accepted Answer · answered Nov 05 '15 at 15:21

0

$$\int_\mathbb{R} f(u)f(\frac{x}{a}-u)du = \int_\mathbb{R} f(\frac{u}{a})f(\frac{x}{a}-\frac{u}{a})du$$

since we're integrating over $\mathbb{R}$

answered Nov 05 '15 at 15:21

Although this doesn't take care of the potential scaling, $du$ or $du/a$...??? – Benjamin Lindqvist Nov 05 '15 at 15:22

score 0 · Answer 2 · answered Jun 19 '19 at 10:40

Consider $g(x)=f(\frac{x}{a})$

\begin{align} f*f(\frac{x}{a})&=f(x)*g(x)\\ &=\int_\mathbb{R}f(u)g(x-u)du\\ &=\int_\mathbb{R}f(u)f(\frac{x-u}{a})du\\ \end{align}

And for the other possibility

\begin{align} f*f(\frac{x}{a})&=f(x)*g(x)\\ &=\int_\mathbb{R}f(x-u)g(u)du\\ &=\int_\mathbb{R}f(x-u)f(\frac{u}{a})du\\ \end{align}

Therefore, the statements in the question are not true.

2 Answers2