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I'm having a bit of difficulty in understanding why the convolution between two functions $f(x)$ and $g(x) \equiv \delta(a x)$ doesn't seem to be commutative.

Specifically,

$(f * g)(y) = \int_{-\infty}^{\infty} f(x) \delta(a y - x) dx = f(a y)$

but

$(g * f)(y) = \int_{-\infty}^{\infty} \delta(a x) f(y - x) dx = \frac{f(y)}{a}$

Is the change of variables during the integration in the second expression where I'm going wrong?

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    $\begingroup$ $\delta(a(y-x))$ in the first integral. $\endgroup$ Commented Aug 9 at 20:03
  • $\begingroup$ Man, I'm being slow today. Thanks. $\endgroup$ Commented Aug 9 at 20:19
  • $\begingroup$ Might want to say if $a =0$ is permitted. $\endgroup$ Commented Aug 10 at 1:33
  • $\begingroup$ @DermotCraddock you should add this as an answer $\endgroup$ Commented Aug 10 at 11:10
  • $\begingroup$ @LL3.14 Right. Done. $\endgroup$ Commented Aug 10 at 11:23

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It's the first expression where the mistake lies. \begin{align} (f \ast g)(y) &= \int_{-\infty}^{+\infty} f(x)g(y-x)\,dx \\ &= \int_{-\infty}^{+\infty} f(x)\delta\bigl(a(y-x)\bigr)\,dx \\ &= \frac{1}{\lvert a\rvert}\int_{-\infty}^{+\infty} f\biggl(\frac{t}{a}\biggr)\delta(ay-t)\,dt \\ &= \frac{1}{\lvert a\rvert}f\biggl(\frac{ay}{a}\biggr) \\ &= \frac{f(y)}{\lvert a\rvert}\,. \end{align}

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  • $\begingroup$ This is not quite correct. You've tacitly assumed that $a>0$. What happens when $a<0$? $\endgroup$ Commented Aug 11 at 20:03
  • $\begingroup$ Indeed, it should be $1/|a|$ instead of $1/a$ $\endgroup$ Commented Aug 12 at 8:52
  • $\begingroup$ Thanks, @MarkViola. Indeed I only thought of positive $a$. blushes $\endgroup$ Commented Aug 12 at 10:17

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