A convolution of two functions $f$ and $g$ is defined as

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

I am interested on an analogous transformation of the form

$$[f\star g] = \int_{0}^{\infty} f(\tau) g\left(\dfrac{t}{\tau}\right) d\tau.$$

Does this transformation has a name?


1 Answer 1


This is a type of "multiplicative convolution", and will sometimes appear just as $\int_0^\infty f(\tau)g({t\over \tau})\;{d\tau\over \tau}$, since the positive reals already form a group, and ${d\tau\over \tau}$ is the multiplication-invariant measure there. In fact, you might find that you'll prefer ${d\tau\over |\tau|}$ in your situation, as well, so that the measure on the whole line (really with $0$ removed) is invariant under multiplication, so that this version of convolution has the expected properties.

  • $\begingroup$ The integration range was a typo, sorry about this. Could you please point out a reference for this? $\endgroup$
    – Combo
    Aug 7, 2013 at 13:08
  • $\begingroup$ Although there is a very general notion of convolution on topological groups, your example can be converted into the usual "additive" convolution by the change of variables $F(x)=f(e^x)$ and $G(y)=g(e^y)$, and then the multiplicative-invariant measure becomes the usual additive-invariant measure on the real line. So all the things you know about "usual" convolution apply to "multiplicative", just by changing variables. $\endgroup$ Aug 7, 2013 at 13:42
  • $\begingroup$ Many thanks for your help. $\endgroup$
    – Combo
    Aug 7, 2013 at 14:15

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