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I am graduate student.

As you know, the convolution operation satisfy the below equation due to commutative law. a(n)*b(n)*c(n) = a(n)*c(n)*b(n)

In addition, the muliplication operation also satisfy the below equation due to commutative law. a(n)b(n)c(n) = a(n)c(n)b(n)

however, to my knowledge, the below equation was not satisfied; a(n)b(n)*c(n) = a(n)*c(n)b(n)

Can I say that "the commutative law between multiplication operation and convolution operation does not satisfied" ?

Furthermore, do you know any reference for this phenomenon?

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  • $\begingroup$ When you say a(n) b(n) * c(n), do you mean (a(n) b(n))*c(n), or a(n) (b(n) * c(n))? Similarly, is a(n)*c(n)b(n) equal to (a(n)*c(n))b(n), or a(n)*(c(n)b(n))? I doubt there's a reference; the fact is demonstrated by looking at nearly any choice of a, b, . $\endgroup$
    – user44191
    Commented Dec 17, 2018 at 5:50
  • $\begingroup$ I mean, (a(n)b(n))*c(n) = (a(n)*c(n))b(n) $\endgroup$
    – PILSU
    Commented Dec 17, 2018 at 6:11
  • $\begingroup$ You almost certainly mean ‘law’, not ‘raw’, throughout. I would call your proposed third law more a type of distributivity (though not really) than of commutativity. $\endgroup$
    – LSpice
    Commented Dec 17, 2018 at 6:12
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    $\begingroup$ @LSpice It is a kind of commutativity (or failure thereof) of operations; the operation $a \rightarrow ab$ doesn't commute with $a \rightarrow a*c$. $\endgroup$
    – user44191
    Commented Dec 17, 2018 at 6:14
  • $\begingroup$ @user44191, I agree. I had not noticed the clarification regarding the placement of parentheses. $\endgroup$
    – LSpice
    Commented Dec 17, 2018 at 7:05

1 Answer 1

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Maybe the simplest counterexample?

Let $\newcommand{\1}{\mathbf 1}\1=1_{[0,1]}$. Then any $f\cdot\1$ is zero outside of $[0,1]$, but $$\1*\1(x)=\int \1(t)\1(x-t)\,dt = \begin{cases}x& 0\le x\le 1\\ 2-x & 1\le x\le 2\end{cases}$$ is not. So $$\begin{eqnarray*}(\1\cdot \1)*\1&\ne&(\1*\1)\cdot \1\quad\text{and}\\ \1\cdot (\1*\1)&\ne&\1*(\1\cdot \1).\end{eqnarray*}$$ However, $$\begin{eqnarray*}\1\cdot (\1*\1)&=&(\1*\1)\cdot \1\quad\text{and}\\ (\1\cdot \1)*\1&=&\1*(\1\cdot \1).\end{eqnarray*}$$

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  • $\begingroup$ Thank you so much. Then, according to your counterexample, can I say that "the commutative law between multiplication operation and convolution operation does not satisfied" ? $\endgroup$
    – PILSU
    Commented Dec 18, 2018 at 1:20
  • $\begingroup$ @PILSU make sure to discuss associativity as well $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Dec 18, 2018 at 1:49

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