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Discrete convolution (between infinite sequences $f$ an $g$) is defined as $$(f*g)(k) = \sum_{j=-\infty}^{\infty}f(j)g(k-j)$$.

It is well known that convolution is associative, that is $(f*g)*h=f*(g*h)$. But I am a little confused with the following situation:

Let be $f=H$ (discrete Heaviside), that is $H(k)=0$ for $k<0$ and $H(k)=1$ for $k\ge 0$. Let be $g$ defined as $g(0)=1$, $g(1)=-1$ and $g(k)=0$ otherwise. And finally (maybe the problem !) let be $h(k)=1$ for any $k\in\mathbb{Z}$. In this situation we easily obtain:

$(f*g)*h = h$ and $f*(g*h) = 0$

Can someone have an explanation for this point and how it could be avoided?

fmoura2005
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    It is "well known?" Have you seen the proof? It will use a certain amount of absolute convergence, I guess. In particular, $f*h$ is undefined, which makes the proof fall apart, I'm betting. – Thomas Andrews Apr 28 '17 at 15:36
  • I do not think $f*h$ is undefined, it is 0. It is very common to convolute with series that are not absolute convergent. One example is the Heaviside. – fmoura2005 Apr 28 '17 at 16:32
  • No, you are wrong. You are correct that $gh=0$, but $(fh)(k) = \sum_{j=0}^{\infty} h(k-j)=\sum_{j=0}^{\infty} 1$. – Thomas Andrews Apr 28 '17 at 16:34
  • Basically, the convolution is what you get by trying to do a "formal multiplication" of $F=\sum f(k)x^{k}$ and $G=\sum g(k)x^k$. When $f,g,h$ are "nice enough," you can do this, but in a lot of cases, you cannot. – Thomas Andrews Apr 28 '17 at 16:40
  • Oh. Yes. You're right. Sorry, and thanks. – fmoura2005 Apr 28 '17 at 16:43

1 Answers1

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Associativity of convolution is essentially changing the order of summation:

$$(f \star (g \star h))(k) = \sum_i f(i) (g*h)(k-i) = \sum_i \sum_j f(i) g(j) h(k-i-j)$$

$$ \eqalign{((f\star g)\star h)(k) &= \sum_j (f \star g)(j) h(k-j) = \sum_j \sum_i f(i) g(j-i) h(k-j) \cr &= \sum_{j'} \sum_i f(i) g(j') h(k-i-j')}$$ where we relabel $j = j'+i$.

These are the same (Fubini's theorem for sums) if the double sum converges absolutely. If it doesn't, they may be different.

Robert Israel
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