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I have seen the convolution operator in different settings, and I was wondering about the following:

Suppose $h=f\ast g$ for an unordered pair of functions $(f,g)$. Does there exist a pair of functions $(f',g')$, different from $(f,g)$, such that $h=f'\ast g'$? "$\ast$" here means the convolution operator.

Now, when I say that $(f,g)$ and $(f',g')$ are different, I also think there should be a restriction such that \begin{cases} ||f-f'||_{L^1}\neq 0, \\ ||g-f'||_{L^1}\neq 0, \end{cases} or \begin{cases} ||f-g'||_{L^1}\neq 0, \\ ||g-g'||_{L^1}\neq 0, \end{cases}

so that the functions cannot be the same almost everywhere.

It would seem a bit strange if the answer to the question was "no", but I cannot come up with an example that works.

Scounged
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1 Answers1

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Yes. Let us consider different notations. The question is, being given $f_1,g_1$, can we find $f_2,g_2$ different from $f_1,g_1$ such that

$$f_1*g_1=f_2*g_2 \ ?$$ Let us recall that, for $C^1$ functions $f$ and $g$ (for which the convolution product makes sense) we have the property of derivation :

$$ (f*g)'=f'*g=f*g' \ \ \ \text{i.e. take} \ \ f_1=f', \ g_1=g, \ f_2=f, \ g_2=g'$$

(now "prime symbol" means "derivation" ; this is why I have changed notations)

Thus it suffices to take $f_2$ equal to a primitive function of $f_1$, and $g_2=g'_1$ (the derivative of $g_1$), still under the condition that the convolution product makes sense (if we are in the framework of distributions, it will always make sense).

Jean Marie
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