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Convolution is commutative:

$f \star g = g \star f$

where $f$ and $g$ are functions, and $\star$ the convolution operator.

Does deconvolution possess this commutative property?

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    $\begingroup$ Multiplication of numbers is commutative. Division is the analog of deconvolution. Is division commutative? $\endgroup$
    – whuber
    Commented Aug 10, 2016 at 13:40
  • $\begingroup$ No need for sarcasm. $\endgroup$ Commented Aug 11, 2016 at 8:12
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    $\begingroup$ That's not sarcasm. Deconvolution is equivalent to division by the fourier transforms, so your knowledge of division will immediately answer your question without any further explanation. $\endgroup$
    – whuber
    Commented Aug 11, 2016 at 12:38

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