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The convolution of $f$ and $g$ is sometimes written as $f\ast g$ and sometimes as $f \otimes g$.

Is convolution denoted by both of these symbols because the operations are related, or is it just historically determined? Wikipedia and Google dont know.

Leo
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    In the differential equations texts which I own, the notation $f*g$ is used for convolution. I have not seen $\otimes$ used for convolution... however, notation varies. – James S. Cook Jan 24 '14 at 12:01
  • @JamesS.Cook Professor Cook, the convolution isn't a example of Tensor? Because exist a isomorphism between the tensor product $V \otimes V$ and the space of all bilinear functions $L(V^{} \times V^{} ; \mathbb{R})$ . Maybe the notation is due to the operation tensor product $f \otimes g$. – M.N.Raia May 29 '19 at 01:44
  • @M.N.Raia perhaps, I'm not well-versed in enough of the abstract analysis needed to sort through the terminology. My previous comment was given from my admittedly limited knowledge about convolution. More recently, I taught a course which touched on abstract Fourier analysis etc. and I see that convolution has a much broader meaning than I realized when I last commented. Is the convolution a tensor product ? Well, it is bilinear in each entry so.. what else do we need ? Tensor product of what exactly ? Functions, I suppose, it is a function-valued tensor then. – James S. Cook May 29 '19 at 03:55

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In the neural networks (e.g. in Tensorflow) there are convolutional layers which use a tensor product. The latter symbol stands for tensor product. It is not applicable in all convolution-scenarios.

Harm Campmans
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