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Jan 4, 2019 at 13:13 history edited Carlo Beenakker CC BY-SA 4.0
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Oct 16, 2013 at 18:08 vote accept Henrique de Oliveira
Oct 12, 2013 at 18:07 comment added Henrique de Oliveira I see. Maybe I can phrase my confusion as follows: It seems possible that the convolution property above is, after all, all you need to characterize the Fourier transform. I cannot see that the paper answers the following two questions: 1-Does this property uniquely define the Fourier transform? 2-What happens when we restrict the set over which we define the operator, for example, if we only consider real-valued functions.
Oct 11, 2013 at 19:17 comment added Carlo Beenakker Corollary 3.3.4 of Kunze actually derives the linearity of the Fourier transform from its definition as a mapping that transforms convolution into multiplication, so I would think that there is no need to invoke the linearity as an additional requirement.
Oct 11, 2013 at 18:26 comment added Henrique de Oliveira Thanks for the reference. I can see that it is related, but I don't immediately see if it answers the question. He seems to define the Fourier transform in terms of preserving convolution. Does that mean that this is the only relevant property for the Fourier transform, besides linearity? Is it immediate that this also applies to the Laplace Transform?
Oct 11, 2013 at 18:15 history edited Carlo Beenakker CC BY-SA 3.0
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Oct 11, 2013 at 18:05 history answered Carlo Beenakker CC BY-SA 3.0