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Consider $L^2(\mathbb R^n, \mathbb R^m)$. There should be a Fourier transform for these functions, like in the case $L^2( \mathbb R^n, \mathbb R )$. I wonder how these can be defined.

The application I have in mind is defining a Fourier transform for differential forms on $\mathbb R^n$.

Willie Wong
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shuhalo
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    Have you tried the obvious? – Glen Wheeler Apr 22 '11 at 10:23
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    @Martin: What if you just do it componentwise? – JT_NL Apr 22 '11 at 11:53
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    A more interesting question is if we replace $\mathbb R^m$ by a separable Banach space! – JT_NL Apr 22 '11 at 13:36
  • Removing the differential geometry tag, because in the actual geometric setting, even the Fourier transform of scalar functions is problematic. – Willie Wong Apr 22 '11 at 14:25
  • Of course you can try the obvious, but maybe there something you overlook and which appears only in higher dimensions. – shuhalo Apr 22 '11 at 16:26
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    It is not obvious that a component-wise approach is appropriate. Nice band-limited components can yield cusps, as in the deltoid. – yasmar Apr 22 '11 at 20:14
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    One proposal is to use Clifford Algebras link, also on Citeseer. – yasmar Apr 22 '11 at 20:15
  • @WillieWong : As long as the only derivatives (of this function), that we are dealing, are first order partial derivatives, then can we say that using the obvious (component-wise) would suffice without any issues? I am not trying to answer the OP question, but this is a question of mine. – Rajesh D Jun 22 '17 at 05:39
  • I am posting it as a separate question, as I am not interested in the differential forms unlike what OP is asking. – Rajesh D Jun 22 '17 at 05:44
  • Related : https://math.stackexchange.com/q/2332005/2987 – Rajesh D Jun 22 '17 at 05:52

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