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Jan 9, 2019 at 14:31 comment added LSpice @SteveHuntsman, OK, I see.
Jan 9, 2019 at 9:54 comment added Steve Huntsman @LSpice 'finite' is taken in the usual sense. IIRC I was trying to emphasize the Banach algebra structure, which more generally requires boundedness as an explicit condition
Jan 9, 2019 at 2:12 comment added LSpice @SteveHuntsman, does 'finite' here mean something other than the usual set-theoretic notion? Otherwise it's hard to know what else a (real- or complex-valued) function on a finite set could be than bounded.
Jan 31, 2012 at 21:57 comment added Victor Dods Thank you Mariano, that's a really neat and apropos result.
Jan 31, 2012 at 21:56 history edited Victor Dods CC BY-SA 3.0
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Jan 30, 2012 at 3:17 comment added Steve Huntsman Re: my comment above, see mathoverflow.net/questions/32667/…
Jan 30, 2012 at 3:15 comment added Steve Huntsman A notion of derivative applies for functions on many finite sets in $\mathbb{R}^n$ (say). Though the Banach algebra of bounded functions on a finite set turns out to carry no nonzero derivations, one can pick a suitable function space for which the interpolation problem has a unique solution and differentiate the interpolant. This is intrinsically non-local.
Jan 30, 2012 at 0:33 comment added Mariano Suárez-Álvarez As for «what's the essential quality of a differential operator», one answer is Peetre's theorem. (Wikipedia's write-up on this is particularly obfuscated, so Googling a bit might be helpful)
Jan 30, 2012 at 0:29 comment added Tom Goodwillie Yes, pseudo-differential operators are defined by convolution and are somehow not local but very close to being local.
Jan 30, 2012 at 0:16 comment added Mariano Suárez-Álvarez A keyword you might search for is pseudo-differential operators, which is usually attached to well-behaved operators in that spirit...
Jan 30, 2012 at 0:04 history asked Victor Dods CC BY-SA 3.0