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I see the statement of "nice functions" in textbooks and the authors usually don't need to give the definition of "nice functions". For example in a book which I read now the authors write

"Morrey spaces is not separable. A version of Morrey space where it is possible to approximate by "nice functions" is vanishing Morrey space."

and don't give the definition of "nice functions" anywhere in the book.

I wonder in here what is the meaning of "nice functions" ?


Is there a fixed definition of "nice functions" ?

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It means a function that has sufficiently many derivatives, and which converges fast enough at infinity, that we don't have to worry about rigour when considering any expression involving derivatives, integrals, or integration by parts. Usually you can replace "nice" by "Schwarz class." But it has no real definition, and the term is intentionally vague. –  Stephen Montgomery-Smith Apr 22 '14 at 12:11
I read "nice" or "well-behaved" as "satisfying all requirements for the appropriate theorems I am using." –  apnorton Apr 22 '14 at 14:01
@anorton You should make that an answer, so I can vote for it (that's the most correct answer I see right now) –  Mario Carneiro Apr 22 '14 at 16:52
@MarioCarneiro Done. –  apnorton Apr 22 '14 at 18:16

3 Answers 3

The terms "nice" and "good" are used in an ad-hoc way throughout mathematics, and so giving them a fixed definition is counterproductive. The idea is to build intuition: we can expect our theory to work when we only consider objects that are not too strange, or we can guarantee the existence of objects satisfying certain properties that are easy to work with. Concrete definitions do not need to be introduced unless a technical discussion is forthcoming, and can inhibit readability otherwise.

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It might mean continuous, it might mean differentiable, it might mean smooth etc. The only common theme is that the author didn't bother explaining it in more detail.

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Comment converted to answer:

I read "nice" or "well-behaved" as "satisfying all requirements for the appropriate theorems I am using."

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