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The Schwartz space is defined as the set of all indefinitely differentiable functions such that the supremum over the free variable of any (order) derivative times any (order) power is finite.

However, a differentiable function is continuous, and a continuous (over the reals) function is bounded on any closed interval. Hence, if the product of a power by a derivative is not finite it can only happen in the limit to infinity. If a continuous function has a finite limit at infinity, then multiplication by a power will produce an infinite limit at infinity. So if any power times a specific derivative has a finite limit, the next power will go to infinity. Hence, the requirement is that the limit is zero.

Assuming no error on my part here, the supremum could be replaced by a limit.

The Wikipedia gives the supremum definition and then uses the limit version informally as the description.

What would change in the theory of Schwartz spaces if one replaced the supremum being finite with the limit to infinity being zero in the definition?

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    $\begingroup$ A bounded function need not go to $0$ at $\infty$. $\endgroup$
    – LSpice
    Commented Nov 20 at 4:37
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    $\begingroup$ I think you could use the limit for the definition. But one wants to put a topology on the Schwartz space and this is achieved by the countable family of seminorms so these suprema are useful to have anyways… $\endgroup$
    – Dirk
    Commented Nov 20 at 9:38
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    $\begingroup$ @LSpice if a function remains bounded after you multiply it by $x$, it need $\endgroup$
    – Fedor Petrov
    Commented Nov 20 at 10:02
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    $\begingroup$ @FedorPetrov, re, right, of course, as is observed in the question itself. Sorry! $\endgroup$
    – LSpice
    Commented Nov 20 at 18:12
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    $\begingroup$ Re, I do not answer questions about specific votes. However, to be clear, my initial comment was not only mathematically foolish but based on careless reading, and I apologise for it. (I'd just delete it, except that it already has one explicit, and your implicit, response.) $\endgroup$
    – LSpice
    Commented Nov 21 at 22:37

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