Definition talk:Differentiability Class

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I was under the impression that the second criterion, "$\dfrac {\mathrm d^{k+1}} {\mathrm dx^{k+1}}$ is not continuous", was "not necessarily continuous". Am I wrong? --GFauxPas (talk) 20:48, 30 September 2012 (UTC)

Yes. Any function $f$ is in one and only one differentiability class. Under your understanding, a function could be in more than one. A polynomial, for example, would be in $C^1, C^2, C^3$ etc. whereas it's not, it's in $C^\infty$ only. --prime mover (talk) 06:19, 1 October 2012 (UTC)
I'm with GFP. A polynomial is $C^1$; whenever I say "Let $f$ be a $C^1$ function" I don't want to exclude that it may be $C^2$ as well. Moreover, restricting the $C$s in this way would destroy the vector space structure they have (since $0$ would only be $C^\infty$, not $C^k$; thus, except for $C^\infty$). --Lord_Farin (talk) 08:42, 1 October 2012 (UTC)
Also consider the extremal case $C^0$, the functions in which we generally describe as continuous. Now it would seem that you advocate cont.diffable functions to be no longer called continuous, which seems ridiculous very awkward. --Lord_Farin (talk) 08:44, 1 October 2012 (UTC)
Yep $C^k \supseteq C^{k+1}$ always. Also, unless I've missed something, the $f$'s are missing, i.e $d^kf/dx^k$ is continuous. --Linus44 (talk) 15:48, 1 October 2012 (UTC)

Under weight of popular opinion I have amended this definition as what you all say makes sense. So differentiability classes are not equivalence classes after all. --prime mover (talk) 20:17, 20 November 2014 (UTC)