Let $f:\mathbb R\to \mathbb R$ and $a\in\mathbb R$. Does it make sense to say that a function is $\mathcal C^k(\{a\})$ for $k\geq 2$ ? I would say no since to take about continuity of $f'$ if it's not defined on a neighborhood of $a$ looks strange... but in the doubt, I prefer to check with you.
Does it has sense to say that a function is $\mathcal C^k(\{a\})$ for $k\geq 2$?
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real-analysis
continuity
1 Answers
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If we want to be able to speak about the derivative at $a$, we require by definition that $f$ is defined in an open neighbourhood of $a$. So we could define $f \in \mathcal{C}^k(\{a\})$ if there exists an open neighborhood $U$ of $a$ such that $f^{(k)}$ is defined and continuous on $U$. This can be of course done analogously for every subset $S$ of $\mathbb{R}$.