Let $\operatorname{f}$ be a real-valued function of one variable that is continuously differentiable on the intervall $(a,b)$, except possibly at $\gamma\in(a,b)$. Let $\operatorname{f}$ be such that $\operatorname{f}(x)\to 0$ when $x\to \gamma$. Let $$\operatorname{F}(x) = \cases{\operatorname{f}(x),\enspace \forall x\in (a,b):x\neq \gamma \\ 0, \enspace x = \gamma}$$ Is $\operatorname{F}$ necessarily continuously differentiable on $(a,b)$? If so, how should I go about proving it? Can the Bolzano–Weierstrass theorem somehow be used to prove it (or disprove it)?
A question regarding continuous differentiability of a function
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calculus
real-analysis
analysis
functions
derivatives
1 Answers
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No $F$ is not necessarily continuously differentiable. Consider for instance the absolute value function $F(x) = |x|$. Here $\gamma = 0$ and the interval can be any positive interval including $0$.
Clearly, $f(x)$ is differentiable at any non-zero point ($f'(x) = \pm 1$), but at $0$ there is no derivative even though $F(0) = 0$.
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0What proprties would I have to Impose on $\operatorname{F}$ in order for it to be continuously differentiable on $(a,b)$? In addition to the aforementioned ones that is. – 2017-01-03
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1The derivative should exist and be continuous. – 2017-01-03
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0@juanarroyo You're just repeating the definition. – 2017-01-11
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0@zhw. lol I know the point is you're probably not going to find many good conditions that don't exclude a lot of $C^1$ functions. – 2017-01-11