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In my bachelor courses, we have defined the space $C^1(\mathbb{R})$ as $$C^1(\mathbb{R}) = \{f: \mathbb{R} \rightarrow \mathbb{R}~|~f \text{ is differentiable}~\land~f' \text{ is continuous}\}.$$ However, I really do not understand why it is required that $f'$ must be continuous. Indeed, if $f$ is differentiable at $a \in \mathbb{R}$, then the limit $$\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$$ does exist. Isn't it equivalent to the fact that our function $f'$ is continuous in $a$ ? I asked my teacher for an example of a function that would be differentiable at a point but whose derivative would not be continuous at this point. He gave me the function $$f(x) = |x| \quad \text{at zero.}$$But I disagree with him, it's not that the derivative is not continuous in zero, it doesn't even exist because these two limits don't agree $$- 1 = \lim_{x \rightarrow 0^-} \frac{f(x) - f(0)}{x - 0} \neq \lim_{x \rightarrow 0^+} \frac{f(x) - f(0)}{x - 0} = 1.$$ Therefore, $f$ is not differentiable at $0$. Could someone explain to me the definition of $C^1(\mathbb{R})$? Is there an example of a differentiable function whose derivative is not continuous? Thank you very much!

Arctic Char
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Falcon
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    $f(x)=x^{2}\sin (\frac 1 x)$ for $x \neq 0$, $f(0)=0$. – Kavi Rama Murthy Oct 29 '20 at 09:33
  • VEry related: https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be – Tito Eliatron Oct 29 '20 at 09:38
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    A counterexample has already been given, but to address your confusion about the continuity, the fact that the limit of $\frac{f(x)-f(a)}{x-a}$ exists means that $\frac{f(x)-f(a)}{x-a}$ is continuous at $a$ (or rather, can be continuously extended to $a$). It does not mean that $f'(x)$ is continuous at $a$, since $f'(x)\neq\frac{f(x)-f(a)}{x-a}$. – Vercassivelaunos Oct 29 '20 at 09:46
  • Yes, thank you very much! – Falcon Oct 29 '20 at 09:54

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