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As I understand, a smooth function is continuously differentiable.

But if I have a function which is continuous AND differentiable, I cannot automatically say that it is smooth. For it has to be so for all its differentials.

So I wonder, what function would be continuous and differentiable, but not continuously differentiable?

I cannot find the answer myself, as I do not clearly understand the difference between continuous AND differentiable, and continuously differentiable....

Context: I ask this because of an arc length contest. The function has to be continuous and differentiable on [0,1]. But does this automatically mean that I may always use the formula for an arc length, which has the condition that the function is smooth... (or, in another book, that it has a continuous derivative)?

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    $\begingroup$ For instance the map $\displaystyle x\mapsto \int \limits_0^x|t|\mathrm dt$. $\endgroup$
    – Git Gud
    Feb 17, 2015 at 18:35
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    $\begingroup$ Continuously differentiable function is $f(x)$ s.t. $f'(x)$ is continuous. But smooth is $f(x)$ s.t. $f^n(x)$ continuous $\forall n \in \mathbb{Z}_+ $. $\endgroup$
    – Mihail
    Feb 17, 2015 at 18:42

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We say that $f$ is a function of class $C^k$ if it has $k$ continuous derivatives. You want to find a differentiable function which is not $C^1$, that is, a differentiable function with discontinuous derivative. Given $k \in \Bbb Z_{> 0}$, the function $$f(x) = \begin{cases} x^k\sin(1/x), & \text{if } x \neq 0 \\ 0, & \text{if }x=0\end{cases}$$ is $C^{k-1}$ but not $C^k$.

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    $\begingroup$ Just thinking about the case $k=2$ makes me think you meant to say the given function's $(k-1)^{th}$ derivative exists on all $\mathbb{R}$, yet the function is not $C^{k-1}$. $\endgroup$
    – Selrach Dunbar
    Oct 3, 2020 at 14:13
  • $\begingroup$ If $x\neq 0$, $f^{\prime}(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x}$. Otherwise, $f^{\prime}(0)=\lim_{x\rightarrow0}\frac{x^2\sin\frac{1}{x}-0}{x-0}=\lim_{x\rightarrow0}x\sin\frac{1}{x}=0$. However, $\lim_{x\rightarrow0}f^{\prime}(x)$ does not exist. Hence, $f^{\prime}(x)$ is discontinuous at 0. That is to say, $f(x)$ is not $C^{2-1}$, i.e., $C^1$. Your description with that example is not correct. $\endgroup$
    – suineg
    Mar 30 at 5:15
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EDIT: there is some confusion as to what is being asked here. I am answering "Does there exist a continuous and differentiable function which isn't smooth?" (mentioned in the title and the question), but I see that "what function would be continuous and differentiable, but not continuously differentiable?" is also asked. To this second question, I recommend looking at @Ian's comment below.

A good example is

$$ f(x) = \begin{cases} x^2 & : x \geq 0 \\ 0 & : x < 0 \end{cases} $$

You can check the right-hand limit of both the function and first derivative are $0$, but the second derivative is discontinuous.

Similarly, $$ f(x) = \begin{cases} x^n & : x \geq 0 \\ 0 & : x < 0 \end{cases} $$

provides an example where the function and first $n-1$ derivatives are continuous, but the $n$th derivative is not.

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    $\begingroup$ Actually, in this case the second derivative is continuous where it is defined, which is on $\mathbb{R} \setminus \{ 0 \}$. Because derivatives always satisfy the intermediate value property on their domain, a derivative which exists even while it is discontinuous requires an "oscillation-type" discontinuity. For instance, $f(x)=\int_0^x \sin(1/y) dy$ has $f'(0)=0$ (one can calculate this directly) but $\lim_{x \to 0^+} f'(x)$ does not exist. $\endgroup$
    – Ian
    Feb 17, 2015 at 18:47
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    $\begingroup$ I agree; in rereading the question, there is some confusion as to what is being asked here. I am answering "Does there exist a continuous and differentiable function which isn't smooth?" (mentioned in the title and the question), but I see that "what function would be continuous and differentiable, but not continuously differentiable?" is also asked (a question to which I agree your answer is better). $\endgroup$
    – BaronVT
    Feb 17, 2015 at 18:50

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