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Hello everyone is the following true :

Assume $F$ is a continuous function $[a,b] \to \mathbb{R}$ and $C^1$ on $]a,b]$ (Think of $\sqrt{x}$ on $[0;1]$. Is it true that the derivative is Lebesgue integrable in $[a,b]$ ?

Edit: changed "integrable" into "Lebesgue integrable"

2 Answers 2

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You can take a modification of @Fred's example: Take $F(x)=x^2\sin(1/x^2)$, $F(0)=0$, show that the fact that the derivative of $F$ is Lebesgue-integrable on $[0,1]$ is equivalent to say that the function $G$: $G(x)=\frac{1}{x}\cos(1/x^2)$, $G(0)=0$ is Lebesgue integrable on $[0,1]$. As if $h(x)$ is Lebesgue integrable, then $|h(x)|$ too, this imply after a change of variable that $\displaystyle \int_1^{+\infty} \frac{|\cos(t)|}{t}dt<+\infty$, and this is not true.

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    Thx for the example – 2017-01-24
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Let

$F(x)=x^{3/2} \sin(1/x)$ for $x \in (0,1]$ and $F(0)=0$.

Show that $F$ is differentiable on $[0,1]$ (with $F'(0)=0$),

but $F'$ is unbounded on $[0,1]$. Hence $F'$ is not Riemann- integrable on $[0,1]$.

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    The derivative here is bounded by $x \to \frac{1}{\sqrt(x)}$ at the vicinity of $0$ so it is lebesgue integrable. I should precise my question, sorry i will edit it. – 2017-01-23