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Suppose $f:[a,b] \rightarrow \mathbb{R}$ is a function which is (i) differentiable at all $x \in (a,b)$ (ii) the right-derivative at $x=a$ exists and the left-derivative at $x=b$ exists.

Does it follow that $f$ is absolutely continuous?

Jordan Becker
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1 Answers1

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I don't have enough reputation for a comment, so I'll write this in answer:

Consider Does the everywhere differentiability of $f$ imply it is absolutely continuous on a compact interval?

I think that answers your question.

Community
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dioid
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  • I am a little bit confused by that answer. What happens at zero? I am requiring the derivative to exist at all points in the interval. – Jordan Becker Jul 24 '14 at 21:58
  • @Jordan: $f(0)=0$ in that example (it is the only definition that makes it continuous) and you can check that $f'(0)=0$. – Jonas Meyer Jul 24 '14 at 22:00
  • I may be confused, but I'm getting $$(x^2 \sin (1/x^4))' = 2x \sin(1/x^4) - 4 (\cos (1/x^4))/x^3$$ which seems to suggest a problem at $x=0$. – Jordan Becker Jul 24 '14 at 22:01
  • @Jordan: The formula you derived makes sense only when $x\neq0$. However, the derivative at $0$ is by definition $\lim\limits_{h\to 0}\frac{f(h)-f(0)}{h}$, and you can directly calculate this limit. – Jonas Meyer Jul 24 '14 at 22:03
  • Aha, got it! Thank you! – Jordan Becker Jul 24 '14 at 22:03
  • For future reference, when you find a question that has already been answered, rather than posting an answer with a link, it is better to use the "flag" button to indicate that the question is a duplicate. – Nate Eldredge Jul 24 '14 at 22:26