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Please think it easy because it is not an assignment.

Question : Let $f\in C^{1}(0,1]\cap C[0,1]$ and $f'\not\in L^{1}(0,1)$. Then, does $f$ oscillate frequently and $f'$ is unbounded at $0$?

When I asked the similar question before, Daniel Fischer taught me that such functions satisfy the above conditions. I accepted at the time but a abstract proof seems to be difficult after thoughts. Of course, it is clear that such function is Riemann integrable by the fundamental theorem of calculus and $f'$ should be unbounded at $0$. The problem is whether $f$ should oscillate at $0$ when thinking in the framework of Lebesgue integral. I think that an image is what the infinite sum of slope diverges due to oscillation but I don't know well how it can prove.

I'm glad if you give me the strategy of proof when you can prove. It's good even only hints.

Thank you in advance.

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user
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1 Answers1

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If $f'$ was bounded near $0$, then since $f'$ is continuous on $(0,1]$, then it would follow that $f'$ is bounded on $[0,1]$ and hence integrable.

Since $\int_x^1 f' = f(1)-f(x)$ for $x>0$, we see that $\lim_{x \downarrow 0} \int_x^1 f'$ exists, hence $f'$ must change sign 'frequently' near zero (if $f'$ does not change sign on $(0,\epsilon)$, then it would be integrable).

The description 'oscillates frequently' is a bit vague. However $f$ cannot be monotone on any interval $[0,\epsilon)$, otherwise $f'$ would not change sign (near zero) and would be integrable.

copper.hat
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  • Thank you for giving answer. I don't understand from hence in the second sentence. Why does $f'$ have to change sign? Could you explain more bit detail? – user Oct 15 '15 at 06:37
  • If $f'$ doesn't change sign on $(0,\epsilon]$, then suppose it is positive. Then $\int_0^\epsilon |f'| = \int_0^\epsilon f' $ is finite and hence $f'$ is integrable. Similarly for negative. – copper.hat Oct 15 '15 at 06:40
  • I'm afraid that $\int_{0}^{\varepsilon}f'$ can be infinite due to the unbounded of $f'$ at $0$ even if $f'$ does not change sign. – user Oct 15 '15 at 06:45
  • It can't. Fundamental theorem of calculus. – copper.hat Oct 15 '15 at 06:46
  • Ahh, I see. I understand. You mean that $\int_{0}^{\varepsilon}|f'|=\int_{0}^{\varepsilon}(f')^{+}-(f')^{-}\to\infty$ if $f'$ changes sign, right? – user Oct 15 '15 at 06:51
  • No, I am saying that $f(\epsilon)-f(0) = \lim_{x \downarrow 0} \int_x^\epsilon f'$, hence $\int_0^\epsilon f'$ exists in a conditional sense. You know that $\int_0^\epsilon |f'| = \infty$ by hypothesis (otherwise $f'$ would be integrable). – copper.hat Oct 15 '15 at 06:53
  • Sorry, I'm mistaken the order. I understood that $\int_{0}^{\varepsilon}f'$ exists. What I wanted to say is that $f\not\in L^{1}(0,1)$, i.e., the divergence of the sum of $(f')^{+}$ and $(f')^{-}$ is given by change of sign of $f'$. Is it such a thing? – user Oct 15 '15 at 07:01
  • You lost me there. You are given that $f \in L^1$. Anyway, I am going to sleep. Goodnight. – copper.hat Oct 15 '15 at 07:02