Let $f:J\rightarrow \mathbb{R}$ be an absolutely continuous. Under what kind of extra condition for $f'$, (not $C$) holds the following relation? $$ \Big | \frac{1}{|I_{1}|}\int_{I_{1}}f'(x)dx- \frac{1}{|I_{2}|}\int_{I_{2}}f'(x)dx\Big|\overset{|J|\rightarrow 0}{\longrightarrow} 0, $$ for any $I_{1}\cap I_{2}=\emptyset$ and $I_{1}, I_{2}\subset J$ intervals.
2 Answers
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Modification after the comments from below.
Let I_1=(a,b) and I_2=(c,d), where a< b<=c< d. Then \frac{1}{|I_1|}\int_{I_1}f'(t)dt=\frac{f(b)-f(a)}{b-a}:=J_1, similarly, \frac{1}{|I_2|}\int_{I_2}f'(t)dt=\frac{f(d)-f(c)}{d-c}:=J_2. Let |J| \to 0, J_1 \to f(b-), while J_2 \to f(b+). So one natural assumption is that f' exists everywhere!
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$\begingroup$ Sure you need extra conditions. Consider the function $f(x)=|x|$, take $J=(-\epsilon,\epsilon)$, $I_1=(0,\epsilon)$, $I_2=(-\epsilon,0)$. $\endgroup$ Oct 31, 2013 at 7:52
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1$\begingroup$ "Coincides with a continuous function"? That's not what one knows from real analysis... $\endgroup$ Oct 31, 2013 at 8:27
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$\begingroup$ To ofer zeitouni and Vladimir Dotsenko: Thanks for reminder. $\endgroup$ Oct 31, 2013 at 9:40
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It seems that this condition forces $f^\prime$ to be continuous. As noted in Changyu Guo's answer, the derivative must exist everywhere in the interior of the interval. Now, suppose WLOG $f$ is defined on $[-1,1]$ and $f^\prime$ is discontinuous at $0$. Then there exist sequences $a_n\to 0$ from the left and $b_n\to 0$ from the right such that $$ \liminf_{n\to \infty} |f^\prime(a_n)-f^\prime(b_n)|>0. $$ Now at any point $c\in(-1,1)$ we can choose a very small interval $I$ containing $c$ so that the average value of $f^\prime$ over $I$ is as close to $f^\prime(c)$ as we like. Doing this at the $a_n$ and the $b_n$ we see the required condition is violated.
