Suppose $f$ is differentiable everywhere on $[0,1]$. Must $f$ be absolutely continuous on $[0,1]$? I know this is true if $f'$ is integrable but I'm not sure in this more general case.
Does the everywhere differentiability of $f$ imply it is absolutely continuous on a compact interval?
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real-analysis
measure-theory
1 Answers
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No. Consider $f(x) = x^2 \sin(1/x^4)$ on $[-1,1]$. Note that for $\beta_n = (\pi n)^{-1/4}$ and $\alpha_n = (\pi (n+1/2))^{-1/4}$ we have $\beta_n - \alpha_n = \Theta( n^{-5/4})$ while $|f(\beta_n) - f(\alpha_n)| = \Theta(n^{-1/2})$.
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1It's worth mentioning that there's a theorem that says that if $f$ is absolutely continuous, then $f'$ is integrable, so you can also show this example works by looking at the formula for $f'$ and seeing it blows up like ${1 \over x^3}$ (in a sinusoidal fashion) as $x$ goes to zero. – 2012-08-22
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0@Robert Israel: How doe we conclude from the above that $f$ is not absolutely continuous ? Sorry if I am missing some obvious step after your explanation. – 2015-06-19
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1Definition of absolute continuity: for any $\epsilon > 0$ there exists $\delta > 0$ such that for any finite sequence of intervals $[\alpha_n, \beta_n]$ with $\sum_j (\beta_j - \alpha_j) < \delta$, $\sum_j |f(\beta_j) - f(\alpha_j)| < \epsilon$. In this case, since $\sum_n n^{-5/4}$ converges but $\sum_n n^{-1/2}$ diverges, for any $\epsilon$ and $\delta$ you can take $(\alpha_n, \beta_n)$ for $M \le n < N$ that gives you $\sum_j (\beta_j - \alpha_j) < \delta$ but $\sum_j |f(\beta_j) - f(\alpha_j)| > \epsilon$. – 2015-06-19
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0What is $f(0)$? – 2016-03-05
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0@Jack: $f(0) = 0$. – 2016-08-02
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0How does $f(x)=x^2\sin(1/x^4)$ have a derivative everywhere on [-1,1]? Surely at $x=0$ if $f'(x) = 2x\sin(\frac{1}{x^4}) - 4\frac{\cos(1/x^4)}{x^3}$ then $f'(0)$ is undefined due to the $\cos(\infty)$ term. I can't think of a good reason why everywhere differentiable does not imply continuously differentiable. $f(x)$ seems to be differentiable ae, and not differentiable everywhere. – 2016-08-05
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1The derivative at $x=0$ does not come from that formula for $f'(x)$, it comes directly from the definition of derivative: $\lim_{x \to 0} \dfrac{f(x)-f(0)}{x} = 0$ because $-x^2 \le f(x) \le x^2$. – 2016-08-05
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0This is reminding me of my class on measure theory. It seems like $f(x)=x^2\sin(1/x^4)$ is a function that is differentiable everywhere but does not have a continuous derivative. I'm currently working on a summary of: "Continuously differentiable $\subseteq $ Lipschitz continuous $\subseteq$ $\alpha$-Hölder cont. $\subseteq$ uniformly continuous $\subseteq$ continuity" on compact intervals, as well as: "Lipschitz $\subseteq$ A.C. $\subseteq$ BV $\subseteq$ diff. a.e. The goal is to add convexity to the framework. Where does everywhere differentiability fit within the model (if at all)? – 2016-08-10
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0For functions on a compact interval, everywhere differentiable $\subset$ uniformly continuous $=$ continuous. it neither implies nor is implied by Lipschitz continuous or $\alpha$-Hölder. – 2016-08-10