# absolutely continuous on $[0,1]$ versus absolutely continuous on $[\varepsilon,1]$ for every $\varepsilon>0$

###### Lemma.

Define $f\colon\mathbb{R}\to\mathbb{R}$ by

 $f(x)=\begin{cases}0&\text{ if }x=0\\ \displaystyle x\sin\left(\frac{1}{x}\right)&\text{ if }x\neq 0.\end{cases}$

Then $f$ is absolutely continuous  (http://planetmath.org/AbsolutelyContinuousFunction2) on $[\varepsilon,1]$ for every $\varepsilon>0$ but is not absolutely continuous on $[0,1]$.

###### Proof.

Let $\varepsilon>0$. Then for all $x\in[\varepsilon,1]$:

$\begin{array}[]{ll}|f^{\prime}(x)|&\displaystyle=\left|\sin\left(\frac{1}{x}% \right)-\frac{1}{x}\cos\left(\frac{1}{x}\right)\right|\\ \\ &\displaystyle\leq\left|\sin\left(\frac{1}{x}\right)\right|+\left|\frac{1}{x}% \right|\cdot\left|\cos\left(\frac{1}{x}\right)\right|\\ \\ &\displaystyle\leq 1+\frac{1}{\varepsilon}\cdot 1\\ \\ &\displaystyle=1+\frac{1}{\varepsilon}\end{array}$

Since $f$ is continuous on $[\varepsilon,1]$ and differentiable on $(\varepsilon,1)$, the mean value theorem (http://planetmath.org/MeanValueTheorem) can be applied to $f$. Thus, for every $x_{1},x_{2}\in(\varepsilon,1)$ with $x_{1}\neq x_{2}$, $\displaystyle\left|\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}\right|\leq 1+\frac{1}% {\varepsilon}$. This yields $\displaystyle|f(x_{2})-f(x_{1})|\leq\left(1+\frac{1}{\varepsilon}\right)|x_{2}% -x_{1}|$, which also holds when $x_{1}=x_{2}$. Thus, $f$ is Lipschitz  on $(\varepsilon,1)$. It follows that $f$ is absolutely continuous on $[\varepsilon,1]$.

On the other hand, it can be verified that $f$ is not of bounded variation  on $[0,1]$ and thus cannot be absolutely continuous on $[0,1]$. ∎

Title absolutely continuous on $[0,1]$ versus absolutely continuous on $[\varepsilon,1]$ for every $\varepsilon>0$ AbsolutelyContinuousOn01VersusAbsolutelyContinuousOnvarepsilon1ForEveryvarepsilon0 2013-03-22 16:12:19 2013-03-22 16:12:19 Wkbj79 (1863) Wkbj79 (1863) 11 Wkbj79 (1863) Example msc 26A46 msc 26B30