# Integrability of the derivative

Given a real-valued function $f \in C([0,1]) \cap C^1((0,1])$ such that $\lim\limits_{t \rightarrow 0+} tf'(t) =0$. Does that imply $f' \in L^1((0,1))$ ? By assumption we can tell that $\lim\limits_{\varepsilon \rightarrow 0+} \int_\varepsilon^1 f'(t) dt$ exists. Further the statement seems true to me in case we would know$\lim\limits_{t \rightarrow 0+} f'(t) = \pm \infty$ since I believe we could use monotone convergence in these cases. What about the general case? The statement would imply that $f$ is even absolutely continuous. Can somebody provide a counterexample or an idea how to prove it?
• I think that you intended $f\in C((0,1]) \cap C^1((0,1])$, otherwise $f$ has maximum and minimum in $[0,1]$ and thus it is integrable in $(0,1)$ – Uskebasi Mar 13 '17 at 14:57
• @QWERTZ - the question is about integrability of the derivative $f'$. – uniquesolution Mar 13 '17 at 15:00
Ok. I believe the following might work as counterexample. For $x \in (0,1]$ define $$f(x):= \frac{\pi}{2} - \int\limits_0^{-\ln(x)}\frac{\sin(t)}{t}\, dt.$$ Then $f \in C((0,1])$ and since $\lim\limits_{x \rightarrow 0+} f(x) = 0$ we can extend it to a continuous function in $[0,1]$. It's derivative is $$f'(x) = \frac{\sin(\ln(x))}{x \ln(x)}$$ for which we have $x |f'(x)| \leq \frac{1}{|\ln(x)|}$ and therefore the limit condition is fulfilled. Last but not least if we tried to calculate the integral for $f'$ we get $$\int\limits_0^1 |f'(x)|dx = \int\limits_{- \infty}^0 \frac{|\sin(t)|}{|t|} \, dt$$ which does not exist.