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Let $f:(0,1] \to \mathbb R$ be a bounded non negative integrable function. Assume the limit $$\lim_{x \to 0+} \frac{1}{|\ln(x)|} \int_x^1 \frac{f(t)}{t} \mathrm dt$$ exists and has a positive value. How to show (or give a counterexample) that $\limsup_{t \to 0+} f(t)$ also has a positive value?

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Since $f$ is non-negative, then $\limsup\limits_{x\to0+}f(x)\geq\liminf\limits_{x\to0+}f(x)\geq 0$. Assume $\limsup\limits_{x\to0+}f(x)\leq 0$. Then $\lim\limits_{x\to0+}f(x)=\liminf\limits_{x\to0+}f(x)=\limsup\limits_{x\to0+}f(x)=0$. Fix $\varepsilon>0$, then there exist $\delta>0$ such that $|f(x)|<\varepsilon$ for all $x\in(0,\delta)$. In this case $$ \left|\int_x^1\frac{f(t)}{t}\right| \leq\left|\int_x^\delta\frac{f(t)}{t}\right|+\left|\int_\delta^1\frac{f(t)}{t}\right| \leq\int_x^\delta\frac{|f(t)|}{t}+\left|\int_\delta^1\frac{f(t)}{t}\right| \leq\int_x^\delta\frac{\varepsilon}{t}+\left|\int_\delta^1\frac{f(t)}{t}\right| \leq\varepsilon(\ln(\delta)-\ln(x))+\left|\int_\delta^1\frac{f(t)}{t}\right| $$ Hence $$ \left|\frac{1}{|\ln(x)|}\int_x^1\frac{f(t)}{t}\right|\leq\varepsilon+\frac{1}{|\ln(x)|}\left(\left|\int_\delta^1\frac{f(t)}{t}\right|+\varepsilon\ln(\delta)\right) $$ and we have the bound $$ \limsup\limits_{x\to0^+}\left|\frac{1}{|\ln(x)|}\int_x^1\frac{f(t)}{t}\right|\leq\varepsilon $$ Since $\varepsilon>0$ is arbitrary we conclude $$ \lim\limits_{x\to0^+}\frac{1}{|\ln(x)|}\int_x^1\frac{f(t)}{t}=0 $$ Сontradiction, so $\limsup\limits_{x\to0+}f(x)>0$

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  • $\begingroup$ I suppose the correct assumption is on the limsup, nor the liminf, right? :D $\endgroup$ – user67133 May 23 '13 at 22:19
  • $\begingroup$ yes, that was a typo $\endgroup$ – Norbert May 23 '13 at 22:20
  • $\begingroup$ of course it was! +1 indeed $\endgroup$ – user67133 May 23 '13 at 22:21
  • $\begingroup$ since we don't know the continuity of $f(t)$, as a function of $x$ can we assume that $\int_x^1 \frac{f(t)}{t} \mathrm dt$ has derivative? $\endgroup$ – Deco May 23 '13 at 22:29
  • $\begingroup$ we can define $f(0)=0$ to make it continuous at $0$. Such extending of $f$ from $(0,1]$ to $[0,1]$ will not affect the integral $\endgroup$ – Norbert May 23 '13 at 22:31
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Here is a calculation you can do if L'Hopital rule applies: $$0 \le M = \lim_{x \to 0} -\frac{1}{\ln x}\int_x^1\frac{f(t)}{t}dt = \lim_{x \to 0}\frac{1}{\ln x}\int_1^x\frac{f(t)}{t}dt = \lim_{x \to 0}\frac{1}{\frac{1}{x}}\frac{f(x)}{x} = \lim_{x \to 0}f(x)$$

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  • $\begingroup$ It is not stated that $f$ has limit at $0$ $\endgroup$ – Norbert May 23 '13 at 22:08
  • $\begingroup$ yes, I should have been more specific, thank you fpr pointing out that $\endgroup$ – user67133 May 23 '13 at 22:16

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