Positiveness of Limit Superior 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?
 A: 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 $\limsup\limits_{x\to0+}f(x)>0$
A: Here is a calculation you can do if L'Hopital rule applies:
$$0  \le M = \lim_{x \to 0} -\frac{1}{\ln x}\int_x^1\frac{f(t)}{t}dt = \lim_{x \to 0}\frac{1}{\ln x}\int_1^x\frac{f(t)}{t}dt = \lim_{x \to 0}\frac{1}{\frac{1}{x}}\frac{f(x)}{x} = \lim_{x \to 0}f(x)$$
