# 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?

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 $\limsup\limits_{x\to0+}f(x)>0$
• 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? – Deco May 23 '13 at 22:29
• 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 – Norbert May 23 '13 at 22:31
Here is a calculation you can do if L'Hopital rule applies: $$0 \le M = \lim_{x \to 0} -\frac{1}{\ln x}\int_x^1\frac{f(t)}{t}dt = \lim_{x \to 0}\frac{1}{\ln x}\int_1^x\frac{f(t)}{t}dt = \lim_{x \to 0}\frac{1}{\frac{1}{x}}\frac{f(x)}{x} = \lim_{x \to 0}f(x)$$
• It is not stated that $f$ has limit at $0$ – Norbert May 23 '13 at 22:08