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Let $f:[1,\infty)\to\mathbb{R}$ be a measurable function with $\lim_{t\to\infty} f(t)=0$. I want to show that the function $x\mapsto \int_1^x \frac{f(t)}{t} dt$ is asymptotically sublogarithmic, i.e. $$\lim_{x\to\infty}\frac{1}{\log x} \int_1^x \frac{f(t)}{t} dt = 0.$$ Although I think I should be able to prove this, I am not.

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Using L'hoptial's rule ... we get lim x->inf (f(x)/x)/(1/x) = lim x->inf f(x) = 0 –  Santosh Linkha Jun 29 '12 at 15:34
    
Not sure so i posted it as comment ... if wrong never mind!! –  Santosh Linkha Jun 29 '12 at 15:34
    
@experimentX That would work if $f$ was continuous. But since it's only measurable, the integral is not necessarily a differentiable function of $x$. –  user31373 Jun 29 '12 at 15:37
    
thanks @LeonidKovalev for correcting me but isn't $ \int f(t)/t$ differentialble?? –  Santosh Linkha Jun 29 '12 at 15:37
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Hint: Suppose $|f(t)|<\delta$ for all $t>t_0$. Show that $$\limsup_{x\to\infty}\frac{1}{\log x} \int_1^x \frac{|f(t)|}{t} dt < \delta$$. Do it by decomposing the interval of integration in to two parts: below $t_0$ and above $t_0$. –  GEdgar Jun 29 '12 at 15:39

2 Answers 2

up vote 4 down vote accepted

Since $\lim_{t\to\infty} f(t) = 0$ you have that for every $\epsilon > 0$ there exists $M > 0$ such that $|f(t)| < \epsilon$ for all $t > M$. For $y > M$ you then have

$$ \frac{1}{\log y} \int_1^y \frac{f(t)}{t}\mathrm{d}t = \frac{1}{\log y} \int_1^M \frac{f(t)}{t} \mathrm{d}t + \frac{1}{\log y} \int_M^y \frac{f(t)}{t}\mathrm{d}t $$

The first integral contributes a constant term (only depending on $M$). The second integral can be bounded by

$$ \lvert\int_M^y \frac{f(t)}{t} \mathrm{d}t\rvert \leq \epsilon \log \frac{y}{M} $$

Hence we have that

$$ \left\lvert \frac{1}{\log y}\int_1^y \frac{f(t)}{t} \mathrm{d}t\right\rvert \leq \frac{C_M}{\log y} + \epsilon $$

By choosing $y > M$ large enough we can make the first term also $< \epsilon$ using that $\log y$ grows unboundedly. This means that for every $\epsilon$ you can choose $Y_0$ sufficienly large such that for every $y > Y_0$,

$$ \left\lvert\frac{1}{\log y}\int_0^y \frac{f}{t} \mathrm{d}t \right\rvert < 2\epsilon $$

as desired.

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For every $\epsilon>0$, $$\frac{1}{\log x}\int_1^x \frac{f(t)}{t}\,dt-\epsilon = \frac{1}{\log x}\int_1^x \frac{f(t)-\epsilon}{t}\,dt$$ and since $f(t)-\epsilon<0$ for all large $t$, you should be able to show that $$\limsup_{t\to\infty}\frac{1}{\log x}\int_1^x \frac{f(t)-\epsilon}{t}\,dt\le 0$$ Hence $$\limsup_{t\to\infty}\frac{1}{\log x}\int_1^x \frac{f(t)}{t}\,dt\le \epsilon$$ and replacing $f$ with $-f$ gives you a bound from the other side..

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