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I'm solving this problem and I guess it shouldn't be too hard. Since $f$ is continuous it is bounded, so one has

$$\left| {x\int\limits_x^1 {\frac{{f\left( t \right)}}{t}dt} } \right| \leq x\int\limits_x^1 {\left| {\frac{{f\left( t \right)}}{t}} \right|dt} \leqslant Mx\int\limits_x^1 {\frac{{dt}}{t}} = - Mx\log x \to 0$$

Where $M=\operatorname{sup}\{|f(x)|:x\in[0,1]\}$

I'm not 100% certain on this, so I want a better, clearer approach.

Then, there is a second problem, similar, which is:

If $f$ is integrable on $[0,1]$ and $\exists\lim\limits_{x\to0}f(x)=L$, find

$$\ell = \mathop {\lim }\limits_{x \to {0^ + }} x\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^2}}}dt} $$

ADD: The second might follow from the first since

$$\mathop {\lim }\limits_{x \to {0^ + }} x\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^2}}}dt} =\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) - xf\left( 1 \right) + x\int\limits_x^1 {\frac{{f'\left( t \right)}} {t}dt} $$

$$ = L + \mathop {\lim }\limits_{x \to {0^ + }} x\int\limits_x^1 {\frac{{f'\left( t \right)}}{t}dt} $$

So, what can I sat about $f'(t)$ given $f(t)$ is integrable on $[0,1]$ that will allow me to apply the first case to the last limit?

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The first argument is fine, except that the first $<$ should be $\le$, and you should say explicitly that $M=\sup\{|f(x)|:x\in[0,1]\}$ and that at the end you’re taking the limit as $x\to 0^+$. – Brian M. Scott Apr 8 '12 at 19:23
@Brian Ok. I thought the $M$ was implicitly defined, but I guess it is appropriate to clarify. Any hints on the second one? – Pedro Tamaroff Apr 8 '12 at 19:24
Not offhand; there might be after I think about it a bit, but someone else is likely to get there first. – Brian M. Scott Apr 8 '12 at 19:26
You can't speak about $f'$ in second problem, since $f$ is only integrable – Norbert Apr 8 '12 at 19:44
@Norbert Is there any way to prove that $x\int\limits_x^1 {\frac{{f'\left( t \right)}}{t}dt} \to 0$? – Pedro Tamaroff Apr 8 '12 at 19:48
up vote 3 down vote accepted

For the first problem, your approach is fine (but the first inequality maybe be an equality when $f$ is non-negative). For the second, denote $L:=\lim_{x\to 0}f(x)$. Fix $\varepsilon>0$. We can find $\delta>0$ such that if $0\leq x\leq \delta$ then $|f(x)-L|\leq \varepsilon$, so for $0\leq x\leq \delta$: $$x\int_x^1\frac{f(t)}{t^2}dt=x\int_x^1\frac{f(t)-L}{t^2}dt+Lx\int_x^1\frac{dt}{t^2}=x\int_x^1\frac{f(t)-L}{t^2}dt+L\left(\frac 1x-1\right)x$$ hence \begin{align*}\left|x\int_x^1\frac{f(t)}{t^2}dt-L\right|&\leq x\int_x^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\ &=x\int_x^\delta\frac{|f(t)-L|}{t^2}dt+ x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\ &\leq x\int_x^\delta\frac{\varepsilon}{t^2}dt+ x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\ &=\varepsilon x\left(\frac 1x-\frac 1{\delta}\right)+x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx|\\ &=\varepsilon-\frac{\varepsilon}{\delta}x+x\int_\delta^1\frac{|f(t)-L|}{t^2}dt+|Lx| \end{align*} so $$\limsup_{x\to 0^+}\left|x\int_x^1\frac{f(t)}{t^2}dt-L\right|\leq \varepsilon$$ and since $\varepsilon$ was arbitrary, $L=\ell$.

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I'm OK with this, but can't you devise an approach that avoids the prediction that the limit is indeed $L$? (See my last edit, where it suffices to show that the integral goes to zero). – Pedro Tamaroff Apr 8 '12 at 19:41 My approach gives prediction for integrabale $f$ :) – Norbert Apr 8 '12 at 19:43
First we work in the case on which $\ell=0$, , using $g=f-\ell$, then we try to generalize. – Davide Giraudo Apr 8 '12 at 19:43

If $f$ is continuous let's use L'Hopital rule $$ \lim\limits_{x\to+0}x\int\limits_{x}^{1}\frac{f(t)}{t}dt= \lim\limits_{x\to+0}\frac{\int\limits_{x}^{1}\frac{f(t)}{t}dt}{x^{-1}}= \lim\limits_{x\to+0}\frac{-\int\limits_{1}^{x}\frac{f(t)}{t}dt}{x^{-1}}= \lim\limits_{x\to+0}\frac{-\frac{f(x)}{x}}{-x^{-2}}= \lim\limits_{x\to+0}xf(x)=0 $$ $$ \lim\limits_{x\to+0}x\int\limits_{x}^{1}\frac{f(t)}{t^2}dt= \lim\limits_{x\to+0}\frac{\int\limits_{x}^{1}\frac{f(t)}{t^2}dt}{x^{-1}}= \lim\limits_{x\to+0}\frac{-\int\limits_{1}^{x}\frac{f(t)}{t^2}dt}{x^{-1}}= \lim\limits_{x\to+0}\frac{-\frac{f(x)}{x^2}}{-x^{-2}}= \lim\limits_{x\to+0}f(x) $$ P.S. Big thanks to David Mitra, he pointed out that requirement for integrals to be divergent is unnecessary!

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Accordingly upvoted. However, I'm looking for an approach following what I proposed. – Pedro Tamaroff Apr 8 '12 at 19:45
Remember that you can only use l'Hopital's rule if the limit is in an indeterminate form (in this case, $\infty/\infty$). So your proof works only in the case that the integral in the numerator goes to infinity. – Greg Martin Apr 8 '12 at 20:11
Ok, I will add this restrictions – Norbert Apr 8 '12 at 20:13
@GregMartin For the infinite limit case, you do not need the numerator to tend to infinity to use L'Hopital's rule, only the denominator. – David Mitra Apr 8 '12 at 20:17
@GregMartin For your example, the limit of the quotient of the derivatives does not exist. L'Hopital doesn't apply (Lopital states if the limit of the quotient of the derivatives exists, then... – David Mitra Apr 8 '12 at 20:27

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