5
$\begingroup$

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?

  • 0
    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 $x\to 0^+$.2012-04-08
  • 0
    @Brian Ok. I thought the $M$ was implicitly defined, but I guess it is appropriate to clarify. Any hints on the second one?2012-04-08
  • 0
    Not offhand; there might be after I think about it a bit, but someone else is likely to get there first.2012-04-08
  • 0
    You can't speak about $f'$ in second problem, since $f$ is only integrable2012-04-08
  • 0
    @Norbert Is there any way to prove that $x\int\limits_x^1 {\frac{{f'\left( t \right)}}{t}dt} \to 0$?2012-04-08
  • 0
    The problem is that $f'(t)$ has no meaning for general integrable funtion. There a lots of integrable functions which are not differentiable. For examplу Reimann function.2012-04-08
  • 0
    You can check out my user page, it has a lot of similar questions that might be of interest to you.2012-04-09

2 Answers 2

3

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, $L=\ell$.

  • 0
    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).2012-04-08
  • 0
    @PeterT.off My approach gives prediction for integrabale $f$ :)2012-04-08
  • 0
    First we work in the case on which $\ell=0$, , using $g=f-\ell$, then we try to generalize.2012-04-08
4

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!

  • 0
    Accordingly upvoted. However, I'm looking for an approach following what I proposed.2012-04-08
  • 1
    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.2012-04-08
  • 0
    Ok, I will add this restrictions2012-04-08
  • 1
    @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.2012-04-08
  • 0
    @DavidMitra I don't understand Greg said that there must be infinity in the numerator2012-04-08
  • 0
    @DavidMitra: try using l'Hopital's rule on $\lim_{x\to\infty} \frac{\sin(x^3)}x$.2012-04-08
  • 0
    @Norbert It's not well-known, but the hypothesis that the numerator tends to infinity is not needed. The [wiki page](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) on L'Hopital's rule mentions this. In some analysis texts, such as Bartle and Sherbert's *Introduction to Real Analysis*, the theorem is stated: "Let $f,g$ be differentiable on $(a,b)$ with $g'(x)\ne 0$ for all $x\in(a,b)$. Suppose that $\lim\limits_{x\rightarrow a^+}g(x)=\pm\infty$. Then if $\lim\limits_{x\rightarrow a^+}{f'(x)\over g'(x)}=L$, it follows that $\lim\limits_{x\rightarrow a^+}{f(x)\over g(x)}=L$.2012-04-08
  • 1
    @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...2012-04-08
  • 0
    Fair enough. Let me instead say that the version of l'Hopital that we often use in practice - that is, just take derivatives of numerator and denominator and see what happens - is not valid unless the original limit has an indeterminate form.2012-04-09