Let $f$ be a continuous function in [0,1] and $\alpha >0$, I'd love your help with finding the following limit: $\lim_{x \to 0}, x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}}$.

First I tried to bound the function since it is continues in a closed interval

$\lim_{x \to 0}, |x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}} |\leq \lim_{x \to 0},| x^{\alpha}\int_{x}^{1}\frac{M}{t^{\alpha+1}}|$ but I get a number depends on $\alpha$ and $M$, and it won't do. So, I assume that the integral is always diverges, since $f$ is continues and it is divided by $t$ with exponent bigger than one, so it is 0 multiplied by $\infty$, maybe we can use L'Hopital somehow?



You said it yourself in the question: use L'Hospital's Rule applied to $$\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}$$ The numerator and denominator each approach $\infty$, given your conditions on $f$.

EDIT: The denominator approaches $\infty$ simply because $\alpha$ is positive.

If $f(0)$ is nonzero, then $f$ is bounded below by some positive $\epsilon$ in a neighborhood of $0$. So the integral is bounded below by $\epsilon\int_x^\delta\frac{1}{t^{\alpha+1}}\,dt+\int_{\delta}^1\frac{f(t)}{t^{\alpha+1}}\,dt$ which diverges to infinity as $x$ approaches $0^+$, since the power of $t$ is greater than $1$.

(And if $f(0)=0$ the numerator might not approach $\infty$. But it still approaches something since $f$ is integrable on $[0,1]$ and $t^{\alpha+1}$ is monotonic. Let's call it $L$. Then L'Hospital's Rule is not needed - the OP's limit is $0\cdot L$, or just $0$. This is consistent with the formula given below when $f(0)\neq0$.)

The result is $$\begin{align}\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}&=\lim_{x\to0^+}\frac{\frac{d}{dx}\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}1/x^{\alpha}}\\ &=\lim_{x\to0^+}\frac{-\frac{d}{dx}\int_1^x\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}x^{-\alpha}}\\ &=\lim_{x\to0^+}\frac{-\frac{f(x)}{x^{\alpha+1}}}{-\alpha x^{-\alpha-1}}\\ &= \lim_{x\to0^+}\frac{f(x)}{\alpha}\\ &=\frac{f(0)}{\alpha} \end{align}$$

  • $\begingroup$ Why do you claim the denominator approaches $\infty$? That is not always true. $\endgroup$ – Pedro Tamaroff Feb 8 '12 at 22:45
  • 1
    $\begingroup$ @Peter The OP specifies that $\alpha$ is positive, so yes, $\lim_{x\to0^+}(1/x^{\alpha})$ is $\infty$. I suppose all my limits should be right-hand limits, but this is implied by the position of $x$ in the integral. I'll edit it anyway though. $\endgroup$ – alex.jordan Feb 9 '12 at 1:34
  • $\begingroup$ @Peter Is your question about the numerator? That's a good point. I'll add the explanation to the answer. $\endgroup$ – alex.jordan Feb 9 '12 at 1:41
  • $\begingroup$ Yes I meant numerator. (!) $\endgroup$ – Pedro Tamaroff Feb 9 '12 at 1:44
  • $\begingroup$ @Peter If the numerator does not approach infinity, the OP's limit is just $0\cdot L$ for some real $L$. $\endgroup$ – alex.jordan Feb 9 '12 at 2:06

Let $\varepsilon>0$ and $\delta$ such that if $|x|\leq\delta$ then $|f(x)-f(0)|\leq\varepsilon$. We have for $0<x<\delta$: \begin{align*}\left|x^{\alpha}\int_x^1\frac{|f(t)-f(0)|}{t^{\alpha+1}}dt\right| &\leq x^{\alpha}\int_x^{\delta}\frac{\varepsilon}{t^{\alpha+1}}dt+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\ &=x^{\alpha}\varepsilon \frac{-1}{\alpha}(\delta^{-\alpha}-x^{-\alpha})+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\ &=\frac{(1-x^{\alpha}\delta^{-\alpha})}{\alpha}\varepsilon+x^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt\\ &\leq \frac{(1+|x|^{\alpha}\delta^{-\alpha})}{\alpha}\varepsilon+|x|^{\alpha}\int_{\delta}^1\frac{|f(0)|}{t^{\alpha+1}}dt, \end{align*} so $\limsup_{x\to 0}\left|x^{\alpha}\int_x^1\frac{|f(t)-f(0)|}{t^{\alpha+1}}dt\right|\leq \varepsilon$ and $\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)-f(0)}{t^{\alpha+1}}dt=0$. Since $$x^{\alpha}\int_x^1\frac{dt}{t^{\alpha+1}}=-\frac 1{\alpha}(1-x^{-\alpha})x^{\alpha}=\frac 1{\alpha}(1-x^\alpha),$$ so finally $$\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)}{t^{\alpha+1}}dt=\frac{f(0)}{\alpha}.$$

  • $\begingroup$ @DavidGiraudo Could you give me some hint on why introducing $f(0)$ in the first expression? You see I produced a proof but this type of proofs aren't at reach for me. I need to be educated on this! $\endgroup$ – Pedro Tamaroff Feb 8 '12 at 21:23
  • $\begingroup$ First I try to guess what the result will be, so I test on the simplest continuous functions: constant one. Then we notice that letting $x\to 0$ we integrate over a set which is near $0$, so it's good to control what happens on $f(t)$ compared to $f(0)$. $\endgroup$ – Davide Giraudo Feb 8 '12 at 21:25
  • $\begingroup$ I understand. Do you see any flaws in my solution? I make use of the UC to take the limit, so it seems the operations are legitimate. $\endgroup$ – Pedro Tamaroff Feb 8 '12 at 21:31
  • $\begingroup$ It's correct, but maybe you have to detail a bit more when you take the limit, for example writing $f(xu)=f(xu)-f(0)+f(0)$. $\endgroup$ – Davide Giraudo Feb 8 '12 at 21:34
  • $\begingroup$ How does that change in notation affect the proof? $\endgroup$ – Pedro Tamaroff Feb 8 '12 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.