# $f$ a continuous function in [0,1], Show: $\lim_{x \to 0}, x^{\alpha}\int_{x}^{1}\frac{f(t)}{t^{\alpha+1}}$, for $\alpha >0$

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?

Thanks!

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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}

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Why do you claim the denominator approaches $\infty$? That is not always true. – Peter Tamaroff Feb 8 '12 at 22:45
@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. – alex.jordan Feb 9 '12 at 1:34
@Peter Is your question about the numerator? That's a good point. I'll add the explanation to the answer. – alex.jordan Feb 9 '12 at 1:41
Yes I meant numerator. (!) – Peter Tamaroff Feb 9 '12 at 1:44
@Peter If the numerator does not approach infinity, the OP's limit is just $0\cdot L$ for some real $L$. – 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 $$\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)}{t^{\alpha+1}}dt=\frac{f(0)}{\alpha}.$$

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 @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! – Peter Tamaroff Feb 8 '12 at 21:23 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 $f(0)$. – Davide Giraudo Feb 8 '12 at 21:25 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. – Peter Tamaroff Feb 8 '12 at 21:31 It's correct, but maybe you have to detail a bit more when you take the limit, for example writing $f(xu)=f(xu)-f(0)+f(0)$. – Davide Giraudo Feb 8 '12 at 21:34 How does that change in notation affect the proof? – Peter Tamaroff Feb 8 '12 at 21:36

Let $t = x\cdot u$

$$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du}$$

If I'm not being too crazy, $f(x u)$ tends to $f(0)$ because of continuity (and since it is continuous it is also unif. cont. in $(0,1)$), so you may ultimately have:

$$f\left( 0 \right)\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{du}}{{{u^{\alpha + 1}}}}} = f\left( 0 \right)\int\limits_1^\infty {\frac{{du}}{{{u^{\alpha + 1}}}} = \frac{1}{\alpha }f\left( 0 \right)}$$

I'll leave this open for correction, I'm not 100% sure on this one.

I worked this out, but still no one gave me green light in my question. I'm not sure on the last inequalities ( $<$ vs $\leq$ )

Let $P$ be the statement that $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} = \frac{{f\left( 0 \right)}}{\alpha }$$

Then $P$ is true if and only if

$$\forall \epsilon > 0\exists \delta > 0$$

Such that if $$\left| x \right| < \delta$$ then $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha + 1}}}}du} - \frac{{f\left( 0 \right)}}{\alpha }} \right| < \epsilon$$

But then

$$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| <$$

$$\left| {\int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du} + \int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( 0 \right)}}{{{u^{\alpha + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| \leqslant$$

$$\varepsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } <$$

And since

$$\frac{{1 - {\delta ^\alpha }}}{\alpha } < \frac{1}{\alpha }$$ $$\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } < \frac{\epsilon }{\alpha } < \epsilon$$

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 My concern is that you apply $\lim_{x\to0}$ to $f(xu)$ inside the integrand, without simultaneously applying it to $1/x$ in the upper limit. I don't feel easy about the swap $\lim_{x\to0}F(x,f(x))=\lim_{x\to0}F(x,\lim_{x\to0}f(x))$. – alex.jordan Feb 9 '12 at 2:04 My defense is ultimately uniform convergence. But it does need some extra polishing. – Peter Tamaroff Feb 9 '12 at 2:08