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

Question

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!

Answer 1

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 $f$.

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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 $f(0)\neq0$.)

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

Answer 2

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 $$\lim_{x\to 0}x^{\alpha}\int_x^1\frac{f(t)}{t^{\alpha+1}}dt=\frac{f(0)}{\alpha}.$$

Answer 3

Let $t = x\cdot u$

Your integral becomes

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

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

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