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Some days ago I answered a question that asked to find

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

given that $f$ is continuous in $[0,1]$

I proceeded as follows:

$$\eqalign{ & t = x\cdot u \cr & dt = x\cdot du \cr} $$

So this is produces:

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

I then thought: "Well, if $f$ is continuous in the closed interval, then it is also uniformly continuous, so I can assume

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

This turned out to be true. However, I wasn't very comfortable with such "move". So now I'm thinking, one can put

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

And then

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

However, this is still insufficient since I need to adress the behaviour of the upper limit too. Can someone show me how to adress both behaviours simultaneously?


Would this work?

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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    Hmm, not sure if this works, but if $x_n$ is any sequence so that $x_n \to 0$, then you can probably prove that $\mathop {\lim }\limits_{n \to \infty} \int\limits_1^{\frac{1}{x_n}} {\frac{{f\left( {x_nu} \right)}}{{{u^{\alpha + 1}}}}du}= \frac{1}{\alpha }f\left( 0 \right)$ by using the Lebegue dominant convergence Theorem... Then, since it holds for any sequence, you should be able to conclude that the limit holds. Seems artificial though....2012-02-10
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    @N.S. can you elaborate it?2012-02-10
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    I'm unfamiliar with Lebesgue's thoerem, or any measure theory topic although I do understand its motivations and some ideas, such as the distance function and the metric it induces in the set. But I'm very far away from that still (I'm reading this [book](http://www.amazon.com/Introduction-Topology-Third-Dover-Mathematics/dp/0486663523))2012-02-10

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You can finish the proof cleanly if you split the integral in your last line to integrate separately over $[0,1/\sqrt x]$ and $[1/\sqrt x,1/x]$. For $x$ small enough, $|f(xu)-f(0)|<\epsilon\alpha/2$ for $0 since $xu\le\sqrt{x}$, and always $|f(xu)-f(0)|\le 2M$ where $M=\max_{0\le u\le 1}|f(u)|$. Then $$ \left(\int_1^{1/\sqrt{x}}+\int_{1/\sqrt{x}}^{1/x} \right) \frac{|f(xu)-f(0)|}{u^{\alpha+1}}du \le \frac{\epsilon\alpha}2 \frac{1-x^{\alpha/2}}\alpha + 2M \frac{x^{\alpha/2}-x^\alpha}\alpha < \frac{\epsilon}2+ 2M\frac{x^{\alpha/2}}\alpha. $$ Now if $x$ is small enough, $2Mx^{\alpha/2}<\epsilon\alpha/2$ and the whole integral is less than $\epsilon$. So indeed it vanishes in the limit $x\to0$ if $\alpha>0$.

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    You still have $x$s after the inequality. See my edit to understand where I'm going.2012-02-13
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    The trouble with the argument you give is that it is not justifies to say that $|f(xu)-f(0)|<\epsilon$ for all $u\in(0,1/x)$, since $xu$ can be any point in $(0,1)$. It's true that my argument was left untidy as $x$ needed to be further restricted --- I'll edit to clean it up.2012-02-18
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    I don't understand your remark. "argument you give is that it is not justifies to say that..." In my view, since $u$ varies from $1$ to $\dfrac{1}{x}$ then $f(xu)$ is at most $f(1)$ and at least $f(x)$, so it has upper bound $f(1)$ and tends to $f(0)$ as $x\to 0$.2012-02-20