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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.... –  N. S. Feb 10 '12 at 1:09
    
@N.S. can you elaborate it? –  leo Feb 10 '12 at 1:55
    
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) –  Pedro Tamaroff Feb 10 '12 at 1:58

1 Answer 1

up vote 1 down vote accepted

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 $0<u\le 1/\sqrt x$ 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. –  Pedro Tamaroff Feb 13 '12 at 3:33
    
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. –  Bob Pego Feb 18 '12 at 3:10
    
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 $x\to 0$. –  Pedro Tamaroff Feb 20 '12 at 21:51

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