Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A=\{(x,y)\in \mathbb R^2: 0<x<1, 0< y < \sqrt{x}\}$ and $f \colon A \to\mathbb R$ a continuous function s.t. $$ \frac{1}{x^2+y^2} \le f(x,y) \le\frac{2}{x^2+y^2} $$ for every $(x,y) \in A$. Determine the set of value $\alpha >0$ such that $$ \iint_Af(x,y)^\alpha dx \, dy < +\infty. $$

The function $t\mapsto t^\alpha$ is monotone (increasing) when $\alpha>0$. Therefore, using the hypothesis we get $$ \frac{1}{(x^2+y^2)^\alpha} \le f(x,y)^\alpha \le\frac{2}{(x^2+y^2)^\alpha} $$ for every $(x,y) \in A$. Integrating we get $$ \iint_A \frac{dx \, dy}{(x^2+y^2)^\alpha} \le \iint_A f(x,y)^\alpha dx \, dy\le\iint_A\frac{2 \, dx \, dy}{(x^2+y^2)^\alpha} $$ If we pass to polar coordinates, we have $$ \iint_A \frac{1}{\rho^{2\alpha-1}} \le \iint_A f(\rho \cos{\vartheta},\rho \sin{\vartheta})^\alpha \rho \, d\rho \, d\vartheta \le \iint_A \frac{2}{\rho^{2\alpha-1}} \, d\rho \, d\vartheta $$

Now we have to write the set $A$ using the polar coordinates, but this is quite difficult. What can we do? I think that the first and the third integrals are improper in $0$ with respect to $\rho$. Therefore I think we should ask at least $2\alpha-1<1$ i.e. $\alpha<1$.

I think $\alpha=1$ doesn't work: indeed, we have $$ \begin{split} \iint_A f(x,y) dx \, dy & \ge \iint_A\frac{\, dx \, dy}{(x^2+y^2)} \\ & = \int_0^1 dx \, \int_0^{\sqrt{x}} \frac{dy}{x^2+y^2} = \\ & = \int_0^1 dx \, \frac{1}{x^2}\int_0^{\sqrt{x}}\frac{dy}{1+(\frac{y}{x})^2} =\\ & = \int_0^1 \frac{1}{x}\arctan{\left( \frac{1}{\sqrt{x}}\right)}dx = +\infty \end{split} $$ because $$ \frac{1}{x}\arctan{\left( \frac{1}{\sqrt{x}}\right)} \sim_{x=0} \frac{c}{x} $$ whose integral in $0$ diverges.

What do you think? Is it correct? How can we prove it formally?

Thanks a lot.

share|cite|improve this question
up vote 1 down vote accepted

If $\alpha <1$ we have $2\alpha<2$ and \begin{equation} \int_{A} f(x,y)^{\alpha} \le \int_{B_1} \dfrac{2}{|X|^{2 \alpha}} < \infty. \end{equation} If $\alpha >1$ we have $2\alpha>2$ and considering $B \subset A$ we have \begin{equation} \int_{A} f(x,y)^{\alpha} \ge \int_{B} \dfrac{1}{|X|^{2 \alpha}} < \infty. \end{equation} To $\alpha=1$ see for one idea. Example that $u\in W^{1,2}$, but $u \notin W^{1,3}$

share|cite|improve this answer
Thanks for your answer, but I cannot understand what happens if $\alpha >1$: would you mind fixing the typo? Thanks a lot. – Romeo Aug 5 '12 at 18:16
Yes, I did a type. – user29999 Aug 5 '12 at 18:39
I think you should edit still one typo: I think you mean that if $\alpha > 1$ then the integral is $>+\infty$, hence it diverges. In other words, the only thing we have to do is discussing what happens when $\alpha=1$. Thanks also for the reference, I'll read it! – Romeo Aug 5 '12 at 19:04
Following the link, I got an interesting idea and I wrote down the case $\alpha=1$ in the OP. Would you like to give a look at it, please? Thanks a lot. – Romeo Aug 5 '12 at 19:25
Only note that $\int_0^{\sqrt{x}} \frac{dy}{x^2+y^2}= \frac{1}{x}\arctan{\frac{1}{\sqrt{x}}}$, $\int_0^{\sqrt{x}} \frac{dy}{x^2+y^2} \neq \frac{1}{x^2}\arctan{\frac{1}{\sqrt{x}}}$ and see $\underline{\mbox{Modern Techniques and Their ApplIcations. Folland corollari 2.52}}.$ – user29999 Aug 5 '12 at 20:00

Hint: for $0 < \rho < 1$, $0 < \vartheta < \pi/4 \implies (\rho, \vartheta) \in A \implies 0 < \vartheta < \pi/2$.

share|cite|improve this answer
So we can write $\iint_A \frac{1}{\rho^{2\alpha-1}}d\rho d\vartheta \le \int_0^1 \frac{d\rho}{\rho^{2\alpha-1}} \int_0^{\frac{\pi}{2}}d\vartheta = \int_0^1 \frac{d\rho}{\rho^{2\alpha-1}}\frac{\pi}{2} < \infty$ iff $\alpha<1$. Am I right? Thanks for your answer! – Romeo Aug 5 '12 at 17:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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