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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.

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2 Answers 2

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 $u \notin W^{1,3}$

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

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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

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