I am working to establish an estimate in $X^{s,b}$ spaces to prove local well-posedness of a certain equation, and I need to consider some sub-cases. In particular, I wish to show that the following integral is bounded in $t$ and $x$, i.e. the following supremum is finite:

$$\sup_{t,x\in \mathbb{R}}\, \frac{\langle x\rangle^{2s}}{\langle t-x^{2\alpha}\rangle^{1-}} \int_{\mathbb{R}^2} \frac{dx_1\;dx_2}{\langle x_1\rangle^{2s} \langle x_2 \rangle^{2s} \langle x-x_1+x_2 \rangle^{2s} \langle t- x_1^{2\alpha}+ x_2^{2\alpha}-(x-x_1+x_2)^{2\alpha} \rangle^{2b}}$$

where the exponent 1- denotes numbers sufficiently close to 1 but less than 1, for $\frac{1-\alpha}{2}< s\leq \frac{1}{2}$, and $\frac{1}{2}<\alpha<1$ and $b>\frac{1}{2}$. Define the notation $\langle z \rangle := (1+z^2)^{1/2}$.

If the above problem is perhaps intractable, I am also happy if the following integral is bounded in $x$ (since then I can do something with the above):

$$\sup_{x\in \mathbb{R}} \,\int_{0<|x_1x_2|\lesssim |x|^{2-2\alpha}} \frac{ \langle x\rangle^{2s}}{\langle x+x_1\rangle^{2s} \langle x+x_2\rangle^{2s} \langle x+x_1+x_2\rangle^{2s}}\; dx_1\,dx_2$$ where similarly $\frac{1-\alpha}{2}< s\leq \frac{1}{2}$, and $\frac{1}{2}<\alpha<1$, and the notation $\langle\cdot\rangle$ as above.

I would be quite happy too if the above holds for other ranges of $s\leq \frac{1}{2}$ for the above $\alpha$.

Thank you.

  • $\begingroup$ Is the region of integration the same as $\{(x_1,x_2): 0<|x_1|\le|x|^{2-2\alpha},\ 0<|x_2|\le|x|^{2-2\alpha}\}$? $\endgroup$ – Matt F. Mar 21 '14 at 11:02
  • $\begingroup$ @Matt F. : No, the region of integration is more like $\{(x_1,x_2): 0<|x_1|\leq \frac{|x|^{2-2\alpha}}{x_2},\, x_2\in \mathbb{R}\}$. $\endgroup$ – digiboy1 Mar 22 '14 at 13:42
  • $\begingroup$ What is $x^{2\alpha}$ for $x<0$ ? $\endgroup$ – fedja Mar 23 '14 at 23:17
  • $\begingroup$ @fedja: for $x<0$, it is $|x|^{2\alpha}$. $\endgroup$ – digiboy1 Mar 25 '14 at 3:32

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.