0

Has been 7 years from my last $L_p$ spaces experiences, now I have an exam about this. I have difficulties with the very first exercise:

For what $p \in [1, + \infty]$ and $a \in R$ the function $u(x)=(1+|x|)^{-a}$ defined on $R^n$ verify $||u||_{L_p}< \infty$?

Can someone explain me the methods for handle this kind of exercise?

Alessar
  • 490
  • 1
    The trick is to integrate in polar coordinates ad then make an estimate on the integrand. –  Jul 12 '19 at 11:44

2 Answers2

2

$\int (1+|x|)^{-ap}dx=C \int_0^{\infty} (1+r)^{-ap} r^{n-1}\, dr$ where $C$ is the measure of $S^{n-1}$. Can you finish now? [Answer: $ap >n$].

Kavi Rama Murthy
  • 311,013
1

Split the cases where $a$ is positive and negative. The second one is trivial, for the first one the standard thing to do is to use spherical coordinates in $\mathbb{R}^n$. I could do the exercise for you, but from here it should be quite simple, as the integral becomes practially one dimensional (the only thing that could explode is the radial part, once you have changed coordinate system).

tommy1996q
  • 3,344
  • My textbook give this definition of $||u(x)||_{L_p} = ( \int_E |u(x)|^p d \mu )^{1/p}$ where $\mu$ is a measure and $E$ is a set. This is confusing me, because it's a new way to see it (in physics I've never saw this). From your comment, seems like a measure could be simply a coordinate change. Am I right or wrong? – Alessar Jul 12 '19 at 12:11
  • I assumed you were working with Lebesgue measure. If you have another measure, it completely changes everything. For instance, if you are in a probability space, then every bounded function belongs to every $L^p$ space. A measure is not a coordinate change, but you can change coordinates when integrating with respect to a measure (there should be some assumotions for this to work, but it is a bit technical and I don’t know). A measure is simply a function $m$ from a sigma algebra (your measurable sets) to $\mathbb{R}^+$ that gives to every set $A$ its measure $m(A)$ – tommy1996q Jul 12 '19 at 12:17
  • The thing is that the exercise do not express a specific measure; it is just that definition and then the next line, this exercise. So you can figure out how I am confused about that. I could apply the definition of course, but I don't know that $d \mu$ what could stand for, how evaluate this thinking about it – Alessar Jul 12 '19 at 12:19
  • If not specified, I think you can be sure it is assumed to be the standard Lebesgue measure. – tommy1996q Jul 12 '19 at 12:22
  • Ok thanks for the clarification... so assuming that it's Lebesgue measure, I could simply study the integral in the case $a=0, a<0, a>0$ and when needed, reduce it to the study of a "ball", meaning the spherical substitution, that in this case it'll be the radius; right? The case $a=0$ is trivial because it reduce the function to a constant right? I beg you pardon if I seems so elementary, I'm restarting slowly – Alessar Jul 12 '19 at 12:25
  • Yes, the cases $a\leq 0$ are indeed trivial. For $a>0$ I don’t get what you mean by “reducing to the study of a ball”, I mean, you are still integrating over the whole space, just in different coordinates. Look what Kavi Rama has written: the integral can be reduced from an $n$ dimensional one (integration in $dx$ to a 1 dimensional by using spherical coordinates. – tommy1996q Jul 12 '19 at 12:32
  • 1
    Bu the way, don’t even try writing down the actual change of variable. The only thing you meed to know is that the jacobian is $\rho ^{n-1}$ times some angular garbage that don’t mess up integrability – tommy1996q Jul 12 '19 at 12:34
  • That garbage is the constant $C$ in the Kavi Rama solution right? – Alessar Jul 12 '19 at 12:51
  • 1
    Yeah, it would be the measure of $S^{n-1}$, that is finite and not relevant for your problem – tommy1996q Jul 12 '19 at 13:05
  • I think I figured out: simply by evaluation, I arrive at $C \int_0^{\infty} (1+r)^{-ap} r^{n-1}, dr < C \int_0^{\infty} (r)^{-ap} r^{n-1}, dr < C \int_0^{\infty} (r)^{-ap} r^{n}, dr < C \int_0^{\infty} (r)^{n-ap} , dr$ it converge iif $n-ap<0$ that is $ap>n$. Right? – Alessar Jul 12 '19 at 13:58