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Why if $p\not = q$ we have $L^p(R^n) \not \subseteq L^q(R^n)$? This is a result present in my books, and I can't figure out really a nice proof about this.

An example say that the function $u(x) = (1+|x|)^{-n/p}$ is in all $L^q(R^n)$ with $q>p$ but not in $L^p(R^n)$.

Another example say that the function $u(x) = 1_B|x|^{-n/p}$ is in all $L^q(R^n)$ ($B=B_1(0)$) with $q<p$ but not in $L^p(R^n)$.

I can't figure out why those are verified. Also, $1_B$ is the ball $B=B_1(0)$? I've never used this notation on a function. Can someone explain me?

Alessar
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    $1_B$ means equal to 1 on the ball and zero anywhere else.

    Think of these functions as two different types.

    The first type $u(x) = (1+|x|)^{-n/p}$ is kind of like the fact that $1/(1+x)$ integrates to infinity over an unbounded domain R+.

    The second type $u(x) = 1_B |x|^{-n/p}$ is like how $1/x$ blows up at zero and is not integrable.

    The inclusions are different for the two types of blow up

     – fGDu94 Jul 15 '19 at 12:31
  • Thanks. But why for the first case $q<p$ and for the second $q>p$? This is my great perplexity (my apology if this seems an incredibly trivial question) – Alessar Jul 15 '19 at 13:03
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    It is because if you had say, $f(x) = 1/x^2$ around $0$, this blows up faster than $1/x$. So in general around the origin, higher powers means not in $L^p$.

    At infinity (or on an unbounded positive domain), functions like $1/(1+x^2)$ do actually integrate, and higher powers than that will be in $L^1$, it is the lower powers that are not in $L^1$.

     – fGDu94 Jul 15 '19 at 13:17
  • So if I understand correctly, for the second case $f(x)=1/x$, beign slower than $f(x)=1/x^2$, its contained in the $L^2$ space. And for the first case, the greater is the exponent the less it explode (due to the negative sign)..? For the equivalent, mean that $L^p$ not in itself (the last part of the two cases) it's because the blow up speed is the same? Or in this case am I missing something? – Alessar Jul 15 '19 at 13:33
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    Let us talk of $L^1$. You can adjust the functions accordingly for $L^p$.

    $f(x) = \mathbb{I}_B 1/x^{0.99999} \in L^1$

    but $f(x) = \mathbb{I}_B 1/x \notin L^1$.

    $f(x) = 1/(1+x^{1.00001}) \in L^1$

    but $f(x) = 1/(1+x^1) \notin L^1$

     – fGDu94 Jul 15 '19 at 13:43
  • Ok, thanks, it's more clear now. It's a way classify the different kind of divergent behavior of the functions according to the $L^p$ where you study it. Thank you George for the patience, probably I'll come up with more questions on the Lp spaces sub in the future. Thank you again – Alessar Jul 15 '19 at 13:56

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$1_B$ is the function defined by $1_B(x)=1$ if $x \in B$ and $0$ otherwise. To prove these facts you can use polar coordinates in $\mathbb R^{n}$. See Rudin's RCA for polar coordinates.

Kavi Rama Murthy
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  • Thanks Kavi; can I have a link for Rudin's RCA? Also, in the $L_p$ spaces, the polar coordinates are always the right way to study the p-norms? – Alessar Jul 15 '19 at 13:59
  • The index in Rudin's book has an entry for polar coordinates. Polar coordinates are useful to decide integrability of functions which depend only on the norm. – Kavi Rama Murthy Jul 15 '19 at 23:09