neitheir $L^p((0, \infty),\mathbb{C}) \subset L^q((0, \infty),\mathbb{C})$ nor $L^q((0, \infty),\mathbb{C}) \subset L^p((0, \infty),\mathbb{C})$

Question

I have to show that for $1\le p < q <\infty$
neitheir
$L^p((0, \infty),\mathbb{C}) \subset L^q((0, \infty),\mathbb{C})$
nor
$L^q((0, \infty),\mathbb{C}) \subset L^p((0, \infty),\mathbb{C})$.

I previously had to show, that there is no $a \in \mathbb{R}$ with which the function f:$(0, \infty) \to \mathbb{R}$ f(x)=x^a
would be lebesgue integrable and there is a hint that i should use this information, but i have no idea how to do so.

I would be very thankfull for any kind of help.

Answer 1

Try this: if $p<q$ pick $p<r<q$ and consider the function $f:(0,+\infty)\to\mathbb{R}$ given by $$ f(x)=x^{-\frac{1}{r}}\chi_{(0,1)}(x) $$ it's easy to show that $f\in L^p(0,+\infty)$ but $f\notin L^q(0,+\infty)$. For the second part you might just take $$ g(x)=x^{-\frac{1}{r}}\chi_{(1,+\infty)}(x) $$ Here $f\in L^q(0,+\infty)$ but $f\notin L^p(0,+\infty)$. This example should show you that in general for $p<q$ you have $L^q\subset L^p$ if the space you are considering have "measure mass condensed in bounded sets" (not formal, but in a suitable sense it works), for example in $[0,1]$. If the mass is "condensed at $\infty$", for example with $\mathbb{N}$ and the counting measure, then is reversed: $\ell^p\subset\ell^q$.