0
$\begingroup$

I have to show that, in $\mathbb{R}^d$ with Lebesgue measure, the $L^p$ spaces are not comparable one another. More precisely, I want to show that given $p$ and $q$ such that $1\leq p<q\leq\infty$, we have $L^p(\mathbb{R}^d) \nsubseteq L^q(\mathbb{R}^d)$ and $L^q(\mathbb{R}^d) \nsubseteq L^p(\mathbb{R}^d)$.

Could you give me some hints on how I could do that??

$\endgroup$
1
$\begingroup$

First think about $\mathbb{R}$. In the context of Riemann integral, improper integrals are usually studied distinguishing between two cases: the case where the interval of integration is unbounded and the case where the interval of integration is bounded but the function is unbounded on that interval. Exploiting this idea you can build the examples that will give the desired noninclusions.

CASE 1: $1\leq p<q<\infty$

Consider the function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x) = \frac{1}{|x|^{1/p}} $ if $|x|>1$ and $f(x)=1$ otherwise. This $f$ is in $L^q(\mathbb{R})$ but not in $L^p(\mathbb{R})$.

Now consider the function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x) = \frac{1}{|x|^{1/q}} $ if $|x|<1$ and $f(x)=1$ otherwise. This $f$ is in $L^p(\mathbb{R})$ but not in $L^q(\mathbb{R})$.

CASE 2: $1\leq p<q=\infty$

The first of the two functions considered previously is in $L^\infty(\mathbb{R})$ but not in $L^p(\mathbb{R})$.

The function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x) = \frac{1}{|x|^{1/2p}} $ if $|x|<1$ and $f(x)=1$ otherwise, is in $L^p(\mathbb{R})$ but not in $L^\infty(\mathbb{R})$.


One way of showing the same noninclusions for $\mathbb{R}^d$ is to consider functions of the form $$f(x_1,x_2,\dots,x_d)=\frac{1}{|x_1|^{\alpha_1}}\frac{1}{|x_2|^{\alpha_2}}\cdots\frac{1}{|x_d|^{\alpha_d}}$$ selecting carefully the exponents $\alpha_1,\alpha_2,\dots,\alpha_d$.

$\endgroup$
  • $\begingroup$ How can we show that in Case 1, $f$ is in $L^q(\mathbb{R})$ but not in $L^p(\mathbb{R})$?? $\endgroup$ – Mary Star Dec 8 '14 at 0:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.