3
$\begingroup$

I know that for finite measure space $(X, \mathcal A ,\mu )$ and $1\leq p< q<\infty $ , the inclusion $\mathcal L^q\subseteq \mathcal L^p\subseteq\mathcal L^1 $ holds true (applying Holder's inequality). I was looking for an example that could prove that this inclusion does not hold true for non-finite measure space. Also I was wondering about the inclusion of the space $\mathcal L^\infty$.

$\endgroup$
  • 4
    $\begingroup$ Those inclusions you say you know are all wrong. The reverse inclusions are true. The inclusions you cite are true not for finite measure spaces, but for measure spaces where there exists $\delta>0$ such that $\mu(E)>0$ implies $\mu(E)\ge\delta$, for example counting measure on some set. $\endgroup$ – David C. Ullrich Jul 12 '15 at 16:26
  • $\begingroup$ Thanks , yes I misplaced the $\mathcal L^1$ . $\endgroup$ – NewB Jul 12 '15 at 16:36
3
$\begingroup$

As a complement to Jack D'Aurizio's answer, here are examples of functions showing that neither inclusion, $L^p \subseteq L^q$ nor $L^p \supseteq L^q$, are true in general on an infinite measure space.

Let $q > p \geq 1$, then $f(x) = x^{-1/p}$ is in $L^q( [1,\infty))$ but not in $L^p( [1,\infty))$, since $$\int_1^{\infty} x^{-q/p} dx = \frac{p}{q-p} \textrm{ and } \int_1^{\infty} x^{-p/p} dx = \infty. $$ Furthermore, $g(x) = \frac{1}{(x-1)^{1/q}} \chi_{[1,2)}(x)$ is in $L^p([1,\infty))$ but not in $L^q([1,\infty))$, since $$\int_1^{\infty} g(x)^q dx = \int_1^2 \frac{dx}{x-1} = \infty \textrm{ and } \int_1^{\infty} g(x)^p dx = \int_1^2 \frac{dx}{(x-1)^{p/q}} = \frac{q}{q-p}. $$ In this last example, I am taking advantage of the fact that $L^q$ is not contained in $L^p$ for the finite measure space $[1,2]$.

$\endgroup$
4
$\begingroup$

$\frac{-\log x}{1+x^2}$ is a $L^1(\mathbb{R}^+)$ function that is essentially unbounded, while $\frac{1}{x+1}$ is a bounded function over $\mathbb{R}^+$, but is it not integrable. Hence over $\mathbb{R}^+$, we do not have $L^1\subseteq L^{\infty}$ neither $L^{\infty}\subseteq L^{1}$. Intermediate cases are similar.

Also notice that if $1\leq p<q\leq \infty$ and the measure of the space is finite, we have: $$ L^1\color{red}{\supseteq} L^p \color{red}{\supseteq} L^q \color{red}{\supseteq} L^{\infty}$$ by Holder's inequality.

$\endgroup$

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.