Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

Given a measure space $(X,\Sigma,\mu)$, when is it the case that $L^p(X,\mu)\subset L^r(X,\mu)$ for $p>r$, or for $p<r$ . Thanks.

share|cite|improve this question

marked as duplicate by Davide Giraudo, Norbert, ncmathsadist, Dennis Gulko, Sasha Mar 2 '13 at 14:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See the accepted answer to – Martin Mar 2 '13 at 10:03

When $\mu(X)<\infty$ then Jensen's inequality applied to the convex function $\varphi(x)=x^{\frac rp}$ ($r\geq p$) ensures that if $g\in L^r(X,\mu)$, then $$ \frac{1}{\mu(X)^{\frac rp}} \|g\|_p^r = \varphi \left( \frac{1}{\mu(X)} \int_X|g|^p\,{\rm d}\mu \right) \leq \frac{1}{\mu(X)} \int_X\varphi(|g|^p)\,{\rm d}\mu = \frac{1}{\mu(X)} \|g\|_r^r $$ i.e. $$ \frac{1}{\mu(X)^{\frac 1p}} \|g\|_p \leq \frac{1}{\mu(X)^{\frac 1r}} \|g\|_r $$ Therefore if $r\geq p$ then $f\in L^r(X,\mu)$ implies $f\in L^p(X,\mu)$ (since from the above inequality it follows that if $\|f\|_r<\infty$ so is $\|f\|_r<\infty$), i.e. $$ L^r(X,\mu)\subseteq L^p(X,\mu) \qquad\text{whenever $p\leq r$ and $\mu(X)<\infty$} $$ Note that (trivially) the inequality holds for $r=\infty$ too (and even more trivially if $\mu(X)=0$, in which case $L^p(X,\mu)=\{0\}$ for every $p$).

The other inclusion does not necessarily hold (nor it need not, as the case $\mu(X)=0$ shows). To prove this you might consider $L^p([0,1],\lambda)$ where $\lambda$ denotes the Lebesgue measure, and find a counterexample there.

Analogously, if $\mu(X)=\infty$ then no inclusion necessary holds (you can easily construct counterexamples in $L^p(\mathbb R,\lambda)$ where $\lambda$ is again the Lebesgue measure).

NB: Jensen's inequality states that if $\varphi:\mathbb R\rightarrow \mathbb R$ is a convex function, then $$ \varphi \left( -\hspace{-10pt}\int_X f\,{\rm d}\mu \right) \leq -\hspace{-10pt}\int_X \varphi(f)\,{\rm d}\mu $$ where $\displaystyle{-\hspace{-10pt}\int_X f\,{\rm d}\mu = \frac{1}{\mu(X)}\int_Xf\,{\rm d}\mu}$

share|cite|improve this answer
@Martin Thanks, I corrected it. – AndreasT Mar 2 '13 at 11:26

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