Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

My question comes from my confusion in studying the Fefferman-Stein inequality which says that for any $f\in L^p$ it holds $\|f\|_p\leq c \|f^\#\|_p$ with $f^\#$ the maximal function. This is an apriori estimation by knowing $f\in L^p$, and I am wondering if the inverse holds: the inequality $\|f^\#\|_p<\infty$ implies $f\in L^p$? This recalls me similar situations like the soblev space $H^1_0(\Omega)$ where we check only the equivalent norm $\int_\Omega |\nabla f|^2<\infty$. So in which situation can we use only the norm to know if $f$ is in a Banach space $(X,\|\cdot\|)$? A counter example is $C^0(\bar\Omega)$ (for some bounded domain $\Omega\subset \mathbb R^n$) equipped with the maximum norm.

share|improve this question
2  
Not clear what you mean. We can always embed isometrically any $(X,\|\cdot\|)$ into a proper superspace $Y$ , and for $f\in Y$ of course $\|f\|<\infty$ does not tell whether $f\in X$. –  Pietro Majer Nov 19 '13 at 18:34
    
Several remarks: (i) there are many different types of maximal function. The Fefferman-Stein one is rather different from the classical Hardy-Littlewood one. (ii) I quote from the original paper of Fefferman and Stein (Acta Mathematica 1972): "The result we prove is that $f^\sharp\in L^p$ implies $f\in L^p$, if $p < \infty$; this may be viewed as the inverse of the corresponding inequality for the [Hardy-Littlewood] maximal function." This, of course, needs to be taken with some consideration of the growth of $f$ at infinity: observe that $f \equiv c$ has $f^\sharp = 0$ and $f$ ... –  Willie Wong Nov 20 '13 at 9:05
    
... is clearly not in $L^p$. Roughly speaking we have that $f^\sharp$ controls local behaviour (see its relation with BMO), and should be complemented with something that controls behaviour in the large (the a priori growth assumptions). (iii) There is still an a priori assumption when you define the Sobolev spaces: firstly for $H^1_0$ you need vanishing on the boundary, which is not given by your "equivalent norm". Secondly you need to be in a space where $\nabla f$ can be defined... –  Willie Wong Nov 20 '13 at 9:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.