1

Fixing an arbitrary $p$, can a function always be found which is $L^p$ but not $L^{p+1}$ or $L^{p-1}$ on $\mathbb{R}_{\geq0}$?

More generally, given a pair $\big(a,b\big)\in\mathbb{Z}_{\geq1}^2$, can a function $f$ always be found which is $L^p$ iff $p\in\big(a,b\big)$ or iff $p\in\big[a,b\big]$?

0-seigfried
  • 389
  • Maybe this https://math.stackexchange.com/questions/66029/lp-and-lq-space-inclusion could answer your question into the negative. – Tito Eliatron Jan 20 '20 at 17:52
  • What is the domain of $L^p$? $\mathbb R$? – Jonas Linssen Jan 20 '20 at 18:02
  • @PrudiiArca Yes, I updated the question. – 0-seigfried Jan 20 '20 at 18:10
  • In general ${ p : f \in L^p }$ for any particular $f$ is an interval, including the possibility of an empty set, a singleton, an open interval, a closed interval, and a half-open interval. This can be shown by developing an interpolation inequality for $| f |{L^r}$ in terms of $| f |{L^p}$ and $| f |_{L^q}$ when $p<r<q$. – Ian Jan 20 '20 at 18:12
  • (Cont.) And in the setting of $[0,\infty)$ with the Lebesgue measure, nothing more can be said: all of those possibilities can be attained. On a finite measure space you can say more, and on a measure space such that ${ m(A) : m(A)>0 }$ is bounded away from zero you can say more. The details for this are in Tito's link. But $[0,\infty)$ with the Lebesgue measure is neither of those. – Ian Jan 20 '20 at 18:20

0 Answers0