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Dec 30, 2016 at 6:04 comment added Bill Johnson The argument works for every infinite dimensional $L_p$ space since every separable $L_p$ space is isometrically isomorphic as a Banach lattice to $X \oplus_p \ell_p^n$ for some $n=0,1,2,\dots, \infty$, where either $X=\{0\}$ or $X=L_p(0,1)$.
Dec 28, 2016 at 18:22 comment added Paata Ivanishvili Do you consider $L^{p}$ with Lebesgue measure? If not then Nelson's hypercontractivity gives the counterexample. Hermite semigroup $e^{-tH}$ maps $L^{p}(d\gamma)$ into $L^{q}(d\gamma)$ for $1 <p<q<\infty$ where $d\gamma$ is the Gaussian measure.
Oct 7, 2014 at 23:55 history answered Bill Johnson CC BY-SA 3.0