Proposition 15.4.2.label Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\cf \subset L^{p}(X; K)$, and let $|\cf|^{p} = \bracsn{\norm{f}^p| f \in \cf}$, then:
- (1)
If $\cf$ is uniformly $p$-integrable, then $|\cf|^{p}$ is uniformly absolutely continuous with respect to $\mu$.
- (2)
If $\sup_{f \in \cf}\norm{f}_{L^p(X; E)}< \infty$ and $|\cf|^{p}$ is uniformly absolutely continuous with respect to $\mu$, then $\cf$ is uniformly $p$-integrable.
Proof. (1): Let $\eps > 0$, then there exists $M \ge 0$ such that
Thus for any $A \in \cm$,
so if $\mu(A) \le M\eps/2$, then $\sup_{f \in \cf}\int_{A}\norm{f}^{p} d\mu \le \eps$.
(2): Let $\eps > 0$, then there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\sup_{f \in \cf}\int_{A} \norm{f}^{p} d\mu < \eps$. By Markov’s inequality,
Thus for sufficiently large $M$, $\sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) < \delta$, and
$\square$
Post a Comment