Exercise :

Let $\Omega \subseteq \mathbb R^n$ be open and bounded, $u_1, u_2 \in L^1(\Omega)$ with $u_1(z) \leq u_2(z)$ almost everywhere in $\Omega$. We let $C$ be the set $C=\{h \in L^1(\Omega) : u_1(z) \leq h(z) \leq u_2(z)\}$. Show that $C \subseteq L^1(\Omega)$ is weakly compact.

Attempt :

I know that if $C$ is uniformly integrable, then, by the Dunford-Pettis Theorem, it will also be relatively weakly compact.

Note that $\Omega$ is bounded. Starting off, if $u_1(z) \leq u_2(z)$ holds for positive values, then $|u_1(z)| \leq |u_2(z)|$. If it holds for negative values, then $|u_1(z)| \geq |u_2(z)|$. In both cases, $|h(z)|$ will be bounded, thus the set $C$ is bounded.

For the case of positive values and for $\varepsilon \geq 0 \; \exists \delta > 0 :$ \begin{align*} |A| < \delta &\Rightarrow \int_A |u_2|\mathrm{d}x < \varepsilon \; \forall u_2 \in \Omega\\ &\Rightarrow\int_A |h|\mathrm{d}x < \varepsilon \; \forall h \in C \end{align*}

For the case of negative values and for $\varepsilon \geq 0 \; \exists \delta > 0 :$ \begin{align*} |A| < \delta &\Rightarrow \int_A |u_1|\mathrm{d}x < \varepsilon \; \forall u_2 \in \Omega\\ &\Rightarrow\int_A |h|\mathrm{d}x < \varepsilon \; \forall h \in C \end{align*}

Thus, in both cases we yield the $\varepsilon-\delta$ definition of uniform integrability and by the Dunford-Pettis Theorem we get that $C$ is relatively weakly compact.

Question : I have failed to show that $C$ is weakly compact and only proved that it is relatively weakly compact. Any hints or elaborations will be greatly appreciated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.