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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.

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