# Showing that $C=\{h \in L^1(\Omega) : u_1(z) \leq h(z) \leq u_2(z)\}$ is weakly compact.

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.