91Ó°ÊÓ

Let \(f \in (U \cap W)^{0}\). Then, by definition, \(f(u) = 0\) for all \(u \in U \cap W\). Now, notice that every element in \(U \cap W\) is also in both \(U\) and \(W\). So, the linear functional \(f\) sends every element of both \(U\) and \(W\) to zero. This means that \(f \in U^{0}\) and \(f \in W^{0}\). Since both annihilators are subspaces of \(V^{*}\), their sum is also in \(V^{*}\); thus, \(f \in U^{0}+W^{0}\). This shows that \((U \cap W)^{0} \subseteq U^{0}+W^{0}\).

Let \(f \in U^{0}+W^{0}\). By definition, there exist linear functionals \(f_{1} \in U^{0}\) and \(f_{2} \in W^{0}\) such that \(f = f_{1} + f_{2}\). Now, we want to show that \(f \in (U \cap W)^{0}\), which means that \(f\) sends every element in \(U \cap W\) to zero. Let \(v \in U \cap W\). Since \(f_{1} \in U^{0}\) and \(f_{2} \in W^{0}\), we have \(f_{1}(v) = 0\) and \(f_{2}(v) = 0\). Then, \(f(v) = f_{1}(v) + f_{2}(v) = 0 + 0 = 0\), so \(f\) sends every element in \(U \cap W\) to zero. This means that \(f \in (U \cap W)^{0}\), and thus \(U^{0}+W^{0} \subseteq (U \cap W)^{0}\). Since we have shown that \((U \cap W)^{0} \subseteq U^{0}+W^{0}\) and \(U^{0}+W^{0} \subseteq (U \cap W)^{0}\), we can conclude that \((U \cap W)^{0} = U^{0}+W^{0}\).