Chapter 13: Problem 10

If \(S\) is any subset of \(\mathcal{H}\), and \(V\) the closed subspace generated by \(S, V=\overline{L(S)}\), show that \(S^{1}=\left\\{u \in \mathcal{H} \mid\\{u|x\rangle=0\right.\) for all \(x \in S\\}=V^{1}\)

Short Answer

Expert verified

The orthocomplement of \(S\), denoted by \(S^{1}\), is in fact equal to the orthocomplement of \(V\), \(V^{1}\), where \(V\) is the closed subspace generated by \(S\). This is shown by first proving that \(S^{1} \subseteq V^{1}\) and then that \(V^{1} \subseteq S^{1}\), which can then be used to conclude that \(S^{1} = V^{1}\).

Step by step solution

Show that \(S^{1} \subseteq V^{1}\)

Start by assuming that \(u \in S^{1}\). This means that for every \(x \in S\), \(u, x = 0\), where ',' represents the inner product. Hence, \langle u, x \rangle = 0 for all \(x \in S\). Because every \(x \in S\) is also in \(V\) (since \(V\) is generated by \(S\)), then \langle u, x \rangle = 0 holds for all \(x \in V\). Therefore, \(u \in V^{1}\), effectively showing that \(S^{1} \subseteq V^{1}\).

Show that \(V^{1} \subseteq S^{1}\)

Again, start by assuming that \(u \in V^{1}\). This means that \langle u, x \rangle = 0 for all \(x \in V\). Now take a sequence \(x_n \in L(S)\) such that \(x_n\) converges to \(x\) in \(V\). Since \(x_n \rightarrow x\), and \(u \in V^{1}\) (meaning that \langle u, x_n \rangle -> 0), it must also be the case that \langle u, x \rangle = 0 for all \(x \in V\), meaning that \(u \in S^{1}\). This shows that \(V^{1} \subseteq S^{1}\).

Conclude that \(S^{1} = V^{1}\)

Having shown in steps 1 and 2 that \(S^{1} \subseteq V^{1}\) and \(V^{1} \subseteq S^{1}\), it can now be concluded that \(S^{1} = V^{1}\), which is the outcome the exercise requested.