91Ó°ÊÓ

Using bounded inverse theorem, prove the closed graph theorem. [Hint: Observe that, if \(A: X_{0} \subseteq X \rightarrow Y\) is a closed operator, where \(X\) and \(Y\) are Banach spaces, then \(A=\pi_{2} \pi_{1}^{-1}\), where \(\pi_{1}\) : \(G(A) \rightarrow X_{0}\) and \(\pi_{2}: X_{0} \times Y \rightarrow Y\) are defined by \(\pi_{1}(x, A x)=x\) \(\pi_{2}(x, y)=y\) for \(\left.x \in X_{0}, y \in Y .\right]\)

Let \(X\) be a Banach space with a Schauder basis \(\left\\{u_{1}, u_{2}, \ldots\right\\} \subseteq X\). For \(x \in X\), let \(\alpha_{j}(x) \in \mathbb{K}\) for \(j \in \mathbb{N}\), be such that \(x=\sum_{j=1}^{\infty} \alpha_{j} u_{j}\) Show that for each \(j \in \mathbb{N}\), the map \(f_{j}: X \rightarrow \mathbb{K}\) defined by \(f_{j}(x)=\) \(\alpha_{j}(x), x \in X\), is a continuous linear functional. [Hint: For \(x \in X\), let \(\|x\|_{*}=\sup _{n \in \mathbb{N}}\left\|\sum_{j=1}^{n} \alpha_{j} u_{j}\right\|\), where \(\left(\alpha_{j}\right)\) is such that \(x=\sum_{j=1}^{\infty} \alpha_{j} u_{j} \in X .\) Show that \(\|\cdot\|_{*}\) is a complete norm on \(X\), and then use the fact that \(\left|f_{j}(x)\right| \leq 2\|x\|_{*}\).]