91Ó°ÊÓ

Let \(Y\) be a subspace of a Banach space \(X\). Show that there exists a one-to- one (in general, nonlinear) isometric map \(\varphi: Y^{*} \rightarrow X^{*}\). Also, \(\left.X^{*}\right|_{Y}=Y^{*}\left(\left.X^{*}\right|_{Y}\right.\) is the set of restrictions to \(Y\) of all \(\left.f \in X^{*}\right)\). Hint: Use the Hahn-Banach theorem to extend functionals on \(Y\).

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that if \(T\) maps bounded closed sets in \(X\) onto closed sets in \(Y\), then \(T(X)\) is closed in \(Y\). Hint: Assume \(T\left(x_{n}\right) \rightarrow y \notin T(X)\). Put \(M=\operatorname{Ker}(T)\), set \(d_{n}=\operatorname{dist}\left(x_{n}, M\right)\) and find \(w_{n} \in M\) such that \(d_{n} \leq\left\|x_{n}-w_{n}\right\| \leq 2 d_{n} .\) If \(\left\\{x_{n}-w_{n}\right\\}\) is bounded, then \(T\left(x_{n}-w_{n}\right) \rightarrow y \in T(X)\), since the closure of \(\left\\{x_{n}-w_{n}\right\\}\) is mapped onto a closed set containing \(y\). If \(\left\|x_{n}-w_{n}\right\| \rightarrow \infty\), then since \(T\left(x_{n}-w_{n}\right) \rightarrow y\), we have \(T\left(\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}\right) \rightarrow 0 .\) By the hypothesis, \(M\) must contain a point \(w\) from the closure of \(\left\\{\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}\right\\}\) since 0 lies in the closure of the image of this sequence. Fix \(n\) so that \(\left\|\frac{x_{n}-w_{n}}{\left\|x_{n}-w_{n}\right\|}-w\right\|<1 / 3\). Then \(\left\|x_{n}-w_{n}-\right\| x_{n}-w_{n}\|w\| \leq \frac{1}{3}\left\|x_{n}-w_{n}\right\|<(2 / 3) d_{n}\) and \(w_{n}+\left\|x_{n}-w_{n}\right\| w \in M\) a contradiction.

Show that \(c_{0}\) is not isomorphic to \(C[0,1]\). Hint: Check the separability of their duals.