91Ó°ÊÓ

Show that a normed space \(X\) is a Banach space if and only if \(\sum y_{n}\) converges whenever \(\left\|y_{n}\right\| \leq 2^{-n}\) for every \(n\). Hint: Use Lemma 1.15. Note that if \(\sum x_{k}\) is absolutely convergent, there are \(N_{n}\) so that \(\sum_{k>N_{n}}\left\|x_{k}\right\| \leq 2^{-k} .\) Set \(y_{n}=\sum_{k=N_{n}+1}^{N_{k+1}} x_{k}\), then \(\left\|y_{n}\right\| \leq 2^{-n}\), and if \(M>N_{m}\), then \(\left\|\sum_{n=1}^{m} y_{n}-\sum_{k=1}^{M} x_{k}\right\| \leq 2^{-m}\).

Let \(Y\) be a closed subspace of a normed space \(X\). Show that if \(Y\) and \(X / Y\) are both Banach spaces, then \(X\) is a Banach space. Note: A property \(\mathcal{P}\) is said to be a three-space property if the following holds: Let \(Y\) be a closed subspace of a space \(X\). If \(Y\) and \(X / Y\) have \(\mathcal{P}\), then \(X\) has \(\mathcal{P}\) (see, e.g., [CaGo]). Thus, the property of being complete is a three-space property in the class of normed linear spaces. Hint: If \(\left\\{x_{n}\right\\}\) is Cauchy in \(X\), there is \(x \in X\) such that \(\hat{x}_{n} \rightarrow \hat{x}\). There are \(\left\\{y_{n}\right\\}\) in \(Y\) such that \(\left\\{x_{n}-x-y_{n}\right\\} \rightarrow 0\). Thus \(\left\\{y_{n}\right\\}\) is Cauchy, so \(y_{n} \rightarrow y\) and \(x_{n} \rightarrow x+y\).

Let \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) be two equivalent norms on a vector space \(X\). Let \(B_{1}\) and \(B_{2}\) be the closed unit balls of \(\left(X,\|\cdot\|_{1}\right)\) and \(\left(X,\|\cdot\|_{2}\right)\), respectively. Show that \(B_{1}\) and \(B_{2}\) are homeomorphic. Recall that two topological spaces \(K\) and \(L\) are called homeomorphic if there exists a bijection \(\varphi\) from \(K\) onto \(L\) such that \(\varphi\) and \(\varphi^{-1}\) are continuous. Such a \(\varphi\) is called a homeomorphism. Hint: Define a map \(\phi\) from \(B_{1}\) onto \(B_{2}\) by \(\phi(0)=0\) and \(\phi(x)=\frac{\|x\|_{1}}{\|x\|_{2}} x\) for \(x \in B_{1} \backslash\\{0\\} .\) Clearly, \(\|\phi(x)\|_{2}=\|x\|_{1}\), and continuity at 0 follows from the equivalence of the norms.

Let \(X\) be a Banach space and \(G\) be a subspace of \(X\) that is a \(G_{\delta}\) set in \(X\). Show that \(G\) is closed in \(X\). Hint: Let \(S=\bar{G} \backslash G\), where \(\bar{G}\) denotes the closure of \(G .\) Since \(G\) is a \(G_{\delta}\) set in \(X, G\) is \(G_{\delta}\) in \(\bar{G}\). Hence \(G=\bigcap G_{n}\), where \(G_{n}\) are open subsets in \(\bar{G}\). Therefore, \(S=\bigcup\left(\bar{G} \backslash G_{n}\right)\) and each \(G_{n}\) is dense in \(\bar{G}\) because it contains \(G .\) Thus, each \(\bar{G} \backslash G_{n}\) is nowhere dense in \(\bar{G}\), so \(S\) is of first category in \(\bar{G}\). We will show that \(S\) is an empty set. If \(x_{0} \in S\), consider the set \(G^{*}=\left\\{x_{0}+x ; x \in G\right\\}\). Note that \(G^{*} \subset S\). Indeed, any point \(x_{0}+x, x \in G\), is in \(\bar{G}\) since \(\bar{G}\) is a linear set. If for some \(x \in G\) we have \(x_{0}+x \in G\), then by linearity of \(G\) we have \(x_{0} \in G_{1}\) a contradiction. Therefore, \(G^{*} \subset S\) and \(G^{*}\) is of first category in \(\bar{G} .\) Thus, the shift \(G\) of \(G^{*}\) is also of first category in \(\bar{G}\). Hence, the whole \(\bar{G}\) as a union of \(G\) and \(S\), which are both of first category in \(\bar{G}\), is of first category in itself. This is a contradiction, since \(\bar{G}\) is a complete metric space.

Let \(X\) be a normed linear space \(X\). Show that if \(X\) is topologically complete in its norm topology (that is, \(X\) is homeomorphic to a complete metric space), then \(X\) is a Banach space. Hint: If \(X\) is topologically complete, then by the Alexandrov theorem, \(X\) is \(G_{\delta}\) in the completion of \(X\), which is a Banach space. Note that this means that a non-complete normed space cannot be homeomorphic to a Banach space.

Find a vector space \(X\) with two norms on it such that both of them are complete norms and they are not equivalent. Hint: Take a vector space \(V\) of linear dimension \(c\), and let \(T_{1}\) and \(T_{2}\) be linear bijections of \(V\) onto \(\ell_{2}\) and \(\ell_{4}\), respectively. Define norms on \(V\) by \(\|x\|_{1}=\left\|T_{1}(x)\right\|_{2}\) and \(\|x\|_{2}=\left\|T_{2}(x)\right\|_{4}\). Then \(\left(V_{1}\|\cdot\|_{1}\right)\) is isomorphic to \(\ell_{2}\) and \(\left(V_{i}\|\cdot\|_{2}\right)\) is isomorphic to \(\ell_{4} .\) Since \(\ell_{2}\) is not isomorphic to \(\ell_{4}\) (see exercises above), \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) are not equivalent.