Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that:
(i) \(T^{*}\) is onto if and only if \(T\) is an isomorphism into \(Y\).
(ii) \(T\) is onto if and only if \(T^{*}\) is an isomorphism into \(X^{*}\).
(iii) \(T(X)\) is closed in \(Y\) if and only if \(T^{*}\left(Y^{*}\right)\) is
closed in \(X^{*}\).
Hint: (i): If \(T^{*}\) is onto, it is an open map (Theorem 2.24), and by
Exercise \(2.29\), there is \(\delta>0\) so that \(\delta B_{X} * \subset
T^{*}\left(B_{Y}\right)\). Then
$$
\begin{aligned}
\|T(x)\|_{Y} &=\sup _{y^{*} \in B_{Y} .} y^{*}(T(x))=\sup _{y^{*} \in B_{Y} *}
T^{*}\left(y^{*}\right)(x)=\sup _{x^{*} \in T^{*}\left(B_{\gamma}
*\right)}\left(x^{*}(x)\right) \\
& \geq \sup _{x^{*} \in \delta B_{x} *}\left(x^{*}(x)\right)=\delta\|x\|_{X}
\end{aligned}
$$
and use Exercise \(1.27\). If \(T\) is an isomorphism into, then \(T^{-1}\) is a
bounded linear operator from \(T(X)\) into \(X\). Given \(x^{*} \in X^{*}\), define
\(y^{*}\) on \(T(X)\) by \(y^{*}(y)=x^{*}\left(T^{-1}(y)\right)\). Clearly, \(y^{*}
\in T(X)^{*}\); extend it to a functional in \(Y^{*}\). Then
\(T^{*}\left(y^{*}\right)=x^{*}\).
(ii): If \(T\) is onto, as in (i) we find \(\delta>0\) such that \(\delta B_{Y}
\subset T\left(B_{X}\right)\); then \(\left\|T^{*}\left(y^{*}\right)\right\|_{X
*} \geq \delta\left\|y^{*}\right\|_{Y *}\) and use Exercise \(1.27\).
Assume \(T^{*}\) is an isomorphism into. By Exercise \(2.29\) and Lemma \(2.23\), it
is enough to find \(\delta>0\) so that \(\delta B_{Y} \subset
\overline{T\left(B_{X}\right)}\). Assume by contradiction that no such \(\delta\)
exists. Then find \(y_{n} \rightarrow 0\) such that \(y_{n} \notin
\overline{T\left(B_{X}\right)}\). The set is closed, so
\(d_{n}=\operatorname{dist}\left(y_{n}, \overline{T\left(B_{X}\right)}\right)>0
.\)
Fix \(n_{1}\) and set \(V_{n}=\bigcup_{y \in
T\left(B_{x}\right)}\left(y+B_{Y}^{O}\left(\frac{d_{n}}{2}\right)\right)\).
Then \(V_{n}\) is an open convex
set and \(y_{n} \notin V_{n}\), so by Corollary \(2.13\) there is \(y^{*} \in
Y^{*}\) such that \(\left|y^{*}\right|<1\) on \(V_{n}\) and
\(y^{*}\left(y_{n}\right)=1\). Since \(T\left(B_{X}\right) \subset V_{n}\), we get
$$
\left\|T^{*}\left(y^{*}\right)\right\|=\sup _{x \in B_{X}}
T^{*}\left(y^{*}\right)(x)=\sup _{x \in B_{X}} y^{*}(T(x))=\sup _{y \in
T\left(B_{X}\right)}\left(y^{*}(y)\right) \leq 1,
$$
so \(\left\|y^{*}\right\|
\leq\left\|\left(T^{*}\right)^{-1}\right\|\left\|T^{*}\left(y^{*}\right)\right\|
\leq\left\|\left(T^{*}\right)^{-1}\right\|, 1=y^{*}\left(y_{n}\right)
\leq\left\|\left(T^{*}\right)^{-1}\right\|\left\|y_{n}\right\|\).
This shows that \(\left\|y_{n}\right\| \geq 1
/\left\|\left(T^{*}\right)^{-1}\right\|\) for every \(n\), contradicting \(y_{n}
\rightarrow 0\).
(iii): If \(T(X)\) is closed, then \(\widehat{T}\) is an operator from \(X /
\operatorname{Ker}(T)\) onto a Banach space \(T(X)\), hence by (ii),
\(\widehat{T}^{*}\) is an isomorphism into, and in particular
\(\widehat{T}^{*}\left(Y^{*} / \operatorname{Ker}\left(T^{*}\right)\right)\) is
closed. By Exercise \(2.35, \widehat{T^{*}}\left(Y^{*} /
\operatorname{Ker}\left(T^{*}\right)\right)=T^{*}\left(Y^{*}\right)\)
is closed. If \(T^{*}\left(Y^{*}\right)\) is closed, consider \(\widehat{T}: X
\rightarrow \overline{T(X)}\). Then \(\widehat{T}^{*}\left(Y^{*} /
\operatorname{Ker}\left(T^{*}\right)\right)=\)
\(\widehat{T^{*}}\left(Y^{*} /
\operatorname{Ker}\left(T^{*}\right)\right)=T^{*}\left(Y^{*}\right)\) is closed
and \(\widehat{T}^{*}\) is one-to-one; hence it is an
isomorphism into. By (ii), \(\widehat{T}\) must be onto, that is,
\(T(X)=\overline{T(X)}\)
