Question

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, \(T(X)=\overline{T(X)}\)

Short answer

(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 \(T^{*}(Y^{*})\) is closed in \(X^{*}\).

Step-by-step solution

- Understand the properties

We need to show three properties involving linear operators and their adjoints between Banach spaces. Banach spaces, denoted as \(X\) and \(Y\), are complete normed vector spaces. Linear operators between these spaces are denoted as \(T\) and their adjoints as \(T^{*}\).

- Property (i): \(T^{*}\) is onto implies \(T\) is an isomorphism into \(Y\)

If \(T^{*}\) is onto, it is an open map. Hence, there exists \(\delta > 0\) such that \(\delta B_{X^{*}} \subseteq T^{*}(B_{Y^{*}})\). For any \(x \in X\):\[\|T(x)\|_{Y} = \sup_{y^{*} \in B_{Y^{*}}} T^{*}(y^{*})(x) = \sup_{x^{*} \in T^{*}(B_{Y^{*}})} x^{*}(x) \geq \sup_{x^{*} \in \delta B_{X^{*}}} x^{*}(x) = \delta \|x\|_{X}\]Thus, \(T\) is bounded below and hence an isomorphism.

- Property (i): \(T\) is an isomorphism implies \(T^{*}\) is onto

Since \(T\) is an isomorphism into \(Y\), \(T^{-1}\) is a bounded linear operator from \(T(X)\) into \(X\). For any \(x^{*} \in X^{*}\), define \(y^{*}\) on \(T(X)\) by \(y^{*}(T(x)) = x^{*}(x)\). Extend \(y^{*}\) to a functional in \(Y^{*}\), yielding \(T^{*}(y^{*}) = x^{*}\). Hence, \(T^{*}\) is onto.

- Property (ii): \(T\) is onto implies \(T^{*}\) is an isomorphism into \(X^{*}\)

If \(T\) is onto, \(T(B_{X})\) includes \(\delta B_{Y}\) for some \(\delta > 0\). Thus, for any \(y^{*} \in Y^{*}\):\[\|T^{*}(y^{*})\|_{X^{*}} \geq \delta \|y^{*}\|_{Y^{*}}\]Hence, \(T^{*}\) is bounded below and is an isomorphism.

- Property (ii): \(T^{*}\) is an isomorphism implies \(T\) is onto

Assume \(T^{*}\) is an isomorphism into. We need \(\delta > 0\) such that \(\delta B_{Y} \subseteq \overline{T(B_{X})}\). Assuming \( \delta\) does not exist, find \(y_n \rightarrow 0\) but \(y_n otin \overline{T(B_{X})}\). Dist \(y_n\) from closure is non-zero; by convexity and separation, find \(y^{*} \in Y^{*}\) s.t. \(\|T^{*}(y^{*})\| \leq 1\) but contradicting small distances. Therefore, \(\delta B_{Y}\) must exist.

- Property (iii): \(T(X)\) is closed implies \(T^{*}(Y^{*})\) is closed

If \(T(X)\) is closed, \(\widehat{T}\) maps \(X / \operatorname{Ker}(T)\) onto \(T(X)\), which is a Banach space. Hence, \(\widehat{T}^{*}\) is an isomorphism and \(\widehat{T}^{*} (Y^{*} / \operatorname{Ker}(T^{*}))\) is closed. Then, \(T^{*} (Y^{*})\) is closed.

- Property (iii): \(T^{*}(Y^{*})\) is closed implies \(T(X)\) is closed

If \(T^{*}(Y^{*})\) is closed, use \(\widehat{T}: X \rightarrow \overline{T(X)}\). Then \(\widehat{T}^{*}(Y^{*} / \operatorname{Ker}(T^{*})) = T^{*}(Y^{*})\) is closed, making \(\widehat{T}^{*}\) one-to-one and thus an isomorphism. Hence, \(\widehat{T}\) is onto denoting \(T(X) = \overline{T(X)}\).

Key concepts

Linear Operators

A linear operator is a fundamental concept in functional analysis and linear algebra. Given two vector spaces, say Banach spaces (which are complete normed spaces) like X and Y, a linear operator T maps elements from X to Y while preserving the operations of addition and scalar multiplication. Mathematically, if T: X → Y, then for any vectors u, v in X and any scalar α, we have

* T(u + v) = T(u) + T(v)
* T(αu) = αT(u)

The study of linear operators is crucial as they help us understand the structure of transformations between different vector spaces. In our exercise, T is such an operator, and its properties (like being onto or an isomorphism) play a central role in determining the behavior of its adjoint operator T*.

Isomorphism

An isomorphism is a special type of linear operator that provides a perfect 'one-to-one' correspondence between two vector spaces. An operator T: X → Y is called an isomorphism if it is both injective (one-to-one) and surjective (onto). This implies that

* T has an inverse, denoted T⁻¹, which is also a linear operator.
* The spaces X and Y are structurally similar or 'isomorphic.' This means their elements and operations correspond in a way that preserves their algebraic and topological properties.

In our exercise, we examine conditions under which T or its adjoint T* are isomorphisms. For example, if T* is onto, then we can show that T is an isomorphism from X into Y. Understanding these relationships helps us generalize results across different Banach spaces.

Adjoint Operator

The adjoint operator, denoted T*, is associated with a given linear operator T and is defined in the context of dual spaces. For an operator T: X → Y, the adjoint operator T* maps elements from the dual space Y* back to the dual space X*. The dual space Y* consists of all continuous linear functionals on Y. The adjoint operator is defined by:

T*(y*)(x) = y*(T(x))

for all y* ∈ Y* and x ∈ X. Essentially, T* captures how the operator T 'transports' functionals backwards from Y to X. The properties of T* are critical in understanding the deeper structure of T. In this exercise, T* helps us explore surjectivity and the closedness of image spaces, thus connecting operator properties in X and Y to those in their respective dual spaces X* and Y*.

Closed Subspaces

In the realm of linear operators, the concept of closed subspaces is essential. A subspace M of a Banach space Y is called closed if it contains all its limit points. In other words, if a sequence {y_n} in M converges to some y in Y, then y must also be in M. Closed subspaces ensure that certain operations, like limits, do not