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Chapter 2: Problem 36

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that \(T\) maps \(X\) onto a dense set in \(Y\) if and only if \(T^{*}\) maps \(Y^{*}\) one-to-one into \(X^{*}\). Also, if \(T^{*}\) maps onto a dense set, then \(T\) is one-to-one. Hint: If \(\overline{T(X)} \neq Y\), let \(f \in Y^{*} \backslash\\{0\\}\) be such that \(f=0\) on \(T(X)\). Then \(T^{*}(f)=0 .\) The other implications are straightforward.

Short Answer

Expert verified
T maps X onto a dense set in Y if and only if T* is injective. If T* maps onto a dense set, T is injective.

Step by step solution

01

Understand the Definitions

Ensure that you are clear about the definitions of Banach spaces, bounded linear operators, and the adjoint operator. A Banach space is a complete normed vector space. An operator \( T \in \mathcal{B}(X, Y) \) is a bounded linear operator from \(X\) to \(Y\). The adjoint operator \( T^{*} \) maps \( Y^{*} \) to \( X^{*} \).
02

Show T Maps X onto a Dense Set in Y

Show that if the closure of \( T(X) \), denoted by \( \overline{T(X)} \), is equal to \( Y \), then for any \( f \in Y^{*} \) such that \( f = 0 \) on \( T(X) \), it follows that \( f = 0 \). This implies that \( T^{*}(f) = 0 \) if and only if \( f = 0 \).
03

Using the Hint

Suppose \( \overline{T(X)} eq Y \). Then there exists a non-zero functional \( f \in Y^{*} \) such that \( f = 0 \) on \( T(X) \). Since \( f \) is not zero but \( T^{*}(f) = 0 \), this would imply that \( T^{*} \) is not one-to-one.
04

Complete the 'If' Direction

Conversely, if \( T^{*} \) is one-to-one, assume \( f \in Y^{*} \) and \( f eq 0 \). Then \( T^{*}(f) eq 0 \), which implies \( f eq 0 \) on \( T(X) \). Therefore, \( T(X) \) must be dense in \( Y \).
05

Show T is One-to-One if T* Maps Onto a Dense Set

Finally, if \( T^{*} \) maps onto a dense set, show that \( T \) is one-to-one. Suppose \( T(x) = 0 \). Then for any \( f \in Y^{*} \), we have \( \langle T^{*}(f), x \rangle = \langle f, T(x) \rangle = 0 \). If \( T^{*} \) maps onto a dense set, then dense functionals separate points, implying \( x = 0 \). Thus, \( T \) is one-to-one.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Banach Spaces
A Banach space is a complete normed vector space. This means that it is a vector space equipped with a norm, and any Cauchy sequence in this space converges to a limit within the space. Understanding Banach spaces is crucial in functional analysis because they provide a structure where we can apply techniques from analysis. Examples of Banach spaces include spaces of continuous functions, denoted as \(C([a,b])\), and the \(L^p\) spaces of integrable functions.

Since the space is complete, we can perform operations on limits of sequences without stepping outside the space. This property is significant because it allows for the study of infinite-dimensional spaces in a rigorous way.
Bounded Linear Operators
A bounded linear operator is a linear transformation between two normed vector spaces that maps bounded sets to bounded sets. In simpler terms, it means that the operator won't 'blow up' values and remains controlled in how it scales inputs. Formally, an operator \(T: X \to Y\) is bounded if there exists a constant \(C\) such that \(\|T(x)\|_Y \leq C \|x\|_X\) for all \(x \in X\).

Bounded linear operators are central in functional analysis because they ensure stability and continuity in transformations. This means that small changes in the input lead to small changes in the output, which is a desirable property in both theoretical and applied contexts.
Adjoint Operator
The adjoint operator is a concept that extends the idea of linear transformations to the dual spaces of the original vector spaces. If \(T\) is a bounded linear operator from \(X\) to \(Y\), then its adjoint operator, denoted \(T^*\), is a map from \(Y^*\) to \(X^*\). Here, \(X^*\) and \(Y^*\) denote the dual spaces of \(X\) and \(Y\), which consist of all bounded linear functionals on these spaces.

The adjoint operator is defined such that for every \(y^* \in Y^*\) and \(x \in X\), we have \( \langle T^*(y^*), x \rangle = \langle y^*, T(x) \rangle \). This condition ensures that the adjoint operator preserves the inner product structure between the original space and its dual. The properties of adjoint operators are key in analyzing the behavior of the original operators, such as determining if they are injective or surjective.
Dense Sets
A subset \(A\) of a topological space \(X\) is said to be dense in \(X\) if every point in \(X\) can be approximated arbitrarily closely by points from \(A\). In mathematical terms, the closure of \(A\), denoted by \(\overline{A}\), is equal to \(X\).

In the context of Banach spaces and bounded linear operators, if \(T: X \to Y\) maps \(X\) onto a dense set in \(Y\), it means that \(\overline{T(X)} = Y\). This property is crucial because it implies that \(T\) is 'almost' surjective; we can get as close as we want to any point in \(Y\) by using points from \(T(X)\). Understanding dense sets helps in grasping how operators behave in infinite-dimensional spaces, indicating their reach and limitations.

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Most popular questions from this chapter

Let \(X\) be a closed subspace of \(C[0,1]\) such that every element of \(X\) is a continuously differentiable function on \([0,1]\). Show that \(X\) is finitedimensional. Hint: Let \(T: X \rightarrow C[0,1]\) be defined for \(f \in X\) by \(T(f)=f^{\prime} .\) Show that the graph of \(T\) is closed: If \(f_{n} \rightarrow f\) uniformly and \(f_{n}^{\prime} \rightarrow g\) uniformly, then \(f^{\prime}=g .\) Therefore \(T\) is continuous by the closed graph theorem. Thus for some \(n \in \mathbf{N}\) we have \(\left\|f^{\prime}\right\|_{\infty} \leq n\) whenever \(f \in X\) satisfies \(\|f\|_{\infty} \leq 1\). Let \(x_{i}=\frac{i}{4 n}\) for \(i=0,1, \ldots, 4 n\). Define an operator \(S: X \rightarrow\) \(\mathbf{R}^{4 n+1}\) by \(S(f)=\left\\{f\left(x_{i}\right)\right\\} .\) We claim that \(S\) is one-to-one. It is enough to show that if \(\|f\|_{\infty}=1\), then for some \(i, S(f)\left(x_{i}\right) \neq 0 .\) Assume that this is not true. If \(f(x)=1\) and \(x \in\left(\frac{i}{4 n}, \frac{i+1}{4 n}\right)\), then by the Lagrange mean value theorem we have \(\left|f(x)-f\left(\frac{i}{4 n}\right)\right|=\left|f^{\prime}(\xi)\right|\left|x-\frac{i}{4 n}\right| \leq n \cdot \frac{1}{4 n}\), a contradiction. Therefore \(\operatorname{dim}(X) \leq 4 n+1\)

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

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

Let \(f\) be a linear functional on a Banach space \(X\). Show that \(f\) is continuous if and only if \(f^{-1}(0)\) is closed. Show that if \(f\) is not continuous, then \(f^{-1}(0)\) is dense in \(X\).

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that if \(Y\) is separable and \(T\) is onto \(Y\), then there is a separable closed subspace \(Z\) of \(X\) such that \(T(Z)=Y\).
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