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Chapter 3: Problem 65

Let \(X\) be a reflexive Banach space, and let \(T\) be a bounded linear operator from \(X\) onto a Banach space \(Y\). Show that \(Y\) is reflexive.

Short Answer

Expert verified
Y is reflexive because the closed range theorem and the surjectivity of T* imply that Y* is reflexive, hence Y is reflexive as well.

Step by step solution

01

Understand Reflexive Spaces

A Banach space X is reflexive if the natural embedding of X into its double dual X** is surjective.
02

Given Information

We are given that X is a reflexive Banach space and that T is a bounded linear operator from X onto a Banach space Y.
03

Use Bounded Linear Operator Property

Since T is a bounded linear operator from X onto Y, for every y in Y, there exists an x in X such that T(x) = y.
04

Analyze Reflexivity and Surjectivity

Given that X is reflexive, there is an isomorphism between X and its double dual X**. Also, T being surjective implies that T has a closed range and maps elements densely in Y.
05

Apply the Closed Range Theorem

The Closed Range Theorem ensures that T* (the adjoint operator of T) maps Y* surjectively onto X*. Since X is reflexive, X** is isomorphic to X, making X* reflexive as well.
06

Show that Y* is Reflexive

Since T* maps Y* onto X* and X* is reflexive, the surjectivity of T* implies that Y* is reflexive.
07

Conclude that Y is Reflexive

Since the dual space Y* is reflexive, the space Y itself must also be reflexive as reflexivity is preserved under dualization.

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Key Concepts

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

Banach space
A Banach space is a complete, normed vector space. Complete means that every Cauchy sequence (a sequence where the elements become arbitrarily close to each other) has a limit that is within the space. The norm provides a way to measure the size or length of vectors in this space.

In simple terms, a Banach space is like a robust playground where every sequence of children playing together eventually gathers into a single group that stays within the playground. This is an important concept in functional analysis because it ensures that various limits and operations we perform will stay within the space.
Bounded linear operator
A bounded linear operator is a function between two normed vector spaces that preserves vector addition and scalar multiplication, and it does not increase the size of vectors by more than a fixed multiple.

Mathematically, if we have a bounded linear operator T from a space X to a space Y, then for all vectors x in X, there exists a constant C such that \(orm{T(x)} \le C orm{x} \). This means that T behaves nicely when scaling vectors鈥攏o exaggeration in lengths happens.

In easier terms, think of a bounded linear operator like a gentle hand that moves objects from one place to another without stretching or changing their shape excessively. It ensures that the essence of the vector, its direction, and proportion are maintained.
Reflexivity
Reflexivity is a property of a Banach space where the space is naturally isomorphic to its double dual. The double dual of a space X is the dual of its dual space \((X^*)^*\). An isomorphism means there is a one-to-one correspondence that preserves the structure between the original space and its double dual.

In a reflexive space, every functional (element of the dual space) corresponds uniquely to an element of the space itself and vice versa. This beautiful symmetry ensures that many functional analysis problems become more manageable because the space and its dual share the same properties.

When we refer to reflexivity in the context of the exercise, it helps to understand why Y must be reflexive if X is reflexive and T is a bounded linear operator. Here, we leverage properties like surjectivity (every element in Y can be mapped from X) and duality principles to show that the reflexive nature is preserved even under mappings by T.

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Most popular questions from this chapter

Let \(X\) be an infinite-dimensional Banach space. Show that \(S_{X}\) is a dense \(G_{\delta}\) set in \(\left(B_{X}, w\right)\). Deduce from this that the norm of an infinitedimensional space is never a weakly continuous function on \(X\).

Show that every Banach space \(X\) in its weak topology is a completely regular space; that is, if \(p \notin C\), where \(C\) is a weakly closed set, then there is a weakly continuous function \(f\) on \(X\) such that \(f(p)=1\) and \(f(x)=0\) for every \(x \in C\)

Define a functional \(f\) on \(\ell_{1}\) by \(f(x)=\sum x_{i}\) for \(x=\left(x_{i}\right)\). Show that \(f\) is norm continuous on \(\ell_{1}\) and that \(f\) is not \(\overline{w^{*}}\) -continuous on \(\ell_{1} \cong c_{0}^{*}\) (that is, \(\left.f \in \ell_{\infty} \backslash c_{0}\right)\)

Show that \(\overline{\operatorname{conv}}\left(\operatorname{Ext}\left(B_{C[0,1]^{*}}\right)\right) \neq B_{C[0,1]^{*}}\) Hint: Show that the Lebesgue measure \(\lambda\) is not in \(\overline{\operatorname{conv}}\left(\operatorname{Ext}\left(B_{C[0,1] *}\right)\right)\). Indeed, if \(\lambda\) is norm-close to \(\sum \alpha_{i} \delta_{t_{i}}\), find \(f \in B_{C[0,1]}\) such that \(f\left(t_{i}\right)=0\) but \(\lambda(f)\) is close to 1 .

Show that there exists a sequence \(\left\\{f_{n}\right\\}_{n=1}^{\infty}\) in the dual of some normed linear space \(X\) such that \(\left\\{f_{n}(x)\right\\}_{n=1}^{\infty}\) is bounded for every \(x \in X\) and yet \(\left\\{\left\|f_{n}\right\|\right\\}\) is unbounded. This shows that the assumption of completeness of \(X\) cannot be dropped in the Banach-Steinhaus theorem.
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