/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 We have \(\|T\|=\left\|T^{*}\rig... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 43

We have \(\|T\|=\left\|T^{*}\right\|\) for a bounded linear operator on a Banach space, so if for a sequence of operators \(T_{n}\) we have \(\left\|T_{n}\right\| \rightarrow 0\), then \(\left\|T_{n}^{*}\right\| \rightarrow\) 0\. Find an example of a sequence of operators \(T_{n}\) on a Banach space \(X\) such that \(\left\|T_{n}(x)\right\| \rightarrow 0\) for every \(x \in X\) but it is not true that \(\left\|T_{n}^{*}\left(x^{*}\right)\right\| \rightarrow 0\) for every \(x^{*} \in X^{*}\). Hint: Let \(T_{n}(x)=\left(x_{n}, x_{n+1}, \ldots\right)\) in \(\ell_{2}\). Then \(T_{n}^{*}(x)=\left(0, \ldots, 0, x_{1}, x_{2}, \ldots\right)\), where \(x_{1}\) is on the \(n\) -th place.

Short Answer

Expert verified
The sequence \(T_n\) defined by \(T_n(x) = (x_n, x_{n+1}, \ldots)\) in \(\ell_2\) meets the criteria where \( T_n(x) \rightarrow 0\) but \( T_n^*(x^*) eq 0\) for some \(x^*\).

Step by step solution

01

Understand the given operators

We are given a sequence of operators, denoted by \(T_n\), acting on a Banach space \(X\), and its dual operators \(T_n^*\). The goal is to find an example where \(\left\|T_{n}(x)\right\| \rightarrow 0\) for every \(x \in X\) but \(\left\|T_{n}^{*}(x^{*})\right\| rightarrow 0\) for some \(x^{*} \in X^{*}\).
02

Utilize the given hint

Use the suggested operators defined by \(T_n(x) = (x_n, x_{n+1}, \ldots)\) in the sequence space \(\ell_2\), and its dual operators given by \(T_n^*(x) = (0, \ldots, 0, x_1, x_2, \ldots)\), where the first \(n\) entries are 0.
03

Verify \(\left\|T_{n}(x)\right\| \rightarrow 0\)

For a fixed \(x = (x_1, x_2, \ldots) \in \ell_2\), we have \(T_n(x) = (x_n, x_{n+1}, \ldots)\). Since the tail of any sequence in \(\ell_2\) converges to 0 as \(n \rightarrow \infty\), it follows that \(\left\|T_{n}(x)\right\| = \sqrt{x_n^2 + x_{n+1}^2 + \ldots} \rightarrow 0\).
04

Examine \(\left\|T_n^*(x^*)\right\|eq 0\) for some \(x^* \in X^*\)

Consider \(x^* = (1, 0, 0, \ldots) \in \ell_2\). Then, \(T_n^*(x^*) = (0, \ldots, 0, 1, 0, 0, \ldots)\), where \(1\) is at the nth position. Therefore, \(\left\|T_n^*(x^*)\right\| = 1\) for all \(n\), thus \(\left\|T_n^*(x^*) \right\| eq 0\) as \(n \rightarrow \infty\).

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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. This means that it is a vector space equipped with a norm, and every Cauchy sequence (a sequence where the elements get arbitrarily close to each other) in the space converges to a point within the space. The requirement of completeness ensures that the space is 'closed' under the limits of sequences of its elements. This property is crucial for analysis, as it guarantees the stability of certain mathematical operations. For example, if we have a sequence of functions that converge in a certain way, their limit will also belong to the Banach space.
Bounded Linear Operators
Bounded linear operators are linear transformations between Banach spaces that respect their bounded nature. Formally, an operator \(T\) is bounded if there exists a constant \(C\) such that for all \(x\) in the domain, \( \|T(x)\| \leq C \|x\| \). Boundedness ensures that the transformation does not excessively distort the input value. In the problem, the operator \(T_n(x) = (x_n, x_{n+1}, \, \ldots)\) is bounded because it takes inputs from the space \(\ell_2\) and produces outputs within that same space without making them disproportionately large.
Sequence of Operators
A sequence of operators involves applying a different operator from a sequence at each step. Here, we are given a sequence of operators \(T_n\) acting on a Banach space. The task is to find an example where the norm of \(T_n(x)\) goes to zero for every \(x\) in the space, but this is not true for the associated dual operators. The hint given guides us to use \(T_n(x) = (x_n, x_{n+1}, \, \ldots)\). This specific transformation filters out the beginning of the sequence, and as \(n\) increases, more initial terms are ignored. Because \(\ell_2\) sequences diminish towards zero over time, the norm \( \|T_n(x)\| \rightarrow 0\).
Dual Operators
Dual operators, denoted as \(T^*\), are operators associated with each bounded linear operator. They act on the dual space, which consists of all continuous linear functionals on the Banach space. In our example, the dual operator \(T_n^*\) was given by \(T_n^*(x) = (0, \, \ldots, \, 0, \, x_1, \, x_2, \, \, \ldots)\), placing zeros up to the \(n\)-th position. Despite \(T_n(x)\) converging to zero, \(\|T_n^*(x^*)\|\) does not necessarily converge for all elements in the dual space. This highlights a fundamental difference in behavior between an operator and its dual, showing that the convergence properties in the dual space can differ significantly from those in the original space.

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Most popular questions from this chapter

Let \(X, Y\) be normed spaces, \(T \in \mathcal{B}(X, Y)\). Consider \(\widehat{T}(\hat{x})=T(x)\) as an operator from \(X / \operatorname{Ker}(T)\) into \(\overline{T(X)}\). Then we get \(\widehat{T}^{*}: \overline{T(X)}^{*} \rightarrow\) \((X / \operatorname{Ker}(T))^{*} .\) Using Proposition \(2.7\) and \(\overline{T(X)}^{\perp}=T(X)^{\perp}=\operatorname{Ker}\left(T^{*}\right)\), we may assume that \(\widehat{T}^{*}\) is a bounded linear operator from \(Y^{*} / \operatorname{Ker}\left(T^{*}\right)\) into \(\operatorname{Ker}(T)^{\perp} \subset X^{*} .\) On the other hand, for \(T^{*}: Y^{*} \rightarrow X^{*}\) we may consider \(\widehat{T^{*}}: Y^{*} / \operatorname{Ker}\left(T^{*}\right) \rightarrow X^{*}\). Show that \(\widehat{T}^{*}=\widehat{T^{*}}\) Hint: Take any \(\hat{y} \in Y^{*} / \operatorname{Ker}\left(T^{*}\right)\) and \(x \in X\). Then, using the above identifications, we obtain $$ \widehat{T}^{*}\left(\widehat{y^{*}}\right)(\hat{x})=\widehat{y^{*}}(\widehat{T}(\hat{x}))=y^{*}(T(x))=T^{*}\left(y^{*}\right)(x)=\widehat{T^{*}}\left(\hat{y^{*}}\right)(\hat{x}) $$

Show that if \(X\) is a finite-dimensional Banach space, then every linear functional \(f\) on \(X\) is continuous on \(X\).

Let \(X\) be a Banach space. Show that: (i) \(\overline{\operatorname{span}}(A)=\left(A^{\perp}\right)_{\perp}\) for \(A \subset X\). (ii) \(\operatorname{span}(B) \subset\left(B_{\perp}\right)^{\perp}\) for \(B \subset X^{*}\). Note that in general we cannot put equality. (iii) \(A^{\perp}=\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp}\) for \(A \subset X\) and \(B_{\perp}=\left(\left(B_{\perp}\right)^{\perp}\right)_{\perp}\) for \(B \subset X^{*}\). Hint: (i): Using the definition, show that \(A \subset\left(A^{\perp}\right)_{\perp}\). Then use that \(B_{\perp}\) is a closed subspace for any \(B \subset X^{*}\), proving that \(\frac{\overline{\operatorname{span}}}(A) \subset\left(A^{\perp}\right)_{\perp}\). Take any \(x \notin \overline{\operatorname{span}}(A)\). Since \(\overline{\operatorname{span}}(A)\) is a closed subspace, by the separation \(\left.f\right|_{A}=0\), hence \(f \in A^{\perp}\); also \(f(x)>0\), so \(x \notin\left(A^{\perp}\right)_{\perp}\) (ii): Similar to (i). (iii): Applying (i) to \(A^{\perp}\), we get \(A^{\perp} \subset\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp} .\) On the other hand, using \(A \subset\left(A^{\perp}\right)_{\perp}\) and the previous exercise, we get \(\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp} \subset A^{\perp} .\) The dual statement is proved in the same way.

We proved the closed graph theorem using the open mapping theorem. Now prove the open mapping principle using the closed graph theorem. Hint: First, prove it for one-to-one maps using the fact that \(\left\\{\left(y, T^{-1}(y)\right)\right\\}\) is closed. For a general case, note that the quotient map is an open map by the definition of the quotient topology.

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.
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