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Chapter 2: Problem 27

Let \(X, Y\) be normed spaces, \(T \in \mathcal{B}(X, Y)\). Show that \(\widehat{T}: X / \operatorname{Ker}(T) \rightarrow\) \(Y\) defined by \(\widehat{T}(\hat{x})=T(x)\) is a bounded linear operator onto \(T(X)\).

Short Answer

Expert verified
\( \widehat{T} \) is a well-defined, bounded, linear operator that maps onto \( T(X) \).

Step by step solution

01

Define the map \( \, \hat{T} \, \)

The operator \( T \, : \, X \rightarrow Y \) is given, and the kernel of \( T \) is \( \operatorname{Ker}(T) \). Define \( \widehat{T} : X / \operatorname{Ker}(T) \rightarrow Y \) by \( \widehat{T}(\hat{x}) = T(x) \), where \( \hat{x} \) is the equivalence class in \( X / \operatorname{Ker}(T) \) corresponding to \( x \).
02

Show that \( \widehat{T} \) is well-defined

To ensure \( \widehat{T} \) is well-defined, check that if \( x_1 \sim x_2 \) (meaning \( x_1 - x_2 \in \operatorname{Ker}(T) \)), then \( T(x_1) = T(x_2) \). Since \( T \) annihilates elements in its kernel, \( T(x_1 - x_2) = 0 \) implies \( T(x_1) = T(x_2) \). Thus, \( \widehat{T} \) is well-defined.
03

Prove that \( \widehat{T} \) is linear

To show linearity, let \( \hat{u}, \hat{v} \in X / \operatorname{Ker}(T) \) and \( \alpha, \beta \in \mathbb{R} \). We need to show \( \widehat{T}(\alpha \hat{u} + \beta \hat{v}) = \alpha \widehat{T}(\hat{u}) + \beta \widehat{T}(\hat{v}) \). Using the definition, we have \( \alpha \hat{u} + \beta \hat{v} = \hat{(\alpha u + \beta v)} \). Then, \( \widehat{T}(\alpha \hat{u} + \beta \hat{v}) \) equals \( T(\alpha u + \beta v) = \alpha T(u) + \beta T(v) = \alpha \widehat{T}(\hat{u}) + \beta \widehat{T}(\hat{v}) \). Hence, \( \widehat{T} \) is linear.
04

Check that \( \widehat{T} \) is bounded

\( \widehat{T} \) being bounded means there exists a constant \( C > 0 \) such that \( \| \widehat{T}(\hat{x}) \|_Y \leq C \| \hat{x} \|_{X/\operatorname{Ker}(T)} \) for all \( \hat{x} \). For \( \hat{x} \in X / \operatorname{Ker}(T) \), choose a representative \( x \in \hat{x} \). We know \( \| T(x) \|_Y \leq \| T \| \| x \|_X \). Define \( C = \| T \| \), then \( \| \widehat{T}(\hat{x}) \|_Y = \| T(x) \|_Y \leq \| T \| \| x \|_X \). Thus, \( \widehat{T} \) is bounded with bound \( \| T \| \).
05

Show \( \widehat{T} \) maps onto \( T(X) \)

Any \( y \in T(X) \) can be written as \( y = T(x) \) for some \( x \in X \). The element \( \hat{x} \in X / \operatorname{Ker}(T) \) maps to \( \widehat{T}(\hat{x}) = T(x) = y \). Since every element of \( T(X) \) has a preimage in \( X / \operatorname{Ker}(T) \) under \( \widehat{T} \), \( \widehat{T} \) maps onto \( T(X) \).

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Key Concepts

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

Normed Spaces
Normed spaces are foundational in functional analysis and are an extension of vector spaces. They come equipped with a norm, which is a function that assigns a non-negative length or size to each vector. The norm satisfies three key properties: the triangle inequality, absolute scalability, and it is zero only for the zero vector. To put it simply, a normed space allows us to measure the 'distance' between elements, making it easier to analyze and compare them.
In mathematical notation, if \(X\) is a normed space with norm \(\| \cdot \|\), for any \(x, y \in X\) and any scalar \(\alpha\), the following properties hold:
  • Triangle Inequality: \(\|x + y\| \leq \|x\| + \|y\|\)
  • Absolute scalability: \(\|\alpha x\| = |\alpha| \cdot \|x\|\)
  • Identity: \(\|x\| = 0\) if and only if \(x = 0\)
Understanding these properties is crucial when dealing with linear mappings and bounded operators, as these concepts often rely on norms to define and analyze their characteristics.
Kernel of a Linear Operator
The kernel (or null space) of a linear operator \(T\) from normed space \(X\) to \(Y\) is a fundamental concept. It consists of all vectors in \(X\) that are mapped to the zero vector in \(Y\) under \(T\). Mathematically, it is defined as \(\operatorname{Ker}(T) = \{ x \in X : T(x) = 0 \}\).
The kernel helps analyze the behavior of \(T\). If the kernel is just the zero vector, \(T\) is injective (one-to-one), meaning it maps distinct vectors in \(X\) to distinct vectors in \(Y\). However, if there are non-zero vectors in the kernel, these vectors are essentially 'lost' as they all map to zero in \(Y\).
This concept plays a crucial role in defining quotient spaces \(X / \operatorname{Ker}(T)\), which are used to 'collapse' the kernel to a single point. As seen in the exercise and solution given in the prompt, \(\widehat{T}\) is defined on this quotient space, making it crucial to understand the kernel to follow through with the exercise.
Linear Mappings
Linear mappings (or linear transformations) between normed spaces are functions that preserve vector addition and scalar multiplication. If \(T: X \to Y\) is a linear operator, for any \(u, v \in X\) and scalars \(\alpha, \beta\), the following must hold:
  • Additivity: \(T(u + v) = T(u) + T(v)\)
  • Homogeneity: \(T(\alpha u) = \alpha T(u)\)
Linear mappings are crucial in many areas of mathematics because they maintain the structure of the space during transformations. We use them to model various physical systems and processes. When defining a new operator \(\widehat{T}\), as seen in the exercise, proving linearity ensures that we preserve the vector space structure when moving between spaces.
Moreover, linear mappings are the backbone of many advanced topics, such as eigenvalues and eigenvectors, linear algebraic equations, and the study of various operators in functional analysis. Understanding their properties helps simplify complex problems by ensuring the transformations do not distort the linear structure of the underlying spaces.
Bounded Operators
A bounded operator between normed spaces is a linear operator whose effect on the size of vectors is limited by a constant. More formally, a linear operator \(T : X \rightarrow Y\) is bounded if there exists a constant \(C\) such that for every \(x \in X\), the inequality \(\|T(x)\|_Y \leq C \|x\|_X\) holds.
In practical terms, this means that a bounded operator does not stretch any vector more than a fixed multiple of its original length. This property is crucial for ensuring the stability and predictability of the operator's behavior.
Boundedness simplifies calculations and proofs in functional analysis because it ensures that the operator behaves in a 'controlled' manner. It allows us to extend the concept of continuity to operators in infinite-dimensional spaces, equipping us with powerful tools to handle various problems. In the exercise provided, showing that \(\widehat{T}\) is bounded is an essential step to conclude that it behaves well over its entire domain, ensuring reliable mappings between the quotient space and the target space \(Y\).

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Most popular questions from this chapter

Let \(N\) be a maximal \(\varepsilon\) -separated set in the unit sphere of a Banach space \(X\). Show that \((1-\varepsilon) B_{X} \subset \overline{\operatorname{conv}}(N)\) Hint: Otherwise, by the separation theorem, we find \(x \in X\) and \(f \in S_{X^{*}}\) with \(\|x\| \leq 1-\varepsilon\) and \(f(x)>\sup _{\operatorname{conv}(N)}(f)>\sup _{N}(f) .\) For \(\delta>0\), choose \(y \in S_{X}\) such that \(f(y)>1-\delta .\) By the maximality of \(N\), there exists \(z \in N\) with \(\varepsilon>\|y-z\| \geq f(y)-f(z)\). Thus \(\sup _{N}(f) \geq f(z)>f(y)-\varepsilon>1-\delta-\varepsilon\) This holds for any \(\delta>0\), so we have \(1-\varepsilon \leq \sup _{N}(f)

Show that \(\ell_{1}\) is not isomorphic to a subspace of \(c_{0}\). Hint: The dual of \(\ell_{1}\) is nonseparable.

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

Let \(X=\mathbf{R}^{2}\) with the norm \(\|x\|=\left(\left|x_{1}\right|^{4}+\left|x_{2}\right|^{4}\right)^{\frac{1}{4}} .\) Calculate directly the dual norm on \(X^{*}\) using the Lagrange multipliers. Hint: The dual norm of \((a, b) \in X^{*}\) is \(\sup \left\\{a x_{1}+b x_{2} ; x_{1}^{4}+x_{2}^{4}=1\right\\} .\) Define \(F\left(x_{1}, x_{2}, \lambda\right)=a x_{1}+b x_{2}-\lambda\left(x_{1}^{4}+x_{2}^{4}-1\right)\) and multiply by \(x_{1}\) and \(x_{2}\), respectively, the equations you get from \(\frac{\partial F}{\partial x_{1}}=0\) and \(\frac{\partial F}{\partial x_{2}}=0\)

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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