/*! 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 35 Let \(X, Y\) be normed spaces, \... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 35

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 $$ \widehat{T}^{*}\left(\widehat{y^{*}}\right)(\hat{x})=\widehat{y^{*}}(\widehat{T}(\hat{x}))=y^{*}(T(x))=T^{*}\left(y^{*}\right)(x)=\widehat{T^{*}}\left(\hat{y^{*}}\right)(\hat{x}) $$

Short Answer

Expert verified
\( \widehat{T}^* = \widehat{T^*} \)

Step by step solution

01

Define the formulation

Given that \( \widehat{T} : X / \operatorname{Ker}(T) \rightarrow \overline{T(X)} \) is defined by \( \widehat{T}(\hat{x}) = T(x) \), we consider its adjoint \( \widehat{T}^* : \overline{T(X)}^* \rightarrow (X / \operatorname{Ker}(T))^* \).
02

Use the given propositions and definitions

From Proposition 2.7 and the fact that \( \overline{T(X)}^\perp = T(X)^\perp = \operatorname{Ker}(T^*) \), we assume \( \widehat{T}^* \) is a bounded linear operator from \( Y^* / \operatorname{Ker}(T^*) \) into \( \operatorname{Ker}(T)^\perp \subset X^* \).
03

Consider the operator \( \widehat{T^*} \)

For the adjoint operator \( T^* : Y^* \rightarrow X^* \), consider \( \widehat{T^*} : Y^* / \operatorname{Ker}(T^*) \rightarrow X^* \).
04

Choose \( \hat{y^*} \in Y^* / \operatorname{Ker}(T^*) \) and \( x \in X \)

Using the given hint, take any \( \hat{y^*} \in Y^* / \operatorname{Ker}(T^*) \) and \( x \in X \).
05

Evaluate step-by-step using identifications

Using the identifications, we obtain \( \widehat{T}^*(\hat{y^*})(\hat{x}) = \widehat{y^*}(\widehat{T}(\hat{x})) = y^*(T(x)) = T^*(y^*)(x) = \widehat{T^*}(\hat{y^*})(\hat{x}) \).
06

Conclude the equality

Thus, \( \widehat{T}^* = \widehat{T^*} \).

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Key Concepts

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

Normed Spaces
When studying Functional Analysis, one of the fundamental concepts you'll encounter is 'Normed Spaces'. A normed space is a vector space equipped with a function called a 'norm'. This norm assigns a non-negative length to every vector in the space. The norm of a vector, denoted as \|x\|, measures its 'size' or 'magnitude'. Imagine it as the length you would measure with a ruler if the vector were a geometric object.

Some key properties:
  • \( \|x\| \geq 0 \, \forall x \in X \) - Norms are always non-negative.
  • \( \|x\| = 0 \Leftrightarrow x = 0 \) - The only vector with zero norm is the zero vector.
  • \( \|\alpha x\| = |\alpha| \|x\| \) - Multiplying a vector by a scalar stretches or compresses its norm by the scalar's absolute value.
  • Triangle Inequality: \( \|x + y\| \leq \|x\| + \|y\| \) - The norm of a sum is at most the sum of the norms.
These properties ensure that the norm behaves intuitively regarding vector lengths and magnitudes. By understanding normed spaces, you establish a solid foundation to explore further topics like bounded linear operators and adjoint operators.

Bounded Linear Operators
Next in line is the concept of 'Bounded Linear Operators'. An operator is essentially a function between two vector spaces that respects the operations of vector addition and scalar multiplication. Specifically, a linear operator from normed space \( X \) to normed space \( Y \) is said to be bounded if there exists a constant \( C \) such that:

\|T(x)\| \leq C \|x\|, \; \forall x \in X

This inequality tells us that the output of the operator, in terms of norm, does not grow faster than a certain multiple of the input norm. If such a \( C \) exists, the operator \( T \) is called bounded.
  • A bounded operator ensures that the transformation it represents is 'controlled' and doesn't wildly increase vector lengths.
  • Bounded operators are continuous, meaning small changes in the input produce small changes in the output.
In our exercise, \( T \) belongs to \( \mathcal{B}(X, Y) \, \) indicating it's a bounded linear operator between normed spaces \( X \) and \( Y \). Recognizing bounded linear operators helps you manage expectations on how transformations affect vectors between spaces.

Adjoint Operators
Lastly, let's dive into 'Adjoint Operators'. The adjoint of an operator \( T \), denoted \( T^* \), is another operator that intuitively represents the 'transpose' of \(T\) in the context of functional analysis. Given a bounded linear operator \( T: X \rightarrow Y \), the adjoint operator \( T^*: Y^* \rightarrow X^* \) satisfies the property:

\( \langle T(x), y^* \rangle = \langle x, T^*(y^*) \rangle \; \forall x \in X, y^* \in Y^\* \)

Here, \( \langle \, \rangle \) denotes the dual pairing between vectors and functionals. This definition arises naturally when we try to 'flip' an operator to act on the dual spaces (spaces of linear functionals).
  • An adjoint operator ensures that inner product relationships are preserved across transformations.
  • In the context of the exercise, we saw operators \( \widehat{T}^\* \) and \( \widehat{T^*} \). Leveraging adjoint operators helps verify that these different perspectives on the transformation \( T \) align.
Understanding adjoint operators is crucial as they frequently appear in the study of Hilbert spaces, optimization problems, and quantum mechanics. Here, they help us reconcile two equivalent formulations of an operator transformation's effect in dual spaces.

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Most popular questions from this chapter

Let \(T \in \mathcal{B}(X, Y)\). Prove the following: (i) \(\underline{\operatorname{Ker}(T)}=T^{*}\left(Y^{*}\right)_{\perp}\) and \(\underline{\operatorname{Ker}\left(T^{*}\right)}=T(X)^{\perp}\). (ii) \(\overline{T(X)}=\operatorname{Ker}\left(T^{*}\right)_{\perp}\) and \(\overline{T^{*}\left(Y^{*}\right)} \subset \operatorname{Ker}(T)^{\perp}\) Hint: (i): Assume \(x \in T^{*}\left(Y^{*}\right)_{\perp} .\) Then for any \(g \in Y^{*}\) we have \(g(T(x))=\) \(T^{*}(g)(\underline{x})=0\), and hence \(T(x)=0\). Thus \(x \in \operatorname{Ker}(T)\). (ii): \(\overline{T(X)}=\overline{\operatorname{span}}(T(X))=\left(T(X)^{\perp}\right)_{\perp}=\operatorname{Ker}\left(T^{*}\right)_{\perp}\)

Let \(L\) be a closed subset of a compact space \(K\). Show that \(C(L)\) is isomorphic to a quotient of \(C(K)\). Hint: Let \(Q: C(K) \rightarrow C(L)\) be defined for \(f \in C(K)\) by \(Q(f)=\left.f\right|_{L}\). Then \(Q\) is onto by Tietze's theorem; then use Corollary \(2.25\).

Let \(T\) be a linear operator (not necessarily bounded) from a normed space \(X\) into a normed space \(Y .\) Show that the following are equivalent: (i) \(T\) is an open map. (ii) There is \(\delta>0\) such that \(\delta B_{Y} \subset T\left(B_{X}\right)\). (iii) There is \(M>0\) such that for every \(y \in Y\) there is \(x \in T^{-1}(y)\) satisfying \(\|x\|_{X} \leq M\|y\|_{Y}\) Hint: (i) \(\Longrightarrow\) (ii): \(T\left(B_{X}^{O}\right)\) is open, so \(y_{0}+\delta B_{Y}^{O} \subset T\left(B_{X}^{O}\right)\) for some \(y_{0} \in Y\), \(\delta>0 . T\left(B_{X}^{O}\right)\) is symmetric and convex, hence every \(b \in \delta B_{Y}^{O}\) belongs to \(T\left(B_{X}^{O}\right)\) since \(b=\frac{1}{2}\left(\left(x_{0}+b\right)-\left(x_{0}-b\right)\right)\) (ii) \(\Longrightarrow(\mathrm{i}): T\left(B_{X}^{O}\right) \supset T\left(\frac{1}{2} B_{X}\right) \supset \frac{\delta}{2} B_{Y} \supset \frac{\delta}{2} B_{X}^{O}\), so \(T\) maps neighborhoods of 0 to neighborhoods of 0 . By linearity, \(T\) is open.

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 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 \(T(X)\).
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