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Chapter 4: Problem 5

This problem provides an alternate proof to Proposition 4.4. Suppose that \(T\) : \(X \rightarrow Y\) is linear, where \(X\) and \(Y\) are normed linear spaces and \(X\) is finite-dimensional. Define \(\|\cdot\|_{\beta}\) on \(X\) by $$ \|x\|_{\beta}=\max \left(\|x\|_{X},\|T x\|_{Y}\right) $$ (a) Check that \(\|\cdot\|_{\beta}\) is a norm on \(X\). (b) Argue that \(T:\left(X,\|\cdot\|_{\beta}\right) \rightarrow Y\) is continuous, and hence that so is \(T: X \rightarrow Y\) in the original norm on \(X\).

Short Answer

Expert verified
\(\|\cdot\|_{\beta}\) is a valid norm on \(X\), and \(T: X \rightarrow Y\) is continuous.

Step by step solution

01

Define Norm Properties

To prove that \(\|\cdot\|_{\beta}\) is a norm, we need to show it satisfies three properties of norms:1. **Positive Definiteness**: \(\|x\|_{\beta} = 0 \Rightarrow x = 0\).2. **Scalar Multiplication**: \(\|\alpha x\|_{\beta} = |\alpha| \|x\|_{\beta}\) for any scalar \(\alpha\).3. **Triangle Inequality**: \(\|x + y\|_{\beta} \leq \|x\|_{\beta} + \|y\|_{\beta}\) for any \(x, y \in X\).
02

Check Positive Definiteness

For any \(x \in X\), if \(\|x\|_{\beta} = 0\), then \(\max(\|x\|_{X}, \|T x\|_{Y}) = 0\). This implies \(\|x\|_{X} = 0\) and \(\|T x\|_{Y} = 0\). Since \(\|x\|_{X} = 0\) is true only when \(x = 0\), \(\|x\|_{\beta} = 0\) implies \(x = 0\). Thus, positive definiteness is satisfied.
03

Check Scalar Multiplication

For any scalar \(\alpha\) and \(x \in X\), we have:\[ \| \alpha x \|_{\beta} = \max(\|\alpha x\|_{X}, \|T(\alpha x)\|_{Y}) = \max(|\alpha| \|x\|_{X}, |\alpha| \|T x\|_{Y}) = |\alpha| \max(\|x\|_{X}, \|T x\|_{Y}) = |\alpha| \|x\|_{\beta}. \]Thus, the scalar multiplication property holds.
04

Check Triangle Inequality

For any \(x, y \in X\), we have:\[ \|x + y\|_{\beta} = \max(\|x + y\|_{X}, \|T(x + y)\|_{Y}) \leq \max(\|x\|_{X} + \|y\|_{X}, \|T x\|_{Y} + \|T y\|_{Y}) \leq \max(\|x\|_{X}, \|T x\|_{Y}) + \max(\|y\|_{X}, \|T y\|_{Y}) = \|x\|_{\beta} + \|y\|_{\beta}. \]Thus, the triangle inequality is satisfied, confirming \(\|\cdot\|_{\beta}\) is a norm on \(X\).
05

Continuity of T with \(\|\cdot\|_{\beta}\)

Since \(X\) is finite-dimensional, all norms on \(X\) are equivalent. Therefore, there exists some constant \(C > 0\) such that for all \(x \in X\), \(\|T x\|_{Y} \leq \|x\|_{\beta} \leq C\|x\|_{Y}\). This means that \(T: (X,\|\cdot\|_{\beta}) \rightarrow Y\) is bounded, hence continuous.
06

Continuity of T with Original Norm

Given that \(T\) is continuous with respect to \(\|\cdot\|_{\beta}\) norm, it must also be continuous with respect to the original norm \(\|\cdot\|_{X}\) due to the equivalence of norms on finite-dimensional spaces.

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Key Concepts

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

Normed Linear Spaces
A normed linear space is a vector space equipped with a function that assigns a non-negative length to each vector. This function is called a **norm**. Norms enable us to quantify the "size" or "length" of vectors in a way that mimics our intuitive understanding of length in physical space. Here are some key properties of norms that make them indispensable in functional analysis:
  • Positive Definiteness: A norm \( \| \cdot \| \) is positive definite if \( \| x \| = 0 \) if and only if \( x = 0 \). This means that the "zero vector" is the only vector with zero length.

  • Scalar Multiplication: The norm is linear with respect to scalar multiplication. In other words, for any scalar \( \alpha \) and vector \( x \), the norm satisfies \( \| \alpha x \| = | \alpha | \| x \| \). This property ensures that the direction and magnitude scale correctly with multiplication.

  • Triangle Inequality: The addition of two vectors does not result in a vector longer than the sum of their norms. Mathematically, \( \| x + y \| \leq \| x \| + \| y \| \). This reflects the intuitive idea that the direct path between two points is shorter than any indirect route via a third point.
These properties play a crucial role when dealing with linear transformations and lead to interesting characteristics when applied to spaces of different dimensions.
Continuity
Continuity in the context of functional analysis refers to the smooth behavior of a function, ensuring that small changes in input result in small changes in output. This concept has parallels with continuity in calculus but is adapted for functions between infinite-dimensional spaces. For a linear transformation \( T: X \rightarrow Y \) between normed spaces, continuity can be understood through some specific criteria:
  • Boundedness: A linear operator \( T \) is continuous if it is bounded, meaning there exists a constant \( C \) such that for all vectors \( x \) in \( X \), \( \| T(x) \|_Y \leq C \| x \|_X \). This implies that the operator does not "stretch" any vector excessively compared to its input size.

  • Equivalence of Norms in Finite Dimensions: In finite-dimensional spaces, all norms are considered equivalent. This equivalence simplifies the analysis since a transformation that is continuous with one norm will be continuous with any other equivalent norm in the same space.

  • Implication for Linear Transformations: Given a norm such as \( \| \cdot \|_{\beta} \), once continuity is established in one normed space, it holds for the original space if dimensions remain finite, demonstrating a robustness in the concept of continuity.
Understanding continuity is essential for controlling how transformations interact with the elements of a space, particularly when ensuring stability and predictability in mathematical models.
Finite-dimensional Spaces
Finite-dimensional spaces are vector spaces that have a finite basis, meaning they can be spanned by a finite number of vectors. Such spaces exhibit particular characteristics that simplify many aspects of functional analysis:
  • Norm Equivalence: Any two norms on a finite-dimensional space are equivalent. This means if you have two different ways of measuring the length of vectors in this space, they will essentially provide consistent results up to a constant factor.

  • Completeness and Compactness: Finite-dimensional normed spaces are automatically complete and compact. Completeness refers to every Cauchy sequence having a limit within the space, while compactness implies that closed and bounded subsets are always compact.

  • Simplification of Continuity: Due to the equivalence of norms, the requirement for a linear transformation to be continuous is not dependent on the choice of the norm. This property massively reduces complexity and aids in establishing continuity for finite-dimensionally driven transformations quickly.
Finite-dimensional spaces are foundational not just for their theoretical elegance but also in practical computations, where they provide manageable frameworks for working with mathematical models.

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Most popular questions from this chapter

Consider the weighted shift operator on \(\ell^{2}\) given by $$ W\left(x_{1}, x_{2}, x_{3}, \ldots\right)=\left(0, x_{1}, \frac{1}{2} x_{2}, \frac{1}{3} x_{3}, \ldots\right) . $$ Show that \(W\) is compact, but \(W\) has no eigenvalues. Find a nontrivial closed invariant subspace for \(W\).

If \(\left\\{e_{n}\right\\}\) is an orthonormal sequence in a Hilbert space \(\mathscr{H}\), and \(T \in \mathscr{B}(\mathscr{H})\) is compact, show that \(T e_{n} \rightarrow 0\).

If \(T \in \mathscr{B}(\mathscr{H})\) for some (complex) Hilbert space \(\mathscr{H}\) and \(\langle T h, h\rangle\) is real for all \(h \in \mathscr{H}\), show that \(T\) is self-adjoint.

Suppose that \(A\) is a compact operator on a Banach space \(X\). Show that if \(A^{2}=\) \(A\), then the range of \(A\) is finite-dimensional.

For an analytic function \(\varphi\) mapping the unit disk into itself, define the linear composition operator \(C_{\varphi}\) by \(C_{\varphi}(f)=f \circ \varphi\) for \(f\) analytic on \(\mathbb{D}\). (a) Show that if \(C_{\varphi}\) is Hilbert-Schmidt on the Bergman space \(L_{a}^{2}(\mathbb{D})\), then $$ \int_{\mathbb{D}} \frac{1}{\left(1-|\varphi(z)|^{2}\right)^{2}} d A(z)<\infty $$ (b) Show that if \(C_{\varphi}\) is bounded on \(L_{a}^{2}(\mathbb{D})\) and $$ \int_{\mathbb{D}} \frac{1}{\left(1-|\varphi(z)|^{2}\right)^{2}} d A(z)<\infty $$ then \(C_{\varphi}\) is Hilbert-Schmidt. (c) Give an example of a compact composition operator on \(L_{a}^{2}(\mathbb{D})\).
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