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Chapter 3: Problem 23

Let \(X=C[0,1]\) in the supremum norm and let \(Y=C^{1}[0,1] \equiv\left\\{f \in C[0,1]: f^{\prime}\right.\) exists and is continuous on \(\left.[0,1]\right\\} .\) Give \(Y\) the supremum norm also. Define \(T: Y \rightarrow X\) by \(T f=f^{\prime}\). Clearly \(T\) is linear. (a) Show that if \(f_{n} \rightarrow f\) and \(T f_{n} \rightarrow g\), then \(g=T f\). (Hint: you need only show \(g(x)=f^{\prime}(x)\) for all \(x \in[0,1]\). Use the fundamental theorem of calculus). (b) Show that \(T\) is not bounded. (c) Why doesn't this contradict the closed graph theorem?

Short Answer

Expert verified
(a) \(g=f'\); (b) \(T\) is not bounded; (c) \(Y\) is not complete, so no contradiction with the closed graph theorem.

Step by step solution

01

Understanding the Setup

We have two spaces, \(X\) and \(Y\), both with the supremum norm. The operator \(T: Y \rightarrow X\) is defined by \(T f = f'\). Our goal is to explore the behavior of this operator under various conditions.
02

Prove Part (a)

To show that if \(f_n \to f\) in \(Y\) and \(T f_n \to g\) in \(X\), then \(g = T f\), we use the fundamental theorem of calculus. According to this theorem,\[ f(x) = f(a) + \int_a^x f'(t) \, dt. \] As \(f_n \to f\), it follows that \[ f'(x) = \lim_{n \to \infty} f_n'(x) = g(x), \] hence \(g(x) = f'(x)\) for all \(x \in [0,1]\), proving that \(g = T f\).
03

Prove Part (b)

To demonstrate that \(T\) is not bounded, consider a sequence \(f_n\) where \(f_n(x) = \frac{1}{n} \sin(nx)\). Notice that each \(f_n\) has \(|f_n(x)| \leq \frac{1}{n}\), so \(\|f_n\|_Y \to 0\) as \(n \to \infty\). However, \(f_n'(x) = \cos(nx)\) means \(\|f_n'\|_X = 1\). Thus, it shows \(\|T f_n\|/\|f_n\|_Y \to \infty\), confirming \(T\) is not bounded.
04

Address Part (c)

The lack of boundedness in \(T\) does not contradict the closed graph theorem because \(Y\) and \(X\) are endowed with the supremum norm, making \(Y\) incomplete. The closed graph theorem applies only to mappings between Banach spaces, and \(C^{1}[0,1]\) with the supremum norm is not a Banach space.

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Key Concepts

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

Supremum Norm
When dealing with functions, especially in functional analysis, the supremum norm is a very useful tool. You might have heard it under its other name, the infinity norm. To understand it, think of it as measuring the "biggest" value a function takes on a specific interval. Here's how it works:

  • For a continuous function, such as those in space \( C[0,1] \), the supremum norm \( \|f\|_\infty \) is defined as the largest absolute value of \( f(x) \) for all \( x \) in the interval \([0, 1]\).
  • Mathematically, it's expressed as: \( \|f\|_\infty = \sup_{x \in [0,1]} |f(x)| \).
  • This norm is crucial in determining convergence and boundedness of function sequences.
As we advance in functional analysis, understanding the supremum norm helps in exploring various properties of spaces such as continuity and differentiability.
Linear Operator
A linear operator is like a function but acts between two spaces, maintaining the structures of these spaces. Here鈥檚 what makes an operator linear:

  • Additivity: \( T(f + g) = T(f) + T(g) \) for all functions \( f \) and \( g \).
  • Homogeneity: \( T(cf) = cT(f) \) for all scalars \( c \) and functions \( f \).
Given the exercise, the operator \( T: Y \rightarrow X \) where \( T f = f' \) means it maps a differentiable function from one space to another, taking its derivative. Since differentiation respects both additivity and homogeneity, \( T \) qualifies as a linear operator. This characteristic often simplifies complex functional problems.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone in calculus linking the concept of differentiation and integration. It helps bridge the two, allowing us to evaluate definite integrals and understand derivatives better. The theorem states:

  • Part 1: If \( f \) is continuous over \([a, b]\) and \( F \) is an antiderivative of \( f \) on \([a, b]\), then \(\int_a^b f(x) \, dx = F(b) - F(a) \).
  • Part 2: For \( F(x) = \int_a^x f(t) \, dt \), the function \( F \) is differentiable and \( F'(x) = f(x) \) for all \( x \) in \([a, b]\).
In the exercise, this theorem shows why if \( f_n \to f \) and \( T f_n \to g \), then \( g \) is indeed \( f' \), verifying the relation between a function and its derivative.
Boundedness in Linear Operators
Boundedness in linear operators is an important property that ensures that operators act nicely between spaces without "blowing up" the values they map. An operator \( T \) is considered bounded if:

  • There exists a constant \( M \) such that for all \( f \) in the domain, \( \|T f\| \leq M \|f\| \).
This means the operator does not stretch functions beyond a certain factor, making it stable and predictable. In the exercise, however, it was shown that \( T \) is not bounded since no such \( M \) can satisfy the condition. The example with \( f_n(x) = \frac{1}{n} \sin(nx) \) leading to \( f_n'(x) = \cos(nx) \) illustrates this clearly as \( \|T f_n\|/\|f_n\|_Y \to \infty \). This example highlights the significance of boundedness in ensuring the feasibility of operators in practice.

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Most popular questions from this chapter

For \(1 \leq p<\infty\) it is a fact that the dual space to \(L^{p}(X, \mu)\), where \((X, \mu)\) is a \(\sigma-\) finite measure space, is \(L^{q}(X, \mu), 1 / p+1 / q=1\), in the following sense: Given \(g \in\) \(L^{q}(X, \mu)\), define \(\Lambda_{g}(f)=\int_{X} f g d \mu .\) This is a bounded linear functional on \(L^{p}(X, \mu)\), and \(\left\|\Lambda_{g}\right\|=\|g\|_{q}\). Conversely, every bounded linear functional on \(L^{p}(X, \mu)\) has this form. Using this and the Hahn-Banach theorem, show $$ \|f\|_{p}=\sup \left\\{\left|\int_{X} f g d \mu\right|: g \in L^{q}(X),\|g\|_{q}=1\right\\} $$ for all \(f \in L^{p}(X)\)

A sequence \(\left\\{h_{n}\right\\}\) of vectors in a Hilbert space \(\mathscr{H}\) is said to be a Bessel sequence if $$ \sum_{n=1}^{\infty}\left|\left\langle h, h_{n}\right\rangle\right|^{2}<\infty $$ for every \(h \in \mathscr{H}\). A sequence \(\left\\{g_{n}\right\\}\) is said to be a Riesz-Fischer sequence if given any \(\left\\{c_{n}\right\\} \in \ell^{2}\) there exists (at least one) vector \(g \in \mathscr{H}\) such that $$ \left\langle g, g_{n}\right\rangle=c_{n} \text { for all } n $$ Note that an orthonormal basis is both a Bessel sequence and a Riesz-Fischer sequence. (a) Show that if \(\left\\{h_{n}\right\\}\) is a Bessel sequence, then there exists \(M<\infty\) so that $$ \sum_{n=1}^{\infty}\left|\left\langle h, h_{n}\right\rangle\right|^{2} \leq M\|h\|^{2} $$ for all \(h \in \mathscr{H}\). Hint: Apply the closed graph theorem to the map \(S: \mathscr{H} \rightarrow \ell^{2}\) defined by \(S h=\left\\{\left\langle h, h_{n}\right\rangle\right\\}\). (b) Show that if \(\left\\{g_{n}\right\\}\) is a Riesz-Fischer sequence, there exists \(m>0\) such that given \(\left\\{c_{n}\right\\} \in \ell^{2}\), the equations in (3.5) hold for at least one solution \(g\) satisfying $$ m\|g\|^{2} \leq \sum_{n=1}^{\infty}\left|c_{n}\right|^{2} . $$ Hint: The closed graph theorem again, applied to the appropriate map $$ T: \ell^{2} \rightarrow \mathscr{H} / N $$ where \(N\) is the orthogonal complement of the closed linear span of the vectors \(g_{n}\).

Show that the quotient \(X / M\) of a Banach space \(X\) by a closed subspace \(M\) is a Banach space. (Begin by showing that $$ \|x+M\| \equiv \inf \\{\|x+m\|: m \in M\\} $$ is a norm on \(X / M\).)

We introduce some terminology for the purpose of this problem: If \(X\) is either a real or complex vector space (meaning that the scalars used in scalar multiplication are real, or, respectively, complex), we say that a real-valued \(\varphi\) is a real-linear functional if \(\varphi(x+y)=\varphi(x)+\varphi(y)\) and \(\varphi(\alpha x)=\alpha \varphi(x)\) holds for all \(x, y \in X\) and \(\alpha\) real. For \(X\) a complex vector space, we say that (a complex-valued) \(\varphi\) is a complexlinear functional if these relationships hold for all \(x, y \in X\) and \(\alpha\) complex. (a) Show that for any complex number \(z, z=\operatorname{Re} z-i \operatorname{Re}(i z)\). (b) Suppose \(X\) is a complex vector space and \(\varphi\) is a complex-linear functional on \(X\). Define \(u: X \rightarrow \mathbb{R}\) by \(u(x)=\operatorname{Re} \varphi(x)\). Show that \(\varphi(x)=u(x)-i u(i x)\) for all \(x \in X\). (c) Suppose \(u\) is a real-linear functional on \(X\). Define \(\varphi: X \rightarrow \mathbb{C}\) by \(\varphi(x)=\) \(u(x)-i u(i x)\). Show that \(\varphi\) is a complex-linear functional on \(X\). (Hint: Check the condition \(\varphi(\alpha x)=\alpha \varphi(x)\) first for \(\alpha\) real, then for \(\alpha=i\), then for \(\alpha\) complex.) (d) Now suppose \(X\) is a normed linear space. For \(\varphi\) and \(u\) related as above, show that \(\|\varphi\|=\|u\|\).

Let \(\mathscr{H}\) be a Hilbert space. Let \(\left\\{x_{n}\right\\}\) be a sequence in \(\mathscr{H}\) with the property that \(\left\langle x, x_{n}\right\rangle \rightarrow 0\) as \(n \rightarrow \infty\) for each vector \(x \in \mathscr{H}\). Show that sup \(\left\\{\left\|x_{n}\right\|: n=\right.\) \(1,2,3, \ldots\\}<\infty\).
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