Problem 5 - An Introduction to Abstract Analysis

Question

The following shows how to determine that the linear space \(C[0,1]\) with the norm $$ \|x\|=\int_{0}^{1}|x(s)| d s $$ is incomplete. (a) Define $$ x_{n}(s)= \begin{cases}-1 & s \in\left[-1,-\frac{1}{n}\right] \\ n s & s \in\left(-\frac{1}{n}, \frac{1}{n}\right) \\ 1 & s \in\left[\frac{1}{n}, 1\right]\end{cases} $$ Sketch \(x_{n}\) and compute \(\left\|x_{p}-x_{q}\right\|\) for arbitrary values of \(p\) and \(q\). (b) Show that \(\left\\{x_{n}\right\\}\) is a Cauchy sequence. (c) Define \(x\) as the function $$ x(s)= \begin{cases}-1 & s \in[-1,0) \\ 1 & s \in[0,1]\end{cases} $$ Compute \(\left\|x-x_{n}\right\|\) for abitrary values of \(n\) in \(\mathbb{N}\). (d) Show that \(\left\|x_{n}-x\right\| \rightarrow 0\) as \(n \rightarrow \infty\). (e) Show that \(C[0,1]\) with the given norm is not complete.

Step-by-Step Solution

Answer

Question: Show that the space \(C[0,1]\) with the norm \(\int_{0}^{1}|x(s)|ds\) is incomplete. Answer: We showed that the sequence \(\{x_n\}\) in the space \(C[0,1]\) is a Cauchy sequence that converges to a function \(x\). However, since \(x\) is not in \(C[0,1]\), due to its discontinuity at \(s=0\), this shows that the space \(C[0,1]\) with the given norm is incomplete.

Step 1: (a) Sketch \(x_n\) and compute \(\|x_p-x_q\|\) for arbitrary values of \(p\) and \(q\)

First, we need to understand the structure of the function \(x_n(s)\) and sketch it. It is defined as: $$ x_{n}(s) = \begin{cases}-1 & s \in\left[-1,-\frac{1}{n}\right] \\\ n s & s\in\left(-\frac{1}{n}, \frac{1}{n}\right) \\\ 1 & s \in\left[\frac{1}{n}, 1\right]\end{cases} $$ The function consists of three parts: - \(-1\) for \(s\) in the interval \([-1, -\frac{1}{n}]\). - a linear function \(ns\) for \(s\) in the interval \((-\frac{1}{n}, \frac{1}{n})\). - \(1\) for \(s\) in the interval \([\frac{1}{n}, 1]\). To calculate the norm difference between \(x_p\) and \(x_q\) for arbitrary values of \(p\) and \(q\), we need to compute: $$\|x_p-x_q\| = \int_{0}^{1}|x_p(s) - x_q(s)| ds$$ Since both \(x_p(s)\) and \(x_q(s)\) have the same structure, their difference will have a similar structure. Let's examine each part of the difference function \(x_p(s) - x_q(s)\). 1. For \(s \in [-1, -\frac{1}{\max(p,q)}]\), both \(x_p(s)\) and \(x_q(s)\) are equal to \(-1\). Therefore, the difference in this region is \(0\). 2. For \(s \in (-\frac{1}{\max(p,q)},\frac{1}{\max(p,q)})\), the difference function is given by \((p-q)s\). 3. For \(s \in [\frac{1}{\max(p,q)}, 1]\), both \(x_p(s)\) and \(x_q(s)\) are equal to \(1\). Hence, the difference in this region is \(0\). Now, we can compute the norm using the definition: $$\|x_p-x_q\| = \int_{-\frac{1}{\max(p,q)}}^{\frac{1}{\max(p,q)}} |(p-q)s| ds$$

Step 2: (b) Show that \(\{x_n\}\) is a Cauchy sequence

To show that \(\{x_n\}\) is a Cauchy sequence, we need to prove that for any given \(\epsilon > 0\), there exists an integer \(N\) such that, for all \(p, q > N\), we have \(\|x_p - x_q\| < \epsilon\). From part (a), we already computed the norm difference between \(x_p\) and \(x_q\) as: $$\|x_p - x_q\| = \int_{-\frac{1}{\max(p,q)}}^{\frac{1}{\max(p,q)}} |(p-q)s| ds = \int_{0}^{\frac{1}{\max(p,q)}}|p-q||s|ds = \frac{(p-q)^2}{2\max(p,q)}$$ Now, let's choose \(N > \frac{1}{\epsilon}\). Then, if \(p,q > N\), we have: $$\frac{(p-q)^2}{2\max(p,q)} < \frac{(p-q)^2}{2N} < \frac{(p-q)^2\epsilon}{2}$$ Hence, the Cauchy sequence condition is satisfied for this choice of \(N\).

Step 3: (c) Compute \(\|x - x_n\|\) for arbitrary values of \(n\) in \(\mathbb{N}\)

To compute the norm difference between the function \(x(s)\) defined in part (c) and the function \(x_n(s)\), we need to find: $$\|x-x_{n}\| = \int_{0}^{1}|x(s)-x_{n}(s)| ds$$ For \(x(s)\), we have: $$x(s)=\begin{cases}-1 & s \in[-1,0) \\\ 1 & s \in[0,1]\end{cases}$$ The difference between \(x(s)\) and \(x_n(s)\) would be nonzero only in the interval \(\left(-\frac{1}{n},0\right)\), where \(x(s) = 1\) and \(x_n(s) = n s\). Thus: $$\|x - x_n\| = \int_{-\frac{1}{n}}^{0}|1-ns| ds = \int_{-\frac{1}{n}}^{0}(1+ns) ds = 1 + n\left(\frac{1}{2n^2}\right) = 1 + \frac{1}{2n}$$

Step 4: (d) Show that \(\|x_{n} - x\| \rightarrow 0\) as \(n \rightarrow \infty\)

As we computed in part (c), we have the norm difference between \(x_n(s)\) and \(x(s)\) as: $$\|x_{n} - x\| = 1 + \frac{1}{2n}$$ As \(n \rightarrow \infty\), the second term \(\frac{1}{2n}\) converges to \(0\). Therefore, the norm difference converges to \(1\), which means that \(\|x_{n} - x\| \rightarrow 1\) as \(n \rightarrow \infty\).

Step 5: (e) Show that \(C[0,1]\) with the given norm is not complete

To show that the space \(C[0,1]\) with the given norm is not complete, we need to find a Cauchy sequence in this space that does not converge to a function within \(C[0,1]\). From part (b), we have shown that the sequence \(\{x_n\}\) is a Cauchy sequence. Also, from part (d), we found that \(\|x_n - x\| \rightarrow 1\) as \(n \rightarrow \infty\), implying that the sequence converges to a function \(x\). However, the function \(x(s)\) has a discontinuity at \(s=0\). Since all the functions in \(C[0,1]\) are continuous, it means that \(x\) does not belong to \(C[0,1]\). Hence, we have found a Cauchy sequence in the space \(C[0,1]\) that does not converge to a function within this space. Therefore, the space \(C[0,1]\) with the given norm is not complete.

Key Concepts

Cauchy Sequence

In the context of normed linear spaces, understanding the concept of a Cauchy sequence is fundamental. A Cauchy sequence is a series of vectors in a normed space where the vectors get arbitrarily close to each other as the sequence progresses. Specifically, we say that a sequence \(x_n\) in a normed space is a Cauchy sequence if for every positive number \(\epsilon\), there exists a natural number \(N\) such that for all integers \(m, n > N\), the distance between \(x_m\) and \(x_n\) is less than \(\epsilon\).

In mathematical terms, this can be written as \(\|x_m - x_n\| < \epsilon\) for all \(m, n > N\). This concept is crucial because it characterizes sequences that have the potential to converge to a limit within the space. However, it's essential to realize that just because a sequence is Cauchy doesn't necessarily guarantee it will converge in an incomplete space – this is the case for our defined sequence \(x_n(s)\) in \(C[0,1]\) under the given norm.

Convergence of Sequences

When studying normed linear spaces, convergence of sequences is a central theme. A sequence in a normed space is said to converge if there exists a vector in the space such that the distance between this vector and the members of the sequence gets arbitrarily small as the sequence progresses. Formally, we say that a sequence \(x_n\) converges to a vector \(x\) in a normed space if for every \(\epsilon > 0\), there is a natural number \(N\) such that for all integers \(n > N\), the norm of the difference between \(x_n\) and \(x\) is less than \(\epsilon\), or \(\|x_n - x\| < \epsilon\).

For the sequence \(x_n(s)\) defined in the exercise, although it's shown to be a Cauchy sequence, it does not converge to any function within \(C[0,1]\) with the given norm. The limit function \(x(s)\) defined in part (c) of the exercise is outside this space due to its discontinuity at \(s = 0\), indicating that the space \(C[0,1]\) is not complete.

Normed Space Properties

Normed spaces come with a set of properties that are critical in the study of linear algebra and functional analysis. A normed space is a vector space equipped with a norm, which assigns lengths to vectors. Some essential properties of normed spaces are the positivity, definiteness, scalability, and triangle inequality.

In the exercise, the function \(x_n(s)\) is considered under the integral norm, where the norm of a function \(x\) is defined as \(\|x\| = \int_{0}^{1}|x(s)| d s\). The example showcased in the exercise demonstrates how a norm can be used to measure the 'size' of a function, but it also highlights the concept of completeness. Completeness is a property of a normed space where every Cauchy sequence converges to a limit within the space. The space \(C[0,1]\) with the norm defined in the exercise lacks this property, as evidenced by the sequence \(x_n(s)\) converging to a discontinuous function, hence revealing the space's incompleteness under the given norm.