Question

Show that there is a linear functional \(L\) on \(\ell_{\infty}\) with the following properties: (1) \(\|L\|=1\); (2) if \(x=\left(x_{i}\right) \in c\), then \(L(x)=\lim _{i \rightarrow \infty}\left(x_{i}\right)\) (3) if \(x=\left(x_{i}\right) \in \ell_{\infty}\) and \(x_{i} \geq 0\) for all \(i\), then \(L(x) \geq 0\); (4) if \(x=\left(x_{i}\right) \in \ell_{\infty}\) and \(x^{\prime}=\left(x_{2}, x_{3}, \ldots\right)\), then \(L(x)=L\left(x^{\prime}\right)\). Hint: For simplicity, we consider only the real scalars setting. Let \(M\) be the subspace of \(\ell_{\infty}\) formed by elements \(x-x^{\prime}\) for \(x \in \ell_{\infty}\) and \(x^{\prime}\) as above. Let 1 denote the vector \((1,1, \ldots)\). We claim that \(\operatorname{dist}(1, M)=1\). Note that \(0 \in M\) and thus \(\operatorname{dist}(1, M) \leq 1\). Let \(x \in \ell_{\infty} .\) If \(\left(x-x^{\prime}\right)_{i} \leq 0\) for any of \(i\), then \(\left\|1-\left(x-x^{\prime}\right)\right\|_{\infty} \geq 1\). If \(\left(x-x^{\prime}\right)_{i} \geq 0\) for all \(i\), then \(x_{i} \geq x_{i+1}\) for all \(i\), meaning that \(\lim \left(x_{i}\right)\) exists. Therefore \(\lim \left(x_{i}-x_{i}^{\prime}\right)=0\), and thus \(\left\|1-\left(x-x^{\prime}\right)\right\| \geq 1\) By the Hahn-Banach theorem, there is \(L \in \ell_{\infty}^{*}\) with \(\|L\|=1, L(1)=1\), and \(L(m)=0\) for all \(m \in M .\) This functional satisfies (1) and (4). To prove (2), it is enough to show that \(c_{0} \subset L^{-1}(0)\). To see this, for \(x \in \ell_{\infty}\) we inductively define \(x^{(1)}=x^{\prime}\) and \(x^{(n+1)}=\left(x^{(n)}\right)^{\prime}\) and note that by telescopic argument we have \(x^{(n)}-x \in M .\) Hence, \(L(x)=L\left(x^{(n)}\right)\) for every \(x \in \ell_{\infty}\) and every \(n\). If \(x \in c_{0}\), then \(\left\|x^{(n)}\right\| \rightarrow 0\) and thus \(L(x)=0 .\) To show (3), assume that for some \(x=\left(x_{n}\right)\) we have \(x_{i} \geq 0\) for all \(i\) and \(L(x)<0 .\) By scaling, we may assume that \(1 \geq x_{i} \geq 0\) for all \(i\). Then \(\|1-x\|_{\infty} \leq 1\) and \(L(1-x)=1-L(x)>1\), a contradiction with \(\|L\|=1\).

Short answer

The linear functional \( L \) on \( \,l_{\infty} \) with the given properties exists by applying the Hahn-Banach theorem and considering the subspace \( M \).

Step-by-step solution

Define the Problem

We need to show the existence of a linear functional \( L \) on \( \,l_{\infty} \) with the following properties:1. \( \|L\| = 1 \)2. For any \( x = (x_i) \in c \), \( L(x) = \lim _{i \rightarrow \infty}(x_i) \)3. For any \( x = (x_i) \in \,l_{\infty} \) with \( x_i \geq 0 \) for all \( i \), \( L(x) \geq 0 \)4. For any \( x = (x_i) \in \,l_{\infty} \) and \( x' = (x_2, x_3, \ldots) \), \( L(x) = L(x') \)

Establish Subspace and Linear Functional

Consider the subspace \( M \) of \( \,l_{\infty} \) formed by elements \( x - x' \) for \( x \in \,l_{\infty} \) and \( x' \) as defined. Note that \(0 \in M\). We claim that \( \operatorname{dist}(1, M) = 1 \), where 1 denotes the vector (1,1,...). This means that the distance between the vector 1 and the subspace \( M \) is at most 1.

Proof by Cases

If \( (x - x')_i \leq 0 \) for any \( i \), then \( \|1 - (x - x')\|_{\infty} \geq 1 \). If \( (x - x')_i \geq 0 \) for all \( i \), \( x_i \geq x_{i+1} \) for all \( i \), meaning that \( \lim \left( x_i \right) \) exists. Hence, \( \|1 - (x - x')\| \geq 1 \). Therefore, \( \operatorname{dist}(1, M) = 1 \).

Apply Hahn-Banach Theorem

Using the Hahn-Banach theorem, there exists \( L \in \,l_{\infty}^{*} \) with \( \|L\| = 1 \), \( L(1) = 1 \), and \( L(m) = 0 \) for all \( m \in M \). This functional satisfies properties (1) and (4).

Prove Property (2)

To show property (2), we need to verify that \( c_0 \subset L^{-1}(0) \). For \( x \in \,l_{\infty} \), define \( x^{(1)} = x' \) and \( x^{(n+1)} = (x^{(n)})' \). By telescoping, \( x^{(n)} - x \in M \). Hence, \( L(x) = L(x^{(n)}) \) for every \( x \in \,l_{\infty} \) and every \(n \). If \( x \in c_0 \), then \( \|x^{(n)}\| \rightarrow 0 \) leading to \( L(x) = 0 \).

Prove Property (3)

For property (3), assume \( x = (x_n) \) with \( x_i \geq 0 \) for all \( i \) and \( L(x) < 0 \). By scaling, assume \( 1 \geq x_i \geq 0 \). Then \( \|1 - x\|_{\infty} \leq 1 \) and \( L(1 - x) = 1 - L(x) > 1 \), contradicting \( \|L\| = 1 \). Therefore, \( L(x) \geq 0 \).

Key concepts

Linear Functional

A linear functional is a function that maps vectors from a vector space to the underlying field of scalars. It holds two main properties: additivity and homogeneity. Additivity means that for any two vectors, the functional of their sum equals the sum of their functionals. Homogeneity means that if you scale a vector, the functional of the scaled vector is the scaled value of the functional. In the context of our exercise, we're working with a linear functional on the space \(\ell_{\text{infty}}\). This functional must satisfy specific properties such as having a norm equal to 1 and performing specific actions on vectors.

Normed Space

A normed space is a vector space equipped with a norm, a function that assigns a non-negative length or size to each vector in the space. For example, the space \(\(l_{\text{infty}}\)\) consists of bounded sequences of real numbers, and the norm of a sequence is defined as the supremum of the absolute values of its elements. The important aspect of a normed space is that it allows us to measure the 'length' of vectors and can be used to discuss convergence and distance. In our problem, we are considering normed spaces to understand how the linear functional operates and to establish the properties needed by leveraging norms.

Subspace Distance

Subspace distance refers to the measurement of the distance between a point and a subspace within a normed space. This is crucially relevant in our exercise when discussing the subspace \(\text{M}\) formed by elements of the form \(x-x'\). Here, we're particularly interested in calculating \(\text{dist}(1, M) = 1\), indicating the distance from the vector \(\text{1}\) (i.e., the vector where all entries are 1) to the subspace \(M\). Proving that this distance is exactly 1 allows us to apply the Hahn-Banach theorem to extend a bounded linear functional from a subspace to the entire space while preserving its properties.

Telescoping Series

A telescoping series is a series whose partial sums eventually only have a small number of terms because most terms cancel out. This concept becomes particularly useful when we use sequences \(x^{(n)}\) derived from \(x\) as defined in the problem. By defining sequences in such a telescoping manner, we ensure that differences like \(x^{(n)} - x \) belong to the subspace \(M\). This allows us to utilize the properties of the linear functional \(\text{L}\) more efficiently and demonstrate desired properties like \(L(x) = 0\) for vectors in \(c_0\), the space of sequences converging to 0.