Question

Define a functional \(f\) on \(\ell_{1}\) by \(f(x)=\sum x_{i}\) for \(x=\left(x_{i}\right)\). Show that \(f\) is norm continuous on \(\ell_{1}\) and that \(f\) is not \(\overline{w^{*}}\) -continuous on \(\ell_{1} \cong c_{0}^{*}\) (that is, \(\left.f \in \ell_{\infty} \backslash c_{0}\right)\)

Short answer

f is norm continuous on \(\ell_{1}\) due to boundedness, but not \(\overline{w^{*}}\)-continuous as it doesn't preserve weak\textsuperscript{*} convergence.

Step-by-step solution

Understand the Functional Definition

The functional is given by the equation \(f(x) = \sum x_i\) for a sequence \(x = (x_i)\) in \(\ell_{1}\). This means it sums the elements of the sequence.

Prove Norm Continuity

To show that \(f\) is norm continuous, consider that any element \(x \in \ell_{1}\) has a finite sum of absolute values, i.e., \(\sum |x_i| < \infty\). The functional \(f\) satisfies the property \(|f(x)| \leq \sum |x_i| = \|x\|_1\). Therefore, \(f\) is bounded above by the norm \(\|x\|_1\), proving it is norm continuous.

Explain Weak\textsuperscript{*} Topology

In the dual space \(\ell_{1} \cong c_{0}^{*}\), the weak\textsuperscript{*} topology, denoted as \(\overline{w^{*}}\), involves convergence based on continuous linear functionals from \(c_{0}\). If \(f\) were \(\overline{w^{*}}\)-continuous, it would map weak\textsuperscript{*} convergent sequences to convergent sequences in \(\mathbb{R}\).

Show f is Not \(\overline{w^{*}}\)-Continuous

Consider the sequence \(e_n\) in \(c_0\) where \(e_n\) has 1 in the n-th position and 0 elsewhere. This sequence weak\textsuperscript{*} converges to 0 in \(\ell_{1}\), but \(f(e_n) = 1\) for all \(n\). Thus, \(f(e_n)\) does not converge to \(f(0) = 0\), demonstrating that \(f\) is not \(\overline{w^{*}}\)-continuous.

Key concepts

Norm Continuity

In functional analysis, a functional is said to be norm continuous if it maps elements in a normed vector space (like \(\ell_{1}\)) to real numbers in a way that respects the norm structure. Specifically, for our functional \(\f(x) = \sum x_i\), we need to show that small changes in the norm lead to small changes in the value of \(\f(x)\).
To do this, observe that any sequence \(\textbf{x} \in \ell_{1}\) has a finite sum of absolute values: \(\sum |x_i| < \infty\). This fact naturally implies that the functional is bounded by the \(\textbf{1}\)-norm, as follows: \(\big|\f(x)\big| \xleq \sum |x_i| = \|\textbf{x}\textbf{\textbf{x}}|_1\).
This inequality indicates the functional can’t ‘blow up’; meaning, as long as the \(\textbf{1}\)-norm of \(\textbf{x}\) is small, \(\f(x)\) remains small too. Thus, the functional \(\f\) is norm continuous.

Weak-Star Topology

Understanding the weak-star topology (or \(\bar{w^{*}}\) topology) requires some setup with dual spaces. The dual space of \(c_0\) (the space of sequences tending to zero) is \(\ell_{1}\). Within this structure, the weak-star topology deals with how functionals in \(\bar{w^{*}}\) converge.
In general, instead of norms, the weak-star topology uses the notion of pointwise convergence based on functionals from \(c_0\). More precisely, a sequence of functionals in \(\ell_{1}\) is said to converge in the weak-star topology if it converges pointwise on all elements of \(c_0\).
However, while pointwise convergence can be weaker than norm convergence, it will still imply that functionals that operate nicely under the weak-star topology must map weak-star convergent sequences to sequences that converge in \(\mathbb{R}\).

Dual Spaces

Dual spaces play a key role in functional analysis. Given a vector space \(\mathbb{V}\), the dual space consists of all continuous linear functionals from \(\mathbb{V}\) into \(\mathbb{R}\). For instance, \(\ell_{1}\) is the dual of \(c_0\) because every continuous linear functional on \(c_0\) corresponds uniquely to an element in \(\ell_{1}\).
In the given problem, \(\f\) as defined corresponds to summing the elements of sequences in \(\ell_{1}\backslash c_{0}\). Importantly, while \(\f\) respects the norm structure of \(\ell_{1}\), mapping norms in \(\ell_{1}\) to norms in \(c_0\), it isn't necessarily weak-star continuous.

• Continuous Linear Functionals: Basic building blocks of dual spaces.
• \ell_{1} as the dual of \(c_0\): Summarizes how dual relationships work in sequence spaces.
• Weak-star Continuity test: Shows limits of such functionals under pointwise convergence.

This context completes the fuller picture behind why the functional \(\f\) from the problem is norm continuous yet not weak-star continuous.