Linked Questions

4
votes
1answer
190 views

How to construct an “explicit” element of $(\ell^\infty(\mathbb N))^* \setminus \ell^1(\mathbb N)$? [duplicate]

Possible Duplicate: Nonnegative linear functionals over $l^\infty$ An explicit functional in $(l^\infty)^*$ not induced by an element of $l^1$? Everything is in the title: How to construct an "...
5
votes
0answers
198 views

An explicit functional in $(l^\infty)^*$ not induced by an element of $l^1$? [duplicate]

Possible Duplicate: Nonnegative linear functionals over $l^\infty$ Setup: Let $l^\infty$ be the set of bounded sequences (with terms in $\mathbb{R}$), and let $l^1$ be the set of sequences of ...
2
votes
0answers
98 views

Existence of a $ l_ {\infty} ^*$ element [duplicate]

Possible Duplicate: Nonnegative linear functionals over $l^\infty$ An explicit functional in $(l^\infty)^*$ not induced by an element of $l^1$? Exercise: Prove there exist a bounded linear ...
11
votes
5answers
9k views

Dual of $l^\infty$ is not $l^1$

I know that the dual space of $l^\infty$ is not $l^1$, but I didn't understand the reason. Could you give me a example of an $x \in l^1$ such that if $y \in l^\infty$, then $ f_x(y) = \sum_{k=1}^{\...
5
votes
2answers
629 views

Positive functionals on $\ell^\infty$

A continuous linear functional $\varphi: \ell^\infty \to \mathbb{R}$ is said to be positive if $$ x \ge 0 \rightarrow \varphi(x) \ge 0 \quad \forall x \in \ell^\infty.$$ If $\varphi$ is in $\ell^1 \...
9
votes
3answers
549 views

Generalized limit in $l_\infty$ (Using: Hahn Banach Extension Theorem)

I am trying to solve the following problem (found in Maddox's book "Elements of Functional Analysis", page 128): So we have the function $p:l_\infty\rightarrow\mathbb{R}$ given by $$ p(x) = \inf\...
2
votes
2answers
914 views

Unbounded and linear transformation

I was thinking about this the other day. Does there exists any unbounded linear transformations from $\ell_\infty(\mathbb{R}) \to \mathbb{R}$ ? Here $\ell_\infty$ represent the infinity norm.
4
votes
1answer
362 views

Extension of the limit operator on $l^\infty$

Let $l^\infty = \{x\in \mathbb{R}^\mathbb{N}\colon \sup_{n\in \mathbb{N}}|x_n|<\infty\}$ and the subspace $C \subseteq l^\infty$ given by the convergent sequences. We consider the linear operator $...
7
votes
0answers
583 views

The Hahn-Banach theorem and the axiom of choice

Let us recall the Hahn-Banach theorem about extensions of linear functionals: Theorem: Let $E$ be a real vector space and $F$ a subspace. If $p:E\to \mathbb{R}$ is a sublinear function, and $g:F\to ...
3
votes
0answers
470 views

continuous linear functional on $l^{\infty}$ space

Let $l_{\infty}$ be the space of all bounded complex-valued sequences equipped with the supremum norm. Consider the natural standard basis $\{e_n\}_{n \in \mathbb{N}}$ of $l_{\infty}$. For any ...
4
votes
1answer
158 views

Exercise in Hahn-Banach Theorem; Finding linear functional $-p(-x)\leq f(x)\leq p(x)$

(The following exercises are in Kreyszig's book 218 page; EXE 10) I want to solve the following exercise : If $X=l^\infty$, let $p(x)=\lim\sup x_i $, which is sublinear. Then find a linear functional $...