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I am new to functional analysis. I wondering what properties of $c_{00}$ have.

In particular, I am wondering:

  1. Why is $c_{00}$ is not closed in $\ell_p$ for $p\in[1,\infty]$.

  2. What does the dual space of $c_{00}$ look like?

  3. Why the dual space of $c_{00}$ is reflexive?

Thank you in advance!

Answer Lee
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    For all this note that $\ell^p(\Bbb N)$ has a Schauder basis given by the sequences $(e_n){k\in\Bbb N} = \delta{kn}$. The linear span of these sequences is $c_{00}$, the closure of the span is $\ell^p(\Bbb N)$. So you have $c_{00}\subset \ell_p$ is dense. The dual of $c_{00}$ (if given $p$-norm) is thus the dual of $\ell^p$. – s.harp Jun 21 '18 at 16:09
  • @s.harp I am sorry I am not familiar with Schauder basis. Can you be more specific on why $c_{00}$ is dense? Thank you! – Answer Lee Jun 21 '18 at 16:56
  • A Schauder basis for a normed linear space $X$ is a linearly independent $B\subset X$ such that for each $v\in X$ there is a $unique$ collection $(s_{v,b}){b\in B}$ of (real or complex) scalars such that $v=\sum{b\in B}b\cdot s_{v,b}.$ If $X$ is infinite-dimensional and separable then $B={b_n: n\in \Bbb N},$ where $b_n\ne b_m$ when $n\ne m,$ and for each $v\in X$ there is a unique sequence $(v_n){n\in \Bbb N} =(s{b_n,v}){n\in \Bbb N}$ of scalars such that $\lim{n\to \infty}|v- \sum_{j=1}^nb_nv_n|=0.$ – DanielWainfleet Jun 22 '18 at 01:02
  • Addendum to my previous comment: For example a Hilbert-space basis for a Hilbert space is a type of Schauder basis. ... Also, if $X$ is a finite-dimensional normed linear space, a Schuader basis is the same thing as a vector-space (Hamel) basis. – DanielWainfleet Jun 22 '18 at 01:08

2 Answers2

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Question 1

The sequence $$ (u_n)=\begin{cases} 1/k^{2/p} & 1\le k \le n\\ 0 & k>n \end{cases}$$ is a sequence of elements of $c_{00}$ that is converging in $\ell_p$ to the sequence $(1/k^{2/p})$ that is not in $c_{00}$. Hence $c_{00}$ is not closed.

For the other questions, see the commentS of s.sharp above.

mathcounterexamples.net
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The purpose of this answer is to provide an elementary discussion about each question that is suitable for other posts to use as a reference.

We use $\mathbb{K}$ to denote a fixed field which is either $\mathbb{R}$ or $\mathbb{C}$. The set $c_{00}$ consists of the set of scalar-valued sequences $x\colon \mathbb{N} \to \mathbb{K}$ where $\{k\in\mathbb{N} : x(k) \neq 0 \}$ is finite. This is a vector space over $\mathbb{K}$. If we define $e_{n}\colon \mathbb{N} \to \mathbb{K}$ for each $n\in\mathbb{N}$ by \begin{equation} e_{n}(k) := \begin{cases} 1 & \text{if } k = n, \\ 0 & \text{if } k \neq n , \end{cases} \end{equation} then $e_{n} \in c_{00}$ for every $n\in\mathbb{N}$. Moreover, $\{e_{n} : n\in\mathbb{N} \}$ is an algebraic basis for $c_{00}$. Hence $c_{00}$ is an infinite-dimensional vector space with a countable algebraic basis.

We also have $c_{00} \subseteq \ell_{p}$ for every $p\in [1, \infty ]$ and $c_{00} \subseteq c_{0}$, where all the inclusions are proper.

For each $p\in [1, \infty ]$ we consider the norm $\|\cdot\|_{p}$ on a subspace $X$ of $\ell_{p}$ given by \begin{equation} \|x\|_{p} := \begin{cases} (\sum_{k=1}^{\infty} |x(k)|^{p} )^{\tfrac{1}{p}} & \text{if } p\in [1, \infty ), \\ \sup\{|x(k)| : k\in\mathbb{N} \} & \text{if } p = \infty . \end{cases} \end{equation}

We naturally equip $\ell_{p}$ with the norm $\|\cdot\|_{p}$ for each $p\in [1, \infty ]$ and $c_{0}$ with the norm $\|\cdot\|_{\infty}$.

Question 1. We have the following.

Proposition 1. The closure of $c_{00}$ in $\ell_{p}$ is $\ell_{p}$ for $p \in [1, \infty )$ and $c_{0}$ for $p = \infty$.

If $p\in [1, \infty )$ and $x\in \ell_{p}$ then define $x_{n} \colon \mathbb{N} \to \mathbb{C}$ for each $n\in\mathbb{N}$ by \begin{equation} x_{n}(k) := \begin{cases} x(k) & \text{if} \; k\in \{1, \ldots , n\} , \\ 0 & \text{if} \; k \in \mathbb{N} \setminus \{1, \ldots , n\} . \end{cases} \end{equation} Then $x_{n}\in c_{00}$ for every $n\in\mathbb{N}$ and it is straightforward to show that the sequence $(x_{n})_{n\in\mathbb{N}}$ converges to $x$ in $(\ell_{p}, \|\cdot\|_{p})$. Hence $c_{00}$ is dense in $(\ell_{p}, \|\cdot\|_{p})$. The case $p = \infty$ is discussed here.

As $c_{00} \neq \ell_{p}$ for all $p\in [1, \infty ]$ and $c_{00} \neq c_{0}$, this implies that $c_{00}$ is not closed in $\ell_{p}$ for any $p\in [1, \infty ]$ by this result. A consequence of this is that $(c_{00}, \|\cdot\|_{p})$ is not a Banach space for any $p\in [1, \infty ]$. In fact, because $c_{00}$ has a countable infinite algebraic basis, this (or this) result shows there is no complete norm on $c_{00}$.

We also have that $c_{00}$ is separable when equipped with any norm. To see this, note that if $\|\cdot\|$ is a norm on $c_{00}$, then $C = \{e_{n} : n\in\mathbb{N} \}$ is a countable subset with ${\rm span}(C) = c_{00}$, so it follows from the proof in this answer that $(c_{00}, \|\cdot\|)$ is separable.

Question 2. You first need to give $c_{00}$ a norm to ask this question. Standard choices for a norm on $c_{00}$ are the $\ell_{p}$-norms for $p\in [1, \infty ]$. One this is done we have the following result.

Theorem 1. Let $p\in [1, \infty ]$ and let $p'$ be the conjugate exponent of $p$. The map $\Phi \colon \ell_{p'} \to (c_{00}, \|\cdot \|_{p})^{*}$ defined by \begin{equation} (\Phi (x))(y) = \sum_{k=1}^{\infty} x(k) y(k) \end{equation} is an isometric isomorphism. Hence $(c_{00}, \|\cdot \|_{p})^{*}$ is isometrically isomorphic to $\ell_{p'}$ for each $p\in [1, \infty ]$.

One way to show Theorem 1 is to take a standard proof of Theorem 2 below and make very small modifications when needed.

Theorem 2. We have the following.

$({\rm i})$ Let $p\in [1, \infty )$. The map $\Phi \colon \ell_{p'} \to \ell_{p}^{*}$ defined by \begin{equation} (\Phi (x))(y) = \sum_{k=1}^{\infty} x(k) y(k) \end{equation} is an isometric isomorphism. Hence $\ell_{p}^{*}$ is isometrically isomorphic to $\ell_{p'}$ for each $p\in [1, \infty )$, where $p'$ is the conjugate exponent of $p$.

$({\rm ii})$ The map $\Phi \colon \ell_{1} \to c_{0}^{*}$ defined by \begin{equation} (\Phi (x))(y) = \sum_{k=1}^{\infty} x(k) y(k) \end{equation} is an isometric isomorphism. Hence $c_{0}^{*}$ is isometrically isomorphic to $\ell_{1}$.

See here for the dual of $c_{0}$, here for the dual of $\ell_{1}$ and here for the dual of $\ell_{p}$ for $p\in (1, \infty )$. Proofs of all the statements in Theorem 2 can also be found in Example 1.10.4 and Theorem C.12 of An Introduction to Banach Space Theory by Megginson.

Another approach to obtain Theorem 1 is to directly combine Theorem 2 and Proposition 1 with the following result.

Lemma 1. Let $X$ be a Banach space and $M$ a dense subspace of $X$. Then $M^{*}$ is isometrically isomorphic to $X^{*}$.

Taking $X$ and $M$ as in the hypothesis of Lemma 1, define $\phi \colon X^{*} \to M^{*}$ by $\phi (f) := f\vert_{M}$. Then $\phi$ is clearly linear and satisfies $\|\phi\| = 1$ because $M$ is dense in $X$. Furthermore, the map $\phi$ is surjective by this result. Hence $\phi$ is an isometric isomorphism.

Question 3. Again, this question depends on the norm $c_{00}$ is provided. We have the following result.

Theorem 3. The dual space $(c_{00}, \|\cdot\|_{p})^{*}$ is reflexive for $p\in (1, \infty )$ and is not reflexive for $p\in \{1, \infty \}$.

If $p\in (1, \infty )$ then $\ell_{p}$ is reflexive by this result. That $\ell_{1}$ and $\ell_{\infty}$ are not reflexive can be seen here and here respectively. (See also Example 1.11.23 and Theorem C.14 of An Introduction to Banach Space Theory by Megginson.) Hence if $p\in [1, \infty ]$, then $\ell_{p}$ is reflexive if and only if $p \in (1, \infty )$. Since isomorphisms preserve reflexivity as can be seen here, the result now follows from Theorem 1.

Dean Miller
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