Prob. 8, Sec. 2.10 in Erwine Kreyszig's INTRODUCTORY FUNCTIONAL ANALYSIS WITH APPLICATIONS: The dual space of $c_0$ is $\ell^1$?

Question

Let $c_0$ be the subspace of $\ell^\infty$ consisting of all sequences of (real or complex ) numbers converging to $0$.

How to prove that the dual space of $c_0$ is (isomorphic to) $\ell^1$?

My effort:

Let $f \in c_0^\prime$. Then $f$ is a bounded linear functional with domain $c_0$.

Then, for any $x \colon= (\xi_j)_{j\in \mathbb{N}} \in c_0$, we have $$x = \sum_{j=1}^\infty \xi_j e_j,$$ where $e_j = (\delta_{ij})$ for each $j \in \mathbb{N}$. Each $e_j \in c_0$. Then using the boundedness (and the consequent continuity) of $f$, we can conclude that $$f(x) = \sum_{j=1}^\infty \xi_j f(e_j).$$ So $$\vert f(x) \vert \leq \sum_{j=1}^\infty \vert \xi_j \vert \ \vert f(e_j) \vert \leq \Vert x \Vert_\infty \sum_{j=1}^\infty \vert f(e_j)\vert, $$ showing that $$\Vert f \Vert \leq \sum_{j=1}^\infty \vert f(e_j)\vert. $$

Is what I've stated so far correct? How to proceed?

Answer 1

Salam brother Mahmud,

The procedure is extremely similar to sec 2.10-7 in the Erwin's book (almost copy and paste with slight modifications for the new space $c_0$) and goes as follows.

A Schauder basis (Sec. 2.3 in the Erwin book) for $c_0$ is $(e_k)$ where $e_k=(\delta_{kj})$ has 1 in the kth place and zeros otherwise;

$\begin{matrix} e_1=(1,0,0,0,\cdots )\\ e_2=(0,1,0,0,\cdots )\\ e_3=(0,0,1,0,\cdots )\\ \vdots \end{matrix}$

Then every $x\in c_0$ has a unique representation

$x= \sum_{k=1}^{\infty} \xi _{k} e_k$

We consider any $f\in c'_0$, where $c'_0$ is the dual space of $c_0$. Since $f$ is linear and bounded

$f(x)= \sum_{k=1}^{\infty} \xi _k \gamma _k \quad\text{where}\quad \gamma _k=f(e_k)\quad\quad\cdots \quad\ (1)$

are uniquely determined by $f$.

Consider $x_n=\left(\xi_k^{(n)}\right)$ with

$\xi_{k}^{(n)}= \left\{\begin{matrix} \left | \gamma_{k} \right | /\gamma_{k} & \text{if}\ \gamma _{k}\neq 0 \ \text{and} \ k\leq n\\ 0 & \text{if}\ \gamma _{k}= 0 \ \text{and} \ k> n \end{matrix}\right.$

By substitution this into (1) we obtain

$\left | f(x_n) \right |=\sum_{k=1}^{\infty} \xi _k^{(n)}\gamma _k = \sum_{k=1}^{n} \left | \gamma _k \right |\quad\quad\cdots \quad\ (2)$

we also have

$\left | f(x_n) \right | \leq \left \| f \right \| \underbrace{\left \| x_n \right \|}_{\max_{k} \left | \xi _{k}^{(n)} \right |=1}\quad\quad\cdots \quad\ (3)$

from (2) and (3) we obtain

$\sum_{k=1}^{n} \left | \gamma _{k} \right | \leq \left \| f \right \|\quad\quad\cdots \quad\ (4)$

Since $n$ is arbitrary, letting $n \to \infty $, we obtain

$\sum_{k=1}^{\infty } \left | \gamma _{k} \right | \leq \left \| f \right \|\quad\quad\cdots \quad\ (5)$

Hence $(\gamma_k)\in l^1$.

On the other hand, for every $b=\left ( \beta_k \right )\in l^1$ we can obtain a corresponding bounded linear functional g on $c_0$. In fact, we may define $g$ on $c_0$ by

$g(x)=\sum_{k=1}^{\infty } \xi _k \beta _k$

where $x=(\xi_k)\in c_0$. Then g is linear and boundedness follows from

$ g(x) = \left |\sum \xi _k \beta _k \right |\leq \sum \left | \xi _k \beta _k \right |\leq \sup_{j}\left | \xi _j \right | \sum \left | \beta _k \right |= \left \| x \right \|\cdot \sum \left | \beta _{k} \right |$

(sum from 1 to $\infty$). Hence $g\in c'_0$.

We finally show that the norm of $f$ is the norm on the space $l^1$. From (1) we have

$\left | f(x) \right | = \left |\sum \xi _k \gamma _k \right |\leq \sum \left | \xi _k \gamma _k \right |\leq \sup_{j}\left | \xi _j \right | \sum \left | \beta _k \right |= \left \| x \right \|\cdot \sum \left | \gamma _{k} \right |$

Taking the supremum over all $x$ of norm 1, we see that

$\left \| f \right \| \leq \sum \left | \gamma _{k} \right |$

From this and (5),

$\left \| f \right \| = \sum \left | \gamma _{k} \right |$,

which the norm on $l^1$. Hence this formula can be written $\left \| f \right \| = \left \| c \right \|_1$ where $c=(\gamma _j)\in l^1$. IT shows that the bijective linear mapping of $c'_0$ on to $l^1$ defined by $f \mapsto c=(\gamma_j) $ is an isomorphism.

Answer 2

What you have done is correct and you have shown $||f||_{c_0'} \le ||\hat f||_{\ell_1}$, where $\hat f$ is the sequence given by $\hat f_i = f(e_i)$. On the other hand, let $g_n\in c_0$ be defined so that

\begin{equation} (g_n)_i = \begin{cases} 1 & \text{ if } f(e_i) \ge 0, i\le n \\ -1 & \text{ if } f(e_i)<0 , i\le n\\ 0 & \text{ if } i >n \end{cases} \end{equation}

Then $||g_n||_{c_0} =1$ and

$$f(g_n) = \sum_{i=1}^n |f(e_i)|. $$

This implies

$$||f||_{c_0'} \ge \sum_{i=1}^n |f(e_i)|$$

for all $n\in \mathbb N$. Take $n\to \infty$ we have the opposite inequality. Thus

$$||f||_{c_0} = ||\hat f||_{\ell_1}.$$

Thus the map $c_0' \to \ell_1$, $f\mapsto \hat f$ is an isometric embedding. Also one can see that this is surjective, thus as a Banach space we have $c_0' \cong \ell_1$.