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

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?

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

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

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 $c_0' \cong \ell_1$.

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.