# The dual space of $c$ is $\ell^1$

Here is what I know/proved so far:

Let $$c_0\subset\ell^\infty$$ be the collection of all sequences that converge to zero. Prove that the dual space $$c_0^*=\ell^1$$.

$$Proof$$: Let $$x\in c_0$$ and let $$y\in\ell^1$$. We claim that $$f_y(x)=\sum_{k=1}^\infty x_ky_k$$ is a bounded linear functional. Clearly $$f_y$$ is bounded since $$|f_y(x)|=\left|\sum_{k=1}^\infty x_ky_k\right|\le\sum_{k=1}^\infty |x_k||y_k|\le||x_k||_\infty\sum_{k=1}^\infty |y_k|=||x_k||_\infty||y||_1.$$ We can also easily see that $$f_y$$ is linear. Let $$x,z\in c_0$$ and $$y\in\ell^1$$ then $$f_y(x+z)=\sum_{k=1}^\infty (x_k+z_k)y_k=\sum_{k=1}^\infty (x_ky_k+z_ky_k)=\sum_{k=1}^\infty x_ky_k+\sum_{k=1}^\infty z_ky_k=f_y(x)+f_y(z)$$ and for $$\alpha\in\mathbb{R}$$ $$f_y(\alpha x)=\sum_{k=1}^\infty \alpha x_k=\alpha\sum_{k=1}^\infty x_k=\alpha f_y(x).$$

Let $$\varepsilon>0$$ and since $$y\in\ell^1$$, we know that $$\sum_{k=1}^\infty |y_k|$$ converges. So there exists $$N\in\mathbb{N}$$ such that whenever $$n>N$$ we have $$\sum_{k=n}^\infty |y_k|<\varepsilon.$$ Now define the following sequence $$x=\{x_k\}_{k=1}^\infty$$ as $$x_k=\begin{cases} \operatorname{sgn}(y_k),&\,k\le N\\ 0, & \,k>N \end{cases}.$$ Thus $$x\in c_0$$ and

\begin{align} \left|f_y(x)-||y||_1\right|&=\left|\sum_{k=1}^\infty x_ky_k-\sum_{k=1}^\infty |y_k|\right|\\ &=\left|\sum_{k=1}^N\operatorname{sgn}(y_k)y_k-\sum_{k=1}^\infty |y_k|\right|=\left|\sum_{k=1}^N |y_k|-\sum_{k=1}^\infty |y_k|\right|=\left|\sum_{k=N+1}^\infty |y_k|\right|<\varepsilon. \end{align}

So we conclude that
$$$$\ell^1\subseteq c_0^*.$$$$ Observe that the above argument also establishes that $$||f_y||_*=||y||_1$$.

Now let $$f$$ be any linear functional on $$c_0$$ and let $$\{e_k\}$$ be the sequence with a 1 in the $$k$$-th position and zero elsewhere. Then for any $$x\in c_0$$ we have $$|f(x)|=\left|f\left(\sum_{k=1}^\infty e_kx_k\right)\right|=\left|\sum_{k=1}^\infty f(e_k)x_k\right|\le\sum_{k=1}^\infty |f(e_k)|\,|x_k|\le||x||_\infty\sum_{k=1}^\infty |f(e_k)|.$$ Since $$f$$ is a bounded functional, we must have $$\sum_{k=1}^\infty |f(e_k)|$$ converging, otherwise $$f(x)$$ would be unbounded. Thus $$\{f(e_k)\}_{k=1}^\infty\in c_0$$ and we conclude that $$$$c_0^*\subseteq\ell^1.$$$$ Thus (1) and (2) tell us that $$c_0^*=\ell^1.$$

QUESTIONS: If we take $$c$$ to be the collection of sequences that converge to some real number and $$c^*$$ to be it dual space, I know that $$c^*=\ell^1$$ as well, but I am not sure how to prove it. Is it enough to observe that if $$x\in c$$ and that $$x\to x'$$ then $$x-x'\in c_0$$, therefore they have the same dual space? I am a little fuzzy here, obviously.

Also, can someone clean up my align environment. I can't figure out how to make it compile properly. The code looks fine on my LaTeX implementation, but it doesn't work here.

The last argument where you say that $f(x)$ would be unbounded does not seem valid since you only have $f(x)\leq \Vert x\Vert_\infty\sum_{k=1}^\infty|f(e_k)|$. If $\sum_{k=1}^\infty |f(e_k)|=\infty$, you don't get any absurd. You could proceed as follows (assuming $f\neq 0$): For every $n$, the sequence $x^n=(\text{sgn}(f(e_1)),\ldots,\text{sgn}(f(e_n)),0\ldots)$ is in $c_0$ and has norm $\leq 1$, so $\sum_{i=1}^n|f(e_i)|=|f(x^n)|\leq\Vert f\Vert$. This show that $(f(e_1),f(e_2),\ldots)\in \ell^1$.

Now, about you next question: First, verify that $c=c_0\oplus\mathbb{R}$. Then $c^*=c_0^*\oplus\mathbb{R}^*=\ell^1\oplus\mathbb{R}=\ell^1$, where the last isomorphism is given by $((x_1,x_2,x_3,\ldots,),\lambda)\mapsto(\lambda,x_1,x_2,x_3\ldots)$.

Luiz is right about your proof. To fix that, consider the following:

Let $f \in c_0^*$, then define $y = [y_1, y_2, ... ,y_n, ...]$ by $y_i = f(e_i), \forall i$. Observe that, if $f \in c_0^*$ then being a bounded linear functional we have that $\sup \{f(x) : x \in c_0^* \text{ and } \|x\| = 1 \} = M < \infty$. In particular, the limsup taken over the family of elements $\{x_n\}_{n=1}^{\infty}$, where $x_n = [\text{sgn}f(e_1) , \text{sgn}f(e_2), ... \text{sgn}f(e_n),\alpha, \alpha, ...]$ is finite. Thus $$\sum_{n=1}^{\infty} |y_i| = \limsup_{x_n} f(x_n) < \infty,$$ Consequently, $y \in \ell^{1}$ and we now have a method of defining a $y \in \ell^{1}$ from an $f \in c_{0}^*$. So that your mapping $\varphi : \ell^{1} \rightarrow c_{0}^*$ now has an inverse. Since you showed it preserves norms, and we now have it has an inverse it must be an isometry. Thus, you created an isometric embedding of $\ell^{1}$ onto $c_0^*$ in order to show the two vector spaces are "equal".

As for your later question, note that if you define $c_{\alpha} = \{ x \in \ell^{\infty} : x_n \rightarrow \alpha\}$ then this a vector space, if and only if $\alpha = 0$ for if $x,y \in c_{\alpha}$ then $x+y \in c_{2\alpha}$.

• Your insights were valuable. Thanks for sharing! Feb 18 '14 at 6:29
• perhaps you meant $0$ instead of $\alpha$ in the definition of $x_n$ Feb 6 '18 at 18:35

I guess you may need to show that

(i) the map $y \mapsto f_y$ is injective, and

(ii) the map $f \mapsto (f(e_k))_{k \geqslant 1}$ is also injective.