13
$\begingroup$

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 $$ f_y(x+z)=\sum_{k=1}^\infty (x_k+z_k)y_k=\sum_{k=1}^\infty (x_ky_k+z_ky_k)=\sum_{k=1}^\infty x_ky_k+\sum_{k=1}^\infty z_ky_k=f(x)+f(z) $$ 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
\begin{equation} \ell^1\subseteq c_0^*. \end{equation} 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 \begin{equation} c_0^*\subseteq\ell^1. \end{equation} 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.

$\endgroup$
5
$\begingroup$

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 $((x_1,x_2,x_3,\ldots,),\lambda)\mapsto(\lambda,x_1,x_2,x_3\ldots)$.

$\endgroup$
3
$\begingroup$

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 $x+y \in c_{2\alpha}$.

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.