The space $c_0$ of sequences that converge to 0 is not reflexive

Question

Consider the Banach space $X$ of null sequence whose elements are complex sequence which converges to $0$. In addition the norm is defined as $$\|(a_1, \dots, a_n)\| := \sup_n |a_n|.$$ Show this space is NOT reflexive. Recall that the dual of $X$ is $l_1$. Also use the following fact. If $X$ is reflexive space and $(x_n)$ is a sequence in $X$ and for all $\psi \in X^*$, the sequence $\psi(x_n)$ has a limit in $\mathbb C$, then $x_n$ converges weakly to some $x \in X$.

Suppose that $X$ is reflexive. Consider the sequence $$u_n = (1, \dots, 1, 0, \dots) \in X,$$ where first $n$ entries are $1$. Then for any $y \in l_1$, we have $$y(u_n) := \sum_n y_n\times u_n = y_1+\dots+y_n \to \sum_n y_n < \infty,$$ since $y$ is absolutely convergent. Now it is only left to show that $u_n$ does not converges weakly to some $u \in X$ which I guess is $(1, 1, \dots)$. That is, for all $y \in l_1$, $$y(u_n) \nrightarrow y(u).$$ However, this seems correct to me. Where did I make mistake, please? Thank you!

Answer 1

Your space $X$ is usually denoted by $c_0$. To show that $c_0$ is not reflexive it is enough to find $f \in c_0^*$ for which $$\|f\|=\sup_{\|\{a_n\}\|=1}|f(\{a_n\})|$$ is not attained for any $\{a_n\}$. (for reflective spaces, such a sup is attained). Consider $f:c_0 \rightarrow \mathbb{C}$ defined by $$f(\{a_n\})=\sum_{n=1}^{\infty} \frac{a_n}{n!}$$ Notice that this function is well defined, since $\left|\frac{a_n}{n!}\right| \leq \left|\frac{1}{n!}\right|$ for large enough $n$, and since $\sum_{n=1}^{\infty} \frac{1}{n!}$ converges absolutely by the ratio test, then $\sum_{n=1}^{\infty} \frac{a_n}{n!}$ converges absolutely by the comparison test. It is clear that $f$ is linear, since

\begin{equation} \begin{split} f(\{a_n\}+\{b_n\}) & = \sum_{n=1}^{\infty} \frac{a_n+b_n}{n!}\\ & = \sum_{n=1}^{\infty} \frac{a_n}{n!}+\sum_{n=1}^{\infty} \frac{b_n}{n!}\\ & = f(\{a_n\})+f(\{b_n\})\\ \end{split} \end{equation}

\begin{equation} \begin{split} f(\alpha \{a_n\}) & = \sum_{n=1}^{\infty} \frac{\alpha a_n}{n!}\\ & = \alpha \sum_{n=1}^{\infty} \frac{a_n}{n!}\\ & = \alpha f(\{a_n\})\\ \end{split} \end{equation}

Notice that for an arbitrary $\{a_n\} \in c_0$ such that $\|\{a_n\}\|=1$, $|a_n| \leq 1$ for all $n$ and $|a_n|<1$ for some $n$ (it is because $\{a_n\}$ converges to $0$). Thus, we get that

\begin{equation} \label{eq:8} \begin{split} |f(\{a_n\})| & = \left|\sum_{n=1}^{\infty} \frac{a_n}{n!}\right|\\ & \leq \sum_{n=1}^{\infty} \left|\frac{a_n}{n!}\right|\\ & < \sum_{n=1}^{\infty}\frac{1}{n!}\\ \end{split} \end{equation}

Since the right side converges by ratio test, $\|f\| \leq \sum_{n=1}^{\infty}\frac{1}{n!}$.

Now let $\epsilon >0$. Since $\sum_{n=1}^{\infty}\frac{1}{n!}$ converges, $$\lim_{k \rightarrow \infty} \sum_{n=k}^{\infty}\frac{1}{n!}=0$$ Therefore, we can find $K$ big enough such that $$\sum_{n=K}^{\infty}\frac{1}{n!}< \epsilon$$ Therefore, we have that $$\sum_{n=1}^{\infty}\frac{1}{n!}=\sum_{n=1}^{K-1}\frac{1}{n!}+\sum_{n=K}^{\infty}\frac{1}{n!}<\sum_{n=1}^{K-1}\frac{1}{n!} + \epsilon$$ Thus, if we consider $\{a_n\}$ defined by $$ a_n = \begin{cases} 1 & n < K \\ 0 & n \geq K \\ \end{cases} $$ then $\{a_n\} \in c_0$ and $\|\{a_n\}\|=1$. Also, we get

$$\left|f(\{a_n\})\right|=\left|\sum_{n=1}^{\infty}\frac{a_n}{n!}\right|=\sum_{n=1}^{K-1}\frac{1}{n!} > \sum_{n=1}^{\infty}\frac{1}{n!}- \epsilon$$

Since $\epsilon$ was arbitrary, this means that we can find $\{a_n\}$ such that $\{a_n\} \in c_0$, $\|\{a_n\}\|=1$ and the norm of $f(\{a_n\})$ is as close to $\sum_{n=1}^{\infty}\frac{1}{n!}$ as we wish. Thus, it has to be that $$\|f\|=\sum_{n=1}^{\infty}\frac{1}{n!}$$ However, we showed that such supremum is not attained for any $\{a_n\} \in c_0$ such that $\|\{a_n\}\|=1$. Thus, $c_0$ is not reflexive.

Answer 2

Without using the notion of weak convergence, one can show directly that the canonical embedding $c_0\to c_0^{**}$ is not surjective. The steps are:

1. Recall or prove that the dual space of $c_0$ is $\ell_1$. A sequence $y\in \ell_1$ is identified with the linear functional $x\mapsto \sum x_k y_k$ (add complex conjugate on $y_k$ if the spaces are over complex scalars.)

2. Observe that the summation functional $S(y) = \sum_k y_k$ is a bounded linear functional on $\ell_1$, thus it is an element of $c_0^{**}$.

3. There is no element $c\in c_0$ that induces $S$. Indeed, suppose $x\in c_0$ is such that $\sum \overline{x_k} y_k = S(y)$ for every $y\in \ell_1$. Applying this to $y = e_n$, a standard basis vector, we get $x_n=1$. But if $x_n=1$ for all $n$, the sequence does not converge to $0$, contradicting $x\in c_0$.

Answer 3

Suppose that $y(u_n)\to y(u)$ for all $y\in\ell^1$ and some $u\in c_0$. For any $m\in\mathbb N$, consider $\delta_m\in\ell^1$, that is $\delta_m(x)=x(m)$, the $m^{\rm th}$ entry of $x$. Then $$ u(m)=\delta_m(u)=\lim_n \delta_m(u_n)=1. $$ But this is a contradiction, since such $u$ wouldn't belong to $c_0$. So we have proven that $u_n$ does not converge weakly in $c_0$.