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!

  • $\begingroup$ The problem is not that $y(u_n)\nrightarrow y(u)$, but the fact that $u\notin X$ ($u$ doesn't decay at infinity). $\endgroup$ – Jose27 Nov 18 '14 at 6:58
  • 1
    $\begingroup$ Since $X^*=\ell_1$, then $X^{**}=\ell_\infty$. Note that $X$ is separable, while $\ell_\infty$ is not separable. Therefore $X$ and $\ell_\infty$ are not isomorphic. $\endgroup$ – Norbert Nov 18 '14 at 10:41

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 $x\in c_0$.

  • $\begingroup$ I am sorry I have a question about 2. We have S is a bounded linear functional on $\ell_1$. But how can we get S is a bounded linear functional on $c_0^{*}$. We only have $\ell_1$ is isometrically isomorphic to $c_0^{*}$. They are not equal. Thank you! $\endgroup$ – Answer Lee Apr 12 '18 at 18:23

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. (proof taken from http://www.polishedproofs.com/non-reflexive-banach-space)


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 $c_0$.


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.