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!
Best Answer
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:
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.)
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 $c_0^{**}$.
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 $x\in c_0$.