First, some definitions:
A sequence $[x_n]$ in a normed linear space $X$ is said to converge weakly to an
element $x$ if for every $\phi \in X^*,\,\, \phi(x_n) \rightarrow \phi (x),$ (where$X^*$ is the conjugate space of $X$.)
As for $c_0$, let $c_0$ denote the space of all real sequences that converge to zero, with the norm $\| x\| = \sup_n |x(n)|.$
Now, I need to show that in order for a sequence $[x_n]$ to converge weakly to an element $x$ in the space $c_0$, it is necessary and sufficient that the sequence be bounded and that (for each $i$) we have $\lim_{n \rightarrow \infty} x_n(i) =x(i).$
Here's my attempt at a proof:
First direction: assume that the sequence is weakly convergent, prove the boundedness and pointwise convergence:
We know that any weakly convergent sequence is bounded. Also, every continuous linear functional $\phi$ in $c_0$ can be written in the following form for some suitable point $\alpha$ in $l^1$. $$\phi(x) = \sum_{i=1}^{\infty } \alpha(i) x(i)$$
Similarly, $$\phi_n (x) = \sum_{i=1}^{\infty } \alpha^\prime (i) x_n(i).$$ Now, due to weak convergence, we have $lim_{n \rightarrow \infty} \phi_n(x) = \phi(x) $. We pick $\alpha(i) = \alpha ^ \prime (i) = e(i) $, where $e(i)$ is the $i$th basis for $c_0$. Thus, we can conclude $lim_{n \rightarrow \infty} x_n(i) = x(i).$
Am I right so far? And if so, I'd appreciate any hint/help on the inverse direction of the proof. Thanks.
Best Answer
For the converse: Let $f \in c_0^*=\ell_1$. Then, there exists $z \in \ell_1$ such that $$f(y)=\sum_{k =1}^\infty y(k) z(k), \ y \in c_0.$$ Let $\varepsilon>0$. Pick a sufficiently large integer $N$ such that $$ \sum_{k>N} |z(k)| < ε/2M$$ where $M=\sup ||x_n||$. Since $x_n(k) \xrightarrow{n \to \infty} x(k)$ for all $k$, we get that $$ \lim_{n\to \infty} \sum_{k=1}^N |x_n(k)-x(k)| \ |z(k)|=0.$$ Therefore, \begin{align*} |f(x_n)-f(x)| &= |f(x_n-x)| \\ &= | \sum_{k=1}^\infty (x_n(k)-x(k)) z(k) | \\ &\leq \sum_{k=1}^N |x_n(k)-x(k)| \ |z(k)| + \sum_{k>N} |x_n(k)-x(k) \ |z(k)|. \end{align*} Conclude that $ \limsup |f(x_n)-f(x) | \leq ε$.