Weak convergence of a sequence Consider the sequence $(x_n)$ in $(c_0,\|.\|_{\infty})$, where $x_n=e_1+e_2+\ldots +e_n;  e_n=(0,0,\ldots,1,0,\ldots)$ for all $n\in \mathbb N$. I want to show that $(f(x_n))$ converges in $\mathbb K$ but $(x_n)$ does not converge weakly in $(c_0,\|.\|)$.
Let $f\in c_0^*$. Since $c_0^*$ is $(c_0,\|.\|)$ isometrically isomorphic to $\ell^1$, therefore there exists $y\in \ell^1$ such that $f(x)=\sum\limits_{n=1}^{\infty}x(n)y(n)$ for all $x\in c_0$. Then $f(e_n)=\sum\limits_{m=1}^{\infty}e_n(m)y(m)=y(1)+\ldots+y(m)\to \sum\limits_{m=1}^{\infty}y(m)$. 
But how to show that $(x_n)$ does not converge weakly in $c_0$? Please suggest anything.
 A: To show that a sequence in some topological space does not converge, one can for example exhibit a property that convergent sequences have, but the sequence in question doesn't. So in a metric space, one could show that the sequence in question is not a Cauchy sequence to show it doesn't converge. However, this strategy doesn't work - at least not without contortions - if the sequence in question converges in a larger space. E.g. to show that a Cauchy sequence in $\mathbb{Q}$ does not converge (in $\mathbb{Q}$) effectively means showing that its limit in $\mathbb{R}$ is irrational.
For the sequence in your question, we are in the latter scenario, that sequence converges in a larger space, but not in $c_0$ (with the weak topology), so it's hard to find a property that all weakly convergent sequences in $c_0$ have but $(x_n)$ doesn't.
So the most obvious angles of attack are a) to directly show that for all $\xi \in c_0$ the sequence $(x_n)$ doesn't converge weakly to $\xi$, or b) to find a larger Hausdorff space $S$ in which the sequence converges to some $s\in S \setminus c_0$. (The space $S$ being Hausdorff guarantees the uniqueness of limits. Since we're looking at sequences, the uniqueness of sequential limits - which is a weaker condition than Hausdorffness - would suffice, but the natural candidate for $S$ is a Hausdorff space, so we can just as well demand that.)
Let's first look at strategy b). Since $(\ell^1)^{\ast} \cong \ell^{\infty}$ and the natural inclusion $c_0 \hookrightarrow \ell^{\infty}$ coincides with the canonical injection $c_0 \hookrightarrow (c_0)^{\ast\ast}$ under the identifications of $\ell^1$ with $(c_0)^{\ast}$ and of $\ell^{\infty}$ with $(\ell^1)^{\ast}$, an obvious candidate for the larger space is $\ell^{\infty}$ in its weak$^{\ast}$ topology. As the weak topology on $c_0$ [$\sigma(c_0, \ell^1)$] coincides with the subspace topology induced on $c_0$ by the weak$^{\ast}$ topology on $\ell^{\infty}$ [$\sigma(\ell^{\infty}, \ell^1)$]; both are generated by the family of seminorms $\{ p_y : y \in \ell^1\}$, where
$$p_y(x) = \Biggl\lvert \sum_{m = 1}^{\infty} y(m)x(m)\Biggr\rvert;$$
and $\sigma(\ell^{\infty}, \ell^1)$ is a Hausdorff topology, this is indeed a suitable space.
It is easy to verify that $x_n \rightharpoonup \mathbb{1}$ in $\bigl(\ell^{\infty}, \sigma(\ell^{\infty}, \ell^1)\bigr)$ - essentially you have that proof in your question - and since $\mathbb{1} \in \ell^{\infty} \setminus c_0$, it follows that $(x_n)$ doesn't converge in $\bigl(c_0, \sigma(c_0, \ell^1)\bigr)$. For if we had $x_n \rightharpoonup \xi$ in $\bigl(c_0, \sigma(c_0, \ell^1)\bigr)$, then also $x_n \to \xi$ in $\bigl(\ell^{\infty}, \sigma(\ell^{\infty}, \ell^1)\bigr)$, and that would contradict the uniqueness of limits in Hausdorff spaces.
Let's now turn to strategy a). Pick an arbitrary $\xi \in c_0$. Then there is an $m_0$ such that $\lvert \xi(m)\rvert \leqslant \frac{1}{2}$ for all $m \geqslant m_0$. For all $y \in \ell^1$ with $y(m) = 0$ for $m < m_0$, we then have
$$\lvert\langle y, \xi\rangle\rvert = \Biggl\lvert \sum_{m = m_0}^{\infty} y(m)\xi(m)\Biggr\rvert \leqslant \sum_{m = m_0}^{\infty} \lvert y(m)\rvert\cdot \lvert \xi(m)\rvert \leqslant \frac{1}{2} \sum_{m = m_0}^{\infty} \lvert y(m)\rvert = \frac{1}{2}\lVert y\rVert_{\ell^1}.$$
Taking in particular
$$y(m) = \begin{cases}\;\; 0 &, m < m_0 \\ 2^{m_0 - m} &, m \geqslant m_0 \end{cases}$$
we have $\lVert y\rVert_{\ell^1} = 2$, and
$$\langle y, x_n\rangle = \sum_{m = m_0}^n 2^{m_0-m} = 2 - 2^{m_0 - n} \geqslant \frac{3}{2}$$
for $n > m_0$. Thus $\lvert\langle y, \xi - x_n\rangle\rvert \geqslant \frac{1}{2}$ for all $n > m_0$, and hence $x_n$ doesn't converge weakly to $\xi$.
Since $\xi \in c_0$ was arbitrary, that means $(x_n)$ doesn't converge weakly in $c_0$.
