Sequence does not converge weakly in $\ell^\infty$ Let $k\in\mathbb{N}$ and define $x^k\in\ell^\infty$ by
$$
x_n^k:=\begin{cases}1, & n\leqslant k\\0, & \textrm{otherwise}\end{cases}.
$$
I am searching for a proof that $(x_n^k)_{k\in\mathbb{N}}$ does not converge weakly in $\ell^\infty$.

I know that there (and also here) is a similar result (using Banach limits) for the sequence where the first $n$ entries are $0$'s and $1$'s otherwise. But here, its vice versa and I am not sure how to perform the proof.
Edit
I wonder, if the following is a valid argument (without referring to the given link at all):
We have $\ell^\infty=(\ell^1)'$ and an isomorphism is given by $T(x)(y)=\sum_{i=1}^\infty x_iy_i$ for $x=(x_i)\in\ell^\infty$ and $y=(y_i)\in\ell^1$.
Thus, for the given sequence $(x_n^k)_k$,
$$
T(x_n^k)(y)=\sum_{n=1}^k y_n\to a< \infty~\quad\textrm{as }k\to\infty
$$
for $y=(y_i)\in\ell^1$.
But, as far as I see, $a$ is not unique here, it depends on $y\in\ell^1$, doesn't it?
Thus, $(x_n^k)_k$ cannot converge weakly in $\ell^\infty$ as there is no (unique) limit with respect to weak* topology. In other words, it does not converge in the weak* topology, however, if it converged weakly this would imply weakly* convergence.

Let $y=(1,1,1,\ldots)$ and let $y^k=y-x^k$. This flips the zeros and ones. Then $(x^k)_k$ is weakly convergent if and only if $(y^k)_k$ is. This will turn your question into the linked question.
