Recall that $\ell^1 (\mathbb N)$ has Schur's property, that every weakly convergent sequence is strongly convergent, i.e. convergent with respect to the norm $||\cdot||_1$. However, the weak topology on an infinite dimensional Banach space is coarser than the strong topology, so there must exist a weakly convergent net that does not converge strongly.
In particular, the weak closure of the unit sphere is the unit ball, so there must exist a net $\{
x^\alpha \}_{\alpha \in I}$ such that $||x^\alpha||_1 = 1$ for all $\alpha \in I$ but $\langle x^\alpha,y \rangle \to 0$ for every bounded sequence $y \in \ell^\infty (\mathbb N)$.
I am interested in constructing an explicit net satisfying the above.
I suspect that descriptive set theory ideas will be helpful, and also that a totally ordered set $I$ won't work. The first idea that I had was spreading the mass of $x^\alpha$ out to infinity and then bringing it back, basically an infinite version of a typewriter,
\begin{align}
&(1, 0, 0, 0 \dots), (0, 1, 0, 0, \dots), (0, 0, 1, 0, \dots), \dots \\
&(\frac12, \frac12, 0, 0, \dots), (0, \frac12, \frac12, 0, \dots), (0, 0, \frac12, \frac12, \dots), \dots \\
&(\frac13, \frac13, \frac13, 0, \dots), (0, \frac13, \frac13, \frac13, \dots), \dots
\end{align}
indexed appropriately by the ordinal $\omega^2$. This converges pointwise to zero, but unfortunately it doesn't converge weakly, as the evaluation of the constant sequence $(1, 1, 1, \dots)$ on this net is constantly $1$. So the next idea was to play around with signs, indexing by $2^{< \mathbb N}$ to determine where to put $+/-$ signs on a sequence. $2^{< \mathbb N}$ has a natural partial ordering by initial segments, however it is not a directed set.
Best Answer
Here's a fairly simple explicit example. Let our index set $I$ be the set of finite partitions of $\mathbb{N}$ ordered by refinement ($P\leq Q$ if every element of $Q$ is a subset of some element of $P$). Given $P\in I$, pick some element $A\in P$ which has at least two elements $i,j$ (if you like, you can do this explicitly, choosing the first non-singleton $A\in P$ when you order the elements of $P$ by their first elements and letting $i$ and $j$ be the first two elements of $A$). Now define $x^P$ as the sequence such that $x^P_{i}=\frac{1}{2}$, $x^P_j=-\frac{1}{2}$, and all other terms are $0$.
Clearly each $x^P$ is in the unit sphere of $\ell^1(\mathbb{N})$. To see it converges weakly to $0$, let $y\in \ell^\infty(\mathbb{N})$ and $\epsilon>0$. Since $y$ is bounded, we can cover its image by finitely many arbitrarily small balls. In particular, there is a finite partition $P$ of $\mathbb{N}$ such that for each $A\in P$, the set $\{y_k\}_{k\in A}$ has diameter at most $\epsilon$. Now suppose $Q\in I$ refines $P$. Then $$\langle x^Q,y\rangle=\frac{1}{2}y_i-\frac{1}{2}y_j$$ for some $i$ and $j$ which are in the same element of $Q$, and thus also in the same element of $P$. But by our choice of $P$, this means $|y_i-y_j|\leq\epsilon$, so we conclude that $$|\langle x^Q,y\rangle|\leq \epsilon/2.$$ Since this holds for every $Q$ refining $P$ and $\epsilon$ is arbitrary, this proves that $\langle x^P,y\rangle\to 0$.