Are sequences in $\ell^1(\mathbb N_0)$ converging uniformly on convex weakly compact subsets of $c_0(\mathbb N_0)$ norm convergent? I think the question as expressed in the title should be clear. I do not know whether there is a known "characterization" of the weakly compact convex sets in $c_0(\mathbb N_0)$ but testing examples has lead me to conjecture that sequences in $\ell^1(\mathbb N_0)$ converging to zero uniformly on these sets also converge to zero in the norm topology. Is this conjecture correct/know/well-know/is there an explicit reference?
 A: The answer is yes.
Proof. Assume to the contrary that a sequence $(x_n)$ in $\ell^1$ converges, say to $0$, uniformly on convex weakly compact subsets of $c_0$, but is not norm convergent and hence not norm convergent to $0$.
Note that the sequence even converges uniformly on all weakly compact subsets of $c_0$, be they convex or not (since the closed convex hull of a weakly compact set is again weakly compact). So if we construct a sequence $(y_n)$ in $c_0$ that converges weakly to $0$ but such that $\langle y_n, x_n \rangle \not\to 0$, then we have a contradiction.
After replacing $(x_n)$ with a subsequence we may assume that $\|x_n\| \ge \varepsilon$ for some $\varepsilon > 0$ and all $n$.
Moreover, since $(x_n)$ converges, in particular, weakly to $0$, it also converges pointswise to $0$. So after replacing $(x_n)$ with yet another subsequence we may assume that, for all $n$, we have $\sum_{k=1}^{n-1} |x_n(k)| < \varepsilon/3$. Hence, there exist numbers $N_n \ge n$ such that $\sum_{k=n}^{N_n} |x_n(k)| \ge \varepsilon / 3$ for each $n$.
Now simply choose $y_n(k)$ to be zero for $k$ outside the set $\{n,\dots, N_n\}$ and to be the complex conjugate of the (complex) sign of $x_n(k)$ for $k$ inside this set. Then we have $\langle y_n, x_n \rangle \ge \varepsilon / 3$ for each $n$. Moreover, the sequence $(y_n)$ in $c_0$ is bounded and converges pointwise to $0$; hence, it also converges weakly to $0$.
A: Please take this post as an appendix to Jochen Glueck's answer. Let $X$ be a Banach space and $(f_n)$ be a sequence in $X^*$. It is fairly routine to show the equivalence of the following.

*

*$\sup\{|f_n(x)|:x\in A\}\to 0$ for every weakly compact $A\subseteq X$.

*$f_n(x_n)\to 0$ for every weakly null sequence $(x_n)$ in $X$.

*$f_n(x_n)\to 0$ for every weakly Cauchy sequence $(x_n)$ in $X$.

*$\sup\{|f_n(x)|:x\in A\}\to 0$ for every weakly precompact     $A\subseteq X$.

Now suppose, in addition, that $X$ does not contain an isomorphic copy of $\ell^1$. Then, the unit ball $B$ of $X$ is weakly precompact. Thus, (1-4) above is also equivalent to

*

*$\|f_n\| = \sup\{|f_n(x)|:x\in B\}\to 0$.

Clearly, $c_0$ is a Banach space that doesn't contain an isomorphic copy of $\ell^1$.
