Concerning the sequential coreflexion $w_{seq}$ of the weak topology on a Banach space $X$ the following characterization can be proved.

**Theorem.** For a Banach space $X$ the following conditions are equivalent:

1) $X$ is reflexive;

2) $(X,w_{seq})$ is a locally convex topological vector space;

3) the addition operation $+:X\times X\to X$ is jointly continuous with respect to the topology $w_{seq}$.

*Proof.* (1)$\Rightarrow$(2) If $X$ is reflexive, then the closed unit ball $B$ of $X$ is compact in the weak topology. Moreover, it is Eberlein compact and hence Frechet-Urysohn, which implies that on each ball $n\cdot B$ the topology $w_{seq}$ induces the weak topology. Since each weakly convergent sequence is bounded, the topology $w_{seq}$ coincides with the topology of the direct limit $\varinjlim nB$ of the sequence $(n B)_{n\in\mathbb N}$, which implies that $(X,w_{seq})$ is a $k_\omega$-space. Now the continuity of the addition $+:nB\times nB\to 2nB$ in the weak topology implies the continuity of the addition in the topology $w_{seq}$ of direct limit $\varinjlim n\cdot B=(X,w_{seq})$. By the same reason, the multiplication map $X\times\mathbb R\to X$ is continuous with respect to the topology $w_{seq}$. So, $(X,w_{seq})$ is a linear topological space. Its local convexity can be proved using the local convexity of the weak topology and the coincidence of the topology $w_{seq}$ with the direct limit topology $\varinjlim nB$ of the sequence of compact convex sets.

(2)$\Rightarrow$(3) is trivial.

(3)$\Rightarrow$(1) Assumining that $X$ is not reflexive, we conclude that the closed unit ball endowed with the weak topology is not compact and hence not sequentially compact (by the classical Eberlian-Smulian Theorem). Consequently, $X$ contains a non-reflexive separable Banach subspace $Y$. Assuming that the addition operation is continuous with respect to the topology $w_{seq}$ on $X$, we conclude that it is continuous with respect to the topology $w_{seq}$ on $Y$ and $(Y,w_{seq})$ is a topological group.

The separability of $Y$ implies that the closed ball $nB_Y$ of radius $n$ is metrizable and separable in the topology $w_{seq}$ which coincides with the weak topology on $nB$. Consequently, the metrizable separable space $(nB,w_{seq})$ has countable base $\mathcal B_n$ of the (weak) topology.
The union $\mathcal B=\bigcup_{n\in\mathbb N}\mathcal B_n$ is a *countable $cs$-network* at zero of the space $(Y,w_{seq})$. The latter means that for any sequence $\{x_n\}_{n\in\omega}\subset X$ that converges to zero in the topology $w_{seq}$ and any neighborhood $U\in w_{seq}$ of zero there exists a set $B\in\mathcal B_n$ such that $0\in B\subset U$ and $B$ contains all but finitely many points $x_n$.

By a result of Banakh and Zdomskyy, a sequential topological group having a countable $cs$-network at zero is either metrizable of contains an open $k_\omega$-subgroup. But $(Y,w_{seq})$ is neighter metrizable nor contains an open $k_\omega$-subgroup. This contradiction shows that $(Y,w_{seq})$ is not a topological group and the addition is discontinuous.

**Remark.** Some topologies near to $w_{seq}$ have been considered in the paper [T.Banakh, *On topological classification of normed spaces endowed with the weak topology or the topology of compact convergence*], published in this book.