Definition 2

— Let u be a bilinear mapping from $E \times F$ into G. We say that u is $\mathfrak{S}$-hypocontinuous if u is separately continuous and if it verifies one of the equivalent conditions a), b), c) of prop. 3.

The condition c) of prop. 3 shows that the notion of $\mathfrak{S}$-hypocontinuous bilinear mapping depends on $\mathfrak{S}$ only through the $\mathfrak{S}$-topology on $\mathscr{L}(E, G)$.

For every set $\mathfrak{T}$ of bounded subsets of F, we define similarly the notion of $\mathfrak{T}$-hypocontinuous mapping, by interchanging the roles of E and F in prop. 3. A separately continuous bilinear mapping u is said to be $(\mathfrak{S}, \mathfrak{T})$-hypocontinuous if it is both $\mathfrak{S}$-hypocontinuous and $\mathfrak{T}$-hypocontinuous.

Every continuous bilinear mapping from $E \times F$ into $G$ is $(\mathscr{S}, \mathscr{T})$-hypocontinuous for every pair $(\mathscr{S}, \mathscr{T})$ of sets of bounded subsets : for every neighbourhood $W$ of 0 in $G$, there exists a neighbourhood $U$ of 0 in $E$ and a neighbourhood $V$ of 0 in $F$ such that $u(U \times V) \subset W$; since every set $M \in \mathscr{S}$ is bounded, there exists $\lambda > 0$ such that $\lambda M \subset V$, and so

The converse is in general false (III, p. 47, exerc. 3).