The box topology on the set $\mathbb R^\infty$ is defined to have sub-basis of sets $U_1 \times U_2 \times \ldots$ for each $U_i \subset \mathbb R$ open. Observe this is different from the product topology which also demands all but finitely many $U_i = \mathbb R$.
The box topology is known to be badly behaved. For example the reasonable-looking function $x \mapsto (x,x,\ldots)$ is not continuous when the codomain carries the box topology. On the other hand functions like $x \mapsto (x,1,1,\ldots)$ are in fact box-continuous.
There is an easy rule to check if any function $F \colon \mathbb R\to \mathbb R^\infty$ is product-continuous. Simply write $F(x) = (f_1(x),f_2(x),\ldots)$ then check all $f_i \colon \mathbb R \to \mathbb R$ are continuous. This condition is necessary and sufficient.
Is there a simple characterisation of exactly which functions are box-continuous? Failing such a characterisation, could someone provide me with a large family of box-continuous functions? That family has to at least include all functions like $G(x) =(g_1(x),g_2(x), \ldots, g_n(x), a_1,a_2,\ldots)$ for $g_i\colon \mathbb R \to \mathbb R$ continuous and $a_i$ constants.