Let $\mathcal{H}$ be an infinite-dimensional separable Hilbert space and let $I: \mathcal{H} \rightarrow \mathcal{H}\otimes \mathcal{H} $ be the (non-linear) map given by the diagonal embedding $h \mapsto h\otimes h, \, h \in \mathcal{H}$.
Is $I$ weakly continuous, i.e. continuous with respect to the weak topologies on $\mathcal{H}$ and $\mathcal{H}\otimes \mathcal{H} $? If not, is $I$ at least weakly Borel measurable, i.e. measurable with respect to the Borel $\sigma$-algebras of the weak topologies on $\mathcal{H}$ and $\mathcal{H}\otimes \mathcal{H} $?
(Above $\otimes$ stands for the standard tensor product of Hilbert spaces.)
$I$ is not continuous relative to the weak topologies. The reason is that the map $$ \varphi _v:h\in H\mapsto \langle h\otimes h, v\rangle \in \mathbb R \tag 1 $$ is discontinuous for some choices of $v$ in $H\otimes H$. One such choice is to take an orthonormal basis $\{e_i\}_{i\in \mathbb N}$ of $H$ and set $$ v=\sum_{i=1}^\infty n^{-1}e_i\otimes e_i. $$ Observe that in this case $$ \varphi _v(h) = \sum_{i=1}^\infty n^{-1}\langle h\otimes h, e_i\otimes e_i\rangle = \sum_{i=1}^\infty n^{-1}\langle h, e_i\rangle ^2. \tag2 $$ Assuming by contadiction that $\varphi _v$ is cotinuous at zero for the weak topology, there would be a weak neighborhood $V$ of zero, such that for every $h$ in $V$, one would have that $$ |\varphi _v(h)|<1. $$ Observe however that every neighborhood of zero in the weak topology contains some finite codimensional subspace, say $K$. If $h$ is any nonzero vector in $K$, then $\lambda h$ lies in $V$ for every $\lambda $ in $\mathbb R$, so $$ \sum_{i=1}^\infty n^{-1}\lambda ^2\langle h, e_i\rangle ^2 < 1, $$ whence necessarily $$ \sum_{i=1}^\infty n^{-1}\langle h, e_i\rangle ^2 =0, $$ from where it follows that $h=0$, conradicting the choice of $h$.
On the other hand, $I$ is clearly norm continuous, hence a measurable map relative to the $\sigma$-algebra of Borel sets associated to the norm topology. However, this $\sigma$-algebra is the same as that of weak Borel sets.