Weak continuity of diagonal embeddings of Hilbert spaces 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.)
 A: $I$ 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.
