Does the square function respect the strong operator topology on self-adjoint operators? A real-valued continuous function $f:\mathbb{R}\to\mathbb{R}$ is referred to as "strongly continuous", if for every Hilbert space $\mathcal H$ and every pointwise convergent net $a_\lambda\to a$ of self-adjoint operators in $\mathcal B(\mathcal H)$, one has $f(a_\lambda)\to f(a)$ pointwise. In other words, applying functional calculus with this function on self-adjoints is continuous in the strong operator topology.
Most of the time this concept pops up only briefly before the proof of Kaplansky's density theorem, in tandem with the fact that any function with at most linear growth (as $t\to\pm\infty$) is strongly continuous. (In the newest edition of Pedersen's book "C*-algebras and their automorphism groups", this is Proposition 2.3.2.)
My question: Is the square function $f(t)=t^2$ strongly continuous?
I am currently in the last stretches of teaching a course on operator algebras, and was recently not able to rigorously answer this question by one of my students. I strongly suspect that the answer is no, but I struggle to come up with a specific counterexample. I have looked through the usual textbooks, but did not see an answer there, either. Can anybody help me come up with a rigorous answer or possibly point me to a reference where this has been done?
 A: No, it's not.
Here is (one of?) the canonical example(s). Note that an example cannot be bounded, since multiplication is jointly sot-continuous on bounded sets.
Fix an orthonormal basis $\{e_n\}$ of $H$. Consider the set $\{\sqrt n\,e_n\}$. If $0\not\in\overline{\{\sqrt n\,e_n\}}^w$, then there exists an open neighbourhood
$$
W=\{x:\ |f_j(x)|<\delta,\ j=1,\ldots,N\}
$$
such that $W\cap\overline{\{\sqrt n\,e_n\}}^w=\varnothing$. Via the Riesz Representation Theorem this means that there exist $y_1,\ldots,y_N\in H$ and indices $j_n\in\{1,\ldots,N\}$ with
$$
\delta\leq |\langle \sqrt n\,e_n,y_{j_n}\rangle|,\qquad n\in\mathbb N. 
$$
Then
$$
\sum_{j=1}^N\|y_j\|^2=\sum_{j=1}^N\sum_{n=1}^\infty|\langle \sqrt n\,e_n,y_{j}\rangle|^2=\sum_{n=1}^\infty\sum_{j=1}^N|\langle \sqrt n\,e_n,y_{j}\rangle|^2\geq\sum_{n=1}^\infty\frac{\delta^2}{n}=\infty, 
$$
a contradiction.
So $0\in\overline{\{\sqrt n\,e_n\}}^w$. This means that there exists a net $\{n_j\}$ of integers such that $\sqrt{n_j}\,e_{n_j}\to0$ weakly.
Define operators $T_j\in B(H)$ by
$$
T_jx=\sqrt{n_j}\,\langle x,e_{n_j}\rangle\,e_{n_j}. 
$$
Then
$$
\|T_jx\|=|\sqrt{n_j}\,\langle x,e_{n_j}\rangle|\to0, 
$$
so $T_j\to0$ sot. Meanwhile, we have $T_j^2x=n_j\,\langle x,e_{n_j}\rangle\,e_{n_j}$. Let $\{n_k\}_{k\in\mathbb N}$ be a set of distinct integers such that, as sets, $\{n_k\}_{k\in\mathbb N}=\{n_j\}_j$. Put $x=\sum_k\tfrac1{n_k}\,e_{n_k}\in H$. Then
$$
\|T_j^2x\|=1
$$
for all $j$, showing that $T_j^2$ does not converge sot to $0$.
