I came across the following technical question, to which I could not - after some time of thinking - find an answer:
Let $\mathcal{U},\mathcal{H}$ be two real (in general infinite dimensional) separable Hilbert spaces. For some linear subspace $\bar{U} \subseteq \mathcal{U}$, let $\Pi_U$ denote the orthogonal projection on this subspace. The question is:
Is the mapping $T \mapsto \Pi_{\operatorname{ker}T}$ continuous from $L(\mathcal{U},\mathcal{H})$ to $L(\mathcal{U})$ when both spaces are endowed with the strong operator topology?
Any hints and thoughts on this are more than appreciated!
For a counterexample take $\mathcal{U} = \mathcal{H} = \mathbb{R}$ and let $T_n: \mathbb{R} \to \mathbb{R}$ be defined by $T_n x = \frac{x}{n}$. Then $T_n \to 0$ in the strong operator topology (and in fact even in operator norm). However $\ker T_n = \{0\}$ for all $n$ so for every $x \in \mathbb{R} \setminus \{0\}$ we have that $0 = \Pi_{\ker T_n} x \not \to \Pi_{\ker 0}x = x$.