Given a bounded self-adjoint operator $H$ and a number $\mu$, consider the spectral projection $P_\mu$ onto the set $\{x|x\leq\mu\}$. I want to check if the map $\mu\mapsto P_\mu$ is strongly/norm continuous but I actually have no clue what a rigorous definition of a strongly/norm continuous map is. What would be the best way to approach this?
Any help/reference is welcome.
OK, so we are talking about a map $$ P:\mu \in {\mathbb R}\mapsto P_\mu \in B(H) $$ from the topological space ${\mathbb R}$ (with its usual topology) into the space $B(H)$ of all operators on the Hilbert space $H$.
Strongly continuous simply means that the map is continuous when $B(H)$ is seen as a topological space equipped with the misterious strong-operator-topology (SOT) (more later).
According to the standard definition, $P$ is continuous at a point $\mu _0\in {\mathbb R}$ iff, for every open set $U$ (relative to the SOT) containing $P_{\mu _0}$, there is an open subset $V\subseteq {\mathbb R}$ cointaining $\mu _0$, such that $$ P_V = \{P_\mu : \mu \in V\} \subseteq U. $$
Everything should be clear by now, except for what the heck is an open set relative to the SOT!
Before answering that, let me say that one of the most common methods to introduce a topology on a set $X$ (in our case $X=B(H)$) is via a generating collection of sets which we later want to be open.
So, suppose we are given a set $X$, and a collection $\mathscr U$ of subsets of $X$.
Should we want to define a topology $\mathscr T$ on $X$, relative to which the members of $\mathscr U$ are open, in other words, such that $$ \mathscr U\subseteq \mathscr T, $$ we are forced to accept that any set of the form $$ U_1\cap U_2\cap \ldots \cap U_n, \tag 1 $$ where the $U_i\in \mathscr U$, is also open.
So we make a first attempt (it will not work) to define $\mathscr T$ to be the collection of all finite intersections of sets from $\mathscr U$, namely all sets of the form (1).
The trouble is that $\mathscr T$ is not a topology: even though it is closed under finite intersections, it is not closed under unions.
In other words, we are also forced to accept as open, any set of the form $$ \bigcup_{\alpha \in A}V_\alpha , \tag 2 $$ where each $V_\alpha $ is of the form (1).
We then define (this time it will work) $\mathscr T$ to be the collection of all (beware of the tongue-twister) arbitrary unions of finite intersections of members of $\mathscr U$, namely, sets of the form (2).
It may now be shown that $\mathscr T$ is a topology, and in fact the smaller topology relative to which $\mathscr U\subseteq \mathscr T$. For that reason $\mathscr T$ is also known as the topology generated by $\mathscr U$.
This said, the SOT is the topology on $B(H)$ generated by the collection $\mathscr U$ of all subsets of the form $$ U_{u, v, r} = \big \{T\in B(H): \|T(u)-v\|<r\big \}, $$ as $u$ and $v$ range in $H$, and $r$ range in $(0,\infty )$.
Exercise. Let $X$ be a topological space and let $T:X\to B(H)$ be a function. Prove that $f$ is SOT-continuous at a point $x_0\in X$, iff, for every $u\in H$, the function $$ x\in X\mapsto T(x)(u)\in H $$ is continuous at $x_0$ (where $H$ has its usual topology defined by the norm).