Question about SOT and compact operators I need some help with functional analysis / Hilbert space theory. If you have a favorite text to recommend, please let me know~
Here is my question:
Given $v_t$ be the "squeeze operator" on $H=L^2[0, 1]$, where $v_t: L^2[0,1] \to L^2[0, \frac{2-t}{2}]$ acts on $f \in L^2[0,1]$ by squeezing the domain of the function. We have that $\{v_t\}$ for $t \in [0, 1]$ is a family of SOT continuous operators. I am wondering why given any $p \in \mathbb{K}(H)$, we have $\{ v_tpv_t^* \}$ is continuous in norm.
I found the following facts on a reference suggested by Wikipedia (Hilbert Space Operators in Quantum Physics), but I am not sure how to prove them or how I may use them...


*

*$T_n \to^{SOT} T$ implies that for any $p \in \mathbb{K}(H)$, we have $T_np \to Tp$ in norm.


Hints or suggestions would be greatly appreciated!
 A: This is essentially the definition of the strong operator topology; the SOT is the topology generated by the evaluation maps $\mathcal{L}(X,Y) \ni T \mapsto Tx \in Y$ (for a fixed $x$). We want all of these maps to be continuous (w.r.t. the topology induced by the norm on $Y$), or equivalently that $T_n \to T$ in this topology on $\mathcal{L}(X,Y)$ if and only if $T_nx \to Tx$ in the topology on $Y$.
Other good references off the top of my head: Folland Real Analysis, ch. 5; Rudin Functional Analysis, maybe Reed and Simon Functional Analysis.
EDIT: ok, I misunderstood the question a bit.
If $T_n \to T$ strongly, and $A$ is compact, we need to show that
$$
\sup_{x \in B_X}\left||(T_nA - TA)x \right|| \to 0
$$
where $B_X = \{x\in X: |x| = 1\}$. Equivalently,
$$
(*):\quad\quad\sup_{y \in A(B_X)}\left||(T_n - T)y \right|| \to 0.
$$
Now, we know that, for each $y$, $||(T_n - T)y|| \to 0$, but things could go wrong if this doesn't happen uniformly in $Y$. But fortunately, $A(B_X)$ is compact, so we have better control of the convergence.
In particular, given a fixed $\epsilon$ for each $y \in A(B_X)$, there is a neighborhood of $y$ and $N_y$ so that $||(T_n - T)\tilde y|| < \epsilon$ when $\tilde y$ is in this neighborhood, and when $n > N_y$. Then, because the set is compact, a finite number of these neighborhoods cover $A(B_X)$, and taking $N = \max\{N_y\}$ among the finite number of $N_y$'s associated to these neighborhoods demonstrates $(*)$.
