Thank you in advance for reading this question, and your thoughts.
I am working with a family of operators $(A_{s,\alpha})_{s\geq 0,\alpha>0}$ in the space $B(\mathcal{H})$ (the space of bounded linear operators on a complex separable Hilbert space $\mathcal{H}$) depending on two parameters $s\geq 0$ and $\alpha>0$. This family is such that for fixed $\alpha>0$, the mapping $[0,\infty)\ni s\mapsto A_{s,\alpha}\in B(\mathcal{H})$ is a strongly continuous semigroup , and $$(\star)\quad \forall\, s>0,\,\forall\, \alpha>0:\quad \big|\big|A_{s,\alpha}\big|\big|_{B(\mathcal{H}}\leq 1\,, $$ i.e. the family of operators $(A_{s,\alpha})_{s\geq 0,\alpha>0}$ lives in the norm-closed unit ball of $B(\mathcal{H})$. Next, the pre-dual of $B(\mathcal{H})$ (up to isometric isomorphism) is $B_1(\mathcal{H})$, the trace-class operators on $\mathcal{H}$. This defines a weak-star topology on $B(\mathcal{H})$, and by the Banach-Alaoglu theorem, the norm-closed unit ball of $B(\mathcal{H})$ is weak-star compact.
Question 1: I think that since $\mathcal{H}$ is separable, it follows that $B_1(\mathcal{H})$ is separable, and hence the norm-closed unit ball of $B(\mathcal{H})$ is weak-star sequentially compact?
Next, assuming an affirmative answer to Question 1, let us fix an arbitrary sequence $\alpha_n\to 0^+$. Then for any fixed $s\geq 0$, $(\star)$ above implies the sequence of operators $(A_{s,\alpha_n})_{n\in \mathbb{N}}$ has a weak-star convergent subsequence $(A_{s,\alpha_{n_k}})_{k\in \mathbb{N}}$. Unfortunately, however, this subsequence may depend on the particular $s\geq 0$.
Question 2: Is it possible to show that there exists a subsequence $(\alpha_{n_j})_{j\in \mathbb{N}}$ of the above $\alpha_n\to 0^+$ so that for all $s\geq 0$ the sequence of operators $(A_{s,\alpha_{n_j}})_{j\in \mathbb{N}}$ converges in the weak-star topology in $B(\mathcal{H})$? If not, are there some "natural" assumptions one can make on the $\alpha$-dependence of the operators such that this works?
