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?

  • 1
    $\begingroup$ The answer to Question 1 is affirmative, here is an outline of the proof: Separability of the space of self-adjoint trace class operators over a separable Hilbert space. For more details about the density of the finite rank operators in the trace-class operators, you can also check out Reed and Simon's Functional Analysis (the first volume of their Methods of Modern Mathematical Physics series), in Chapter VI when they talk about trace-class operators. $\endgroup$
    – Bruno B
    Nov 21 at 20:59


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