Let $H$ be some infinite dimensional $Hilbert$ space. Now, let $B(H)$ be the set of all bounded linear operator over $H$ and let $seq :=\{\rho_{n}\}_{n=1}^{\infty}$ be a sequence of trace class operators in $B(H)$, the trace class operators form an ideal over $B(H)$.
-My first question is this. Under what conditions, if any, does the sequence $seq$ converge to a multiplication operator under the trace norm $\| A\|_{1} : Tr(|A|)= Tr(\sqrt{A^{\dagger}A} )$.
-My second question is a more specific version of the first. If the dynamics are generated by a contracting semigroup, i.e.
$$\rho_{n} = L_{n}\rho_{0}$$ where $L_{n}$ are contracting linear maps,
does the limit $\lim_{n\rightarrow \infty}\|\rho_{n}\|_{1}$ exists always? If so, what is the limiting operator and is it still trace class?
-My third question is essentially the second but for a particular case.
I have been working with the following operator. For $\psi(x) \in H = L^{2}(\mathbb{R})$ and $\sigma_{t}\in B(H)$ is defined as follows.
$$\sigma_{t}\psi(x) := \int_{\mathbb{R}} e^{-t(x-y)^{2}}K(x,y)\psi(y)dy$$.
Where $K(x,y) \in L^{2}(\mathbb{R}^{2})$ is a $Hilbert-Schmidt$ kernel. Note tha for $t=0$ this is a very tame integral transform. I am worried about the behviour as $t\rightarrow \infty$ in the trace norm sense. i.e. if the limit of $\sigma_{t}$ exists under $\| \|_{1}$, say $\sigma_{\infty}$, what is it? I am guessing that it should be some multiplication operator. $\lim_{t\rightarrow \infty}\|\sigma_{t}\|_{1} = ? .$
Thank you very much for your help.