$\{T = A^*A - AA^*: A \in \mathcal B(\mathcal H)\}$ is closed in the norm topology but $\{T = A^*A-AA^*: A \in \mathcal K(\mathcal H)\}$ is not I would like to show that:

The set of all self-commutators is closed under the norm-topology, while the set of all self-commutators of compact operators is not so.

in the setting of infinite-dimensional, separable Hilbert spaces. Recall that an operator $T \in \mathcal B(\mathcal H)$ is said to be a self-commutator if there exists an operator $A \in \mathcal B(\mathcal H)$ such that $$T = [A^*,A] = A^*A- AA^*$$
Also, $[A^*,A]$ is called the self-commutator of $A$.
The norm topology is (clearly) metrizable, and so it is enough to check sequential closedness.

*

*For the first claim, i.e., $\{T = A^*A - AA^*: A \in \mathcal B(\mathcal H)\}$ is closed in the norm topology - consider sequences $\{T_n\}_{n\ge 1}$ and $\{A_n\}_{n\ge 1}$ in $\mathcal B(\mathcal H)$, such that $T_n = [A_n^*,A_n]$, and $T_n \xrightarrow{n\to\infty} T$ in the norm topology, i.e., $\|T_n - T\| \xrightarrow{n\to\infty} 0$. What is a suitable candidate $A \in \mathcal B(\mathcal H)$ that may satisfy $T = [A^*,A]$? Even though the sequence $\{T_n\}_{n\ge 1}$ is assumed to converge in norm, it is not clear that $\{A_n\}_{n\ge 1}$ converges also.


*For the second claim, it is enough to find a counterexample, i.e., specific sequences $\{T_n\}_{n\ge 1}$ and $\{A_n\}_{n\ge 1}$ of compact operators, such that $T_n = [A_n^*,A_n]$ such that $T_n \xrightarrow{n\to\infty} T$ in the norm topology, but $T$ is not a self-commutator.
I haven't made much progress beyond what's stated above, and I'll appreciate any help. Thanks!
 A: One of the conditions in Radjavi's 1966 Indiana J. of Math (formerly J. Math. Mech.) paper is that a selfadjoint operator $T$ is a self-commutator if and only if the essential spectrum of $T$ has a non-negative and a non-positive element ($0$ satisfies both conditions, which in particular proves that all selfadjoint compact operators are self-commutators).
Suppose that a sequence of selfadjoint operators $\{T_n\}$ converges to $T$, and assume that $T$ is not a self-commutator. This means by the above that there exists $c>0$ such that $\sigma_{\rm ess} (T)\subset[c,\infty)$  or $\sigma_{\rm ess} (T)\subset(-\infty,-c]$. By replacing $T$ with $-T$ if necessary, we may assume without loss of generality that we are in the first case; that is, $\sigma_{\rm ess} (T)\subset[c,\infty)$. If we denote by $\pi$ the quotient map onto the Calkin algebra $\mathcal C$, this is the same as saying that $\pi(T-cI)\geq0$.
Let $n_0\in\mathbb N$ such that $\|T-T_n\|<c/2$ for all $n\geq n_0$; since $\pi$ is contractive, we also have $\|\pi(T)-\pi(T_n)\|<c/2$. Working on any faithful representation of $\mathcal C$, for a unit vector $x$ we have
\begin{align}
\langle \pi(T_n)x,x\rangle&=\langle\pi(T)x,x\rangle+\langle(\pi(T_n)-\pi(T))x,x\rangle\\[0.3cm]
&\geq\langle c\,x,x\rangle-\langle (c/2)\,x,x\rangle\\[0.3cm]
&=\langle (c/2)\,x,x\rangle.
\end{align}
This shows that for all $n\geq n_0$ we have $\pi(T_n)\geq\frac c2\,I$, which means that $\sigma_{\rm ess}(T_n)\subset[\frac c2,\infty)$. By Radjavi's criterion, $T_n$ is not a self-commutator. Thus we have shown that a norm-limit of self-commutators is a self-commutator.
In the case of self-commutators of compact operators, it may be the case that a limit of self-commutators of compact operators is not a self-commutator of compact operators. This follows from Theorem 1 in Fan-Fong (Proc. AMS 80, 1, 58-60), where a necessary and sufficient condition for a selfadjoint compact operator to be a self-commutator of compact operators is that the positive and negative parts have the same trace. For instance consider diagonal operators
$$
K_n=\big(-1,\overbrace{\frac1n,\ldots,\frac1n}^{n\ \text{ times}},0,\ldots\big).
$$
Then each $K_n$ is a selfadjoint finite-rank operator, with $\operatorname{Tr}(K_n^+)=1=\operatorname{Tr}(K_n^-)$, so it is a self-commutator by Theorem 1 in Fan-Fong. But $K_n\to K$ in norm, where
$$
K=\big(-1,0,0,\ldots\big)
$$
and $\operatorname{Tr}(K^+)=1\ne0=\operatorname{Tr}(K^-)$, so $K$ is not a self-commutator again by Theorem 1 in Fan-Fong.
