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Let $A$ and $B$ be two bounded symmetric positive operators in Hilbert space, such that $A-B$ is trace class. If needed, $A$ and $B$ may be assumed reasonably "small", let's say, Hilbert-Schmidt.

Does there exist another symmetric operator $C$, $0 \le C \le A$, $0 \le C \le B$, such that both $A-C$ and $B-C$ are trace class?

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up vote 8 down vote accepted

Let $P=\left[\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right]$ and $Q(\phi)=\left[\begin{matrix} \cos^2(\phi) & \cos(\phi)\sin(\phi) \\ \cos(\phi)\sin(\phi) & \sin^2(\phi) \end{matrix}\right]$. Then $P$ and $Q$ are orthogonal projections and if $\phi\neq 0$, then the only operator $T$ which satisfies $0 \leq T \leq P$ and $0 \leq T \leq Q(\phi)$ is zero.

Now, set $A := \oplus_{n \in \mathbb N} n^{-1} P$ and $B:= \oplus_{n \in \mathbb N} n^{-1} Q(\pi/n)$.

Then, $A,B$ are Hilbert-Schmidt (and not trace-class), $A-B$ is trace-class and the $2$-dimensional phenomenon from above shows that every operator $C$ with $0 \leq C \leq A$ and $0 \leq C \leq B$ must be zero. In particular, $A-C$ and $B-C$ are not trace-class. Thus, the answer to your question is no.

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