Let $H\_i, K\_i$ for $i=1,2,3$ be Hilbert spaces with horizontal exact sequences. Assume $T\_1, T\_2, T\_3$ have dense range, that $T\_1, T\_2, T\_3$ are trace class operators and that the squares commute. Assume that $S \colon H\_2 \to K\_2$ is a compact operator with dense range making the squares commute. 


$\require{AMScd}$
\begin{CD}
0 @>>> H\_1 @>f\_1>> H\_2 @>f\_2>> H\_3 @>>> 0\\
@V VV @V T\_1 VV @V T\_2 VV @V T\_3 VV @V VV \\
0 @>>> K\_1 @>>g\_1> K\_2 @>>g\_2> K\_3 @>>> 0
\end{CD}


Is it the case that there is a $C > 0$ such that $C \mathrm{Tr} |T\_2| \geq \mathrm{Tr} |S|$?


This is related to the question: [Are nuclear operators closed under extensions?](https://mathoverflow.net/questions/361225/are-nuclear-operators-closed-under-extensions)



Extensions of trace class operators
The answer is no. Take $H\_2 = K\_2$ to be an infinite dimensional Hilbert space and take $H\_1 = K\_1$ to be an infinite dimensional subspace whose orthocomplement $H\_3 = K\_3$ is also infinite dimensional. Let $f\_1 = g\_1$ be the inclusion and $f\_2 = g\_2$ the orthogonal projection. Let $T\_1: H\_1 \to K\_1$ and $T\_3: H\_3 \to K\_3$ be any trace class operators with dense range such that ${\rm ker}(T\_3)$ is infinite dimensional. Let $T\_2 = T\_1 \oplus T\_3$.


Let $S\_0: H\_2 \to K\_2$ be any operator which vanishes on ${\rm ker}(T\_3)^\perp$ and takes ${\rm ker}(T\_3)$ into $K\_1$. Then $S = S\_0 + T\_2$ will make the diagram (with $T\_1$ and $T\_3$ unaltered, I am sure you meant) commute. So clearly the trace of $|S|$ can be arbitrarily large. Also, ${\rm ran}(S)$ contains both ${\rm ran}(T\_1)$ and ${\rm ran}(T\_3|\_{{\rm ker}(T\_3)^\perp}) = {\rm ran}(T\_3)$, so the range of $S$ is dense.


(In the first version of this answer I didn't notice that the ranges had to be dense.)


