Let $F\subset B(H)$ be a finite dimensional subvector space of the space of all bounded operators on a Hilbert space.


**Question:** Is there an upper bound for $$\{|tr(T)| \text{where} \quad T\in F\quad \text{is a trace class operator of unit norm}\}$$


An indirect but relevant motivation for this question is mentioned in the "Motivation" part of this post:


[Irrational closed orbits of vector fields on $S^2$(Limit cycles and trace formula)](https://mathoverflow.net/questions/332979/irrational-closed-orbits-of-vector-fields-on-s2limit-cycles-and-trace-formul)


**Remark:** I have already learned from a specialist that the answer is "negative" if in this question we replace "trace class operators" with "Fredholm operators" and "trace" with "Fredholm index".



Trace class operators in the unit ball of a finite dimensional subvector space of $B(H)$
If I understand the question correctly, the answer is yes, there is an upper bound. The set of trace class operators $TC(H)$ is itself a linear subspace of $B(H)$, so $F' = F \cap TC(H)$ is a finite dimensional linear subspace of $TC(H)$, and we are interested in $\sup\{|{\rm tr}(T)|: T \in F', \|T\| = 1\}$. Since $F'$ is finite dimensional, the restrictions of the trace norm and the operator norm to $F'$ are equivalent. This means that there is a constant $C$ such that $|{\rm tr}(T)| \leq C\cdot \|T\|$ for all $T \in F'$.


