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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)

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".

Ali Taghavi
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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 $T \in F'$.

Nik Weaver
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