Let $A$ be a trace class operator on a Hilbert space. I wonder if there is an estimate of the form $$ |\log \det (I + A)| \le C\|A\|_1, $$ for some constant $C$, where the norm on the right is the trace norm?
If there is a short proof I'm of course interested in it, otherwise a reference would be great.
For general $A$, it is not true even in finite dimension in the $1\times1$ case: because that would imply the inequality $|\log(1+t)|\leq C|t|$, and we can take $t=-1+\varepsilon$ to get the (false) inequality $|\log\varepsilon|\leq C|1-\varepsilon|$ for all $\varepsilon>0$.
For $A$ positive, though, if the eigenvalue sequence of $A$ is $\{\lambda_j\}$, with $\lambda_j\geq0$ for all $j$, $$ |\log\det(I+A)|=\sum_j\log(1+\lambda_j)\leq\sum_j\lambda_j=\|A\|_1. $$
If you know that $\|A\|<\delta<1$, then (using Plemelj's formula) $$ |\log\det(I+A)|=\left|\sum_{n=1}^\infty(-1)^{n-1}\frac{\mbox{Tr}(A^n)}n\right|\leq\sum_{n=1}^\infty\frac{|\mbox{Tr}(A^n)|}n \leq\sum_{n=1}^\infty\frac{\|A\|_1\,\|A\|^{n-1}}n\\ \leq\|A\|_1\,\sum_{n=1}^\infty\frac{\delta^{n-1}}n=\left(-\frac1\delta\,\log(1-\delta)\right)\,\|A\|_1. $$