Let us define the 1-norm or nuclear norm by the following expressions where $\dagger$ represents the conjugate transpose of a matrix.
$$||A||_{1} = \text{Tr}(\sqrt{A^\dagger A})$$
The one norm represents the sum of the absolute eigenvalues of $A$.
Now, choose $\delta > 0$. We find two positive definite matrices $\rho$ and $\sigma$ that satisfy
$$||\rho – \sigma||_1 \leq \delta$$
Can one say anything about $||\log(\rho) – \log(\sigma)||_1$? Intuitively, it seems that for any $\epsilon > 0$, I should be able to find a $\delta$ such that if $||\rho – \sigma||_1 \leq \delta$, then $||\log(\rho) – \log(\sigma)||_1 \leq \epsilon$. However, I'm not sure if this is true or how to proceed with a proof.
Best Answer
It is impossible to find such a $\delta$ that is independent of $\rho$ or $\sigma$ since the matrix logarithm fails to be uniformly continuous over its domain.
For instance: if we take $$ \rho = e^{-t} I, \quad \sigma = e^{-2t} I $$ for some $t > 0$, then we compute $\|\rho - \sigma\|_1 = \frac{1 - e^{-t}}{e^t}n$. On the other hand, we compute $\|\log(\rho) - \log(\sigma)\|_1 = nt$. It follows that $\frac{\|\log(\rho) - \log(\sigma)\|_1}{\|\rho - \sigma\|_1}$ can be made arbitrarily large.
On the other hand, I believe that if we restrict the domain of the logarithm (for instance, by imposing a positive lower bound on the real part of the eigenvalues of $A$), then we can ensure uniform continuity over this restricted domain.