Let $A$ and $B$ be positive semidefinite finite dimensional matrices with unit trace such that $|A-B|_1\leq \varepsilon$ where one can choose any nonzero $\varepsilon$.
Let $\lambda_1\geq \lambda_2 \geq ... \geq \lambda_n$ be the eigenvalues of $A$ in decreasing order and let $\mu_1\geq \mu_2 \geq ... \geq \mu_n$ be the eigenvalues of $B$. Is there a upper bound one can place on
$$|\lambda_i - \mu_i|$$
for all choices of $i$? I can show the bound $1/2$ for small enough $\varepsilon$ but it feels not tight.
I assume that $A$ and $B$ are symmetric real matrices (or complex self-adjoint matrices). I assume that $|\cdot|_1$ denotes the nuclear norm.
Let $\rho_1 \geq \cdots \geq \rho_n$ denote the eigenvalues of $A-B$. Weyl's inequalities tell us that for all $j = 1,\dots, n,$ we have $$ \mu_j + \rho_n \leq \lambda_j \leq \mu_j + \rho_1 \implies\\ \rho_n \leq \lambda_j - \mu_j \leq \rho_1. $$ However, we have $$ |A-B|_1 = |\rho_1| + \cdots + |\rho_n| \leq \epsilon, $$ which implies that $|\rho_j| \leq \epsilon$ for all $j$, so that $\rho_n \geq - \epsilon$ and $\rho_1 \leq \epsilon$.
With that, we conclude that $|\lambda_j - \mu_j| \leq \epsilon$. This bound is tight, as can be seen by considering certain diagonal matrices $A$ and $B$.