Suppose I have column vectors $u, v$ in a Hilbert space. I can define rank-1 matrices $uu^*$ and $vv^*$, where $^*$ denotes the transpose conjugate. If it is the case that
$$\|uu^* - vv^*\|_1 \leq \varepsilon,$$
where $\|\cdot \|_1$ is the trace norm or nuclear norm or Schatten 1-norm defined as $\|A\|_1 = \text{Tr}(\sqrt{A^*A})$, can one say something about an upper bound for
$$\|u - v\|?$$
Here $\|\cdot \|$ is the Euclidean 2-norm for vectors.
In my case, I have the additional knowledge that $uu^*$ and $vv^*$ are positive semidefinite with unit trace if that helps with the proof.
The most you can say is $\|u-v\|\le 2$.
First of all, notice that $Tr(uu^*)=1\implies \|u\|=1$, and the same holds for $v$. As a consequence, $$\|u-v\|\le \|u\| + \|v\| = 2. $$
Take now $v = -u$. We have that $$ vv^* =uu^* \implies \|vv^*-uu^*\| = 0 \le \varepsilon $$ for any $\varepsilon>0$, but $$ \|u-v\| = \|2u\| = 2. $$