**I edited a few small comments to the question in order to make it perhaps more comprehensible.**

Today I came across the following question in quantum mechanics.

In Quantum mechanics it is common to use different so-called pictures, which I thought were obviously equivalent, but I am not so sure anymore. So please let me introduce the framework: Let $X$ be a Hilbert space.

The Schrödinger dynamics teaches that in the Hilbert space norm $\left\lVert e^{itH}\psi-\psi \right\rVert_{X} \rightarrow 0$ as $t$ goes to zero.

For density operators one uses the nuclear norm (positive trace-class operators with unit trace) and has $$\left\lVert e^{itH}\rho e^{-itH}-\rho \right\rVert_{\text{nuclear}} \rightarrow 0.$$

I would like to understand whether these two converges agree in certain situations, i.e. assume that $\rho(\psi)=\langle \bullet, \psi \rangle \psi$ is the projection onto $\psi.$

Then one can ask whether uniform convergence is still equivalent:

$$\sup_{\left\lVert \psi \right\rVert=1} \left\lVert e^{itH}\psi-\psi \right\rVert_{X} \rightarrow 0 \Leftrightarrow \sup_{\left\lVert \psi \right\rVert=1} \left\lVert e^{itH}\rho e^{-itH}-\rho \right\rVert_{\text{nuclear}}\rightarrow 0,$$

i.e. both pictures should agree, is that true? Or does only one of the implications hold?

Clearly uniform convergence in the Schrödinger picture holds true iff $H$ is bounded. What about the other picture? Clearly continuity is sufficient, but is it also necessary?