For dissipation in harmonic oscillators one often takes the $L_k$ to be annihilation operators for modes $k$. These are not bounded operators!

Now, do you think $\dot{\rho}$ should be bounded? What if I write a state $$\rho(t)=\begin{pmatrix}e^{-t/\tau}&0\\0&1-e^{-t/\tau}\end{pmatrix}?$$ Then $$\dot{\rho}=-\frac{1}{\tau}\begin{pmatrix}e^{-t/\tau}&0\\0&-e^{-t/\tau}\end{pmatrix}.$$ I can choose any $\tau$ that I want and this will be a valid state, so I can make $\tau$ infinitesimally small, so I can make $\dot{\rho}$ arbitrariliy large. This seems to imply that one need not bound the operators $L_k$.

For dissipation in harmonic oscillators one often takes the $L_k$ to be annihilation operators for modes $k$. These are not bounded operators!

Now, do you think $\dot{\rho}$ should be bounded? What if I write a state $$\rho(t)=\begin{pmatrix}e^{-t/\tau}&0\\0&1-e^{-t/\tau}\end{pmatrix}?$$ Then $$\dot{\rho}=-\frac{1}{\tau}\begin{pmatrix}e^{-t/\tau}&0\\0&-e^{-t/\tau}\end{pmatrix}.$$ I can choose any $\tau$ that I want and this will be a valid state, so I can make $\tau$ infinitesimally small, so I can make $\dot{\rho}$ arbitrariliy large (I might need to set something like $t=0$ or $t=0.1\tau$ too but that's no problem; there is some time for which the derivative can be arbitrarily large). This seems to imply that one need not bound the operators $L_k$.