Displacment field for linear dielectric Consider the following statement:

In the case of a linear dielectric of arbitrary shape, when no free charges are present, in an external applied electric field $\vec E_0$, then  $\vec D=0$.

Is this statement always true (considering that $\nabla \times \vec D= \nabla \times \vec P$) and can it be proved? If it is false can you please provide a counter example.
Can a similar statement be made for the auxiliary field $H$ in terms of the magnetization $M$?  
 A: No, this is essentially never true. In a linear dielectric, $D=\varepsilon E$, so $D=0$ can only be true if $E=0$. In other words, the statement is saying that any linear dielectric always perfectly screens a uniform external electric field. That's not true. Conductors perfectly screen electric fields in their interior, but dielectrics don't (unless the permittivity goes to infinity).
A particularly simple example is the limit $\varepsilon \rightarrow \varepsilon_0$, in which case the dielectric acts like a vacuum and has no effect on the external field. So obviously D inside is nonzero.
Another simple example is a parallel-plate capacitor in which a dielectric slab fills most of the space (but doesn't touch the plates). That's equivalent to being in a uniform electric field. As you'll find in textbooks, the (E or D) field inside the dielectric is not zero in this situation.
A: *

*Adding a counter example for a finite volume dielectric solved in detail in these notes on BOUNDARY VALUE PROBLEMS WITH “CLASS-A”/LINEAR DIELECTRICS  (starting on pg.4 bottom through pg.10): a spherical (linear) dielectric in uniform external field. 

*Also adding an interesting bit on why dielectrics cannot perfectly screen electric fields.
In the absence of any free charges, bound charges in the dielectric's volume $\Omega$ are also null, $\rho_\text{free} = \rho_\text{bound} = 0$. Then from ${\vec \nabla} \cdot \vec{D} =0$, ${\vec \nabla} \times \vec{D} =0$ it follows that the potential within the dielectric must be harmonic, $\Delta V = 0$. Since free charges are absent on the boundary too, the potential must be continuous there, $V_{in} = V_\text{out}\Big|_{\partial\Omega}$, along with its normal derivatives, $\left(\frac{\partial V_\text{in}}{\partial n} - \frac{\partial V_\text{out}}{\partial n}\right)\Big|_{\partial\Omega} = \sigma_\text{free}\Big|_{\partial\Omega} = 0$. But if the displacement $\vec{D}$ were null in $\Omega$, so would the field $\vec{E}$, and the potential's normal derivative would have to vanish on the interior of the boundary. This would leave $\left(-\frac{\partial V_\text{out}}{\partial n}\right)\Big|_{\partial\Omega} = 0$ for a non-vanishing  field outside the boundary, which cannot be. 
The result is that there must be a non-vanishing bound/polarization charge density on the dielectric's boundary, $\sigma_\text{bound} = \vec{P}\cdot\vec{n}\neq 0$, and a non-vanishing displacement in the volume, $\vec{D} \neq 0$.
