Cohen-Macaulayness of the direct image of the canonical sheaf Let $Y$ be a normal projective variety and let $f:X\to Y$ be a desingularization. Define $\mathcal K_X=f_*\omega_X$, the Grauert--Riemenschneider canonical sheaf of $X$. It is independent of the resolution. Does this have some good properties? 
In particular, are there mild assumptions that ensure it is Cohen--Macaulay?  
 A: Let me expand Donu's answer a bit.  Say that $Y$ is normal for simplicity.  Then we always have:
$$f_* \omega_X \subseteq \omega_Y.$$
The sheaf $\omega_Y$ is Cohen-Macaulay if $Y$ is Cohen-Macaulay, but in general, it is not.  
On the other hand, $\omega_Y$ is always S2 (see for instance Kollár-Mori).  Furthermore, if $Y$ is normal, then $f_* \omega_X \subseteq \omega_Y$ is an isomorphism outside a set of codimension 2.  Hence 
Observation For normal $Y$, $f_* \omega_X$ will be S2 if and only if $f_* \omega_X = \omega_Y$.  This equality is equivalent to rational singularities if $Y$ is Cohen-Macaulay.
In other words, the only reasonable way to guarantee that $f_* \omega_X$ is Cohen-Macaulay is to require that $Y$ has rational singularities (at least if $Y$ is normal).
A: Yes, it behaves like the canonical sheaf for smooth varieties in many respects. For example, it satisfies Kodaira vanishing $H^i(Y, \mathcal{K}_Y\otimes L)=0$ when $L$ is ample when $i>0$.
To prove this, use  Grauert-Riemenschneider vanishing to rewrite this as $H^i(X,\omega_X\otimes f^*L)$, and then apply Kawamata-Viehweg to show that this is zero.
Regarding the last question, if $\mathcal{K}_X$ equals the (shifted) dualizing complex, then $X$ will be CM, and in fact it will have rational singularities. I'm not sure if this is the kind of answer you're looking for.
