Can the pushforward of a very ample invertible sheaf under a birational morphism be reflexive? Let $f: Y \to X$ be a birational  morphism of  projective varieties.  Let $\mathcal{M}$ be a very ample invertible sheaf on $Y$.  Suppose also that:


*

*$f^{-1}$ is defined away from a single point $x \in X$.

*$f_* \mathcal{O}_Y = \mathcal{O}_X$.
Two questions:
(1) If $V$ is a set of global sections of $\mathcal{M}$ that generate $\mathcal{M}$, consider the induced evaluation map
$ V \otimes \mathcal{O}_X \to f_* \mathcal{M}$. 
Let $\mathcal{N}$ be the image of this map.  Is it possible for $\mathcal{N}$ to be reflexive?  (Full disclosure:  I would like the answer to be no.)
If we know that $X$ is smooth at $x$, or more generally that $(f_* \mathcal{M})^{\vee\vee}$ is invertible, then the answer   is no.  Let $E$ be the exceptional locus of $f$.  Then $E$ is positive-dimensional, so any hyperplane section meets $E$ nontrivially, and thus any section of $f_* \mathcal{M}$ must vanish at $x$.  But if $(f_* \mathcal{M})^{\vee\vee}$ is only reflexive, then I don't see how to generalise this argument.  
(2) 
Is it possible for $f_* \mathcal{M}$ to be reflexive?  If so, what is the weakest possible condition that will guarantee that $f_* \mathcal{M}$ is not reflexive?
My chief interest in (2) is that a negative answer to (2) would be a cheap way to get a negative answer to (1).  
 A: Hi Sue, I think it can be reflexive.
Let's consider $X = \text{Proj} k[x,y,u,v,t]/\langle xy - uv \rangle$.  This has only an isolated singularity at $x=y=u=v=0$ (it's the simplest non-Q-factorial singularity I know of).  Fix $U$ to be the regular locus of $X$.
(Blow up a divisor:) Set $\pi: Y \to X$ to be the blowup of the prime divisor $D = V(x,u)$.  This is a small resolution of $X$ (and also contains $U$ as an open set).  Lets define $F$ on $Y$ to be such that $O_Y(-F) = O_X(-D) \cdot O_Y$.  In other words, $F$ is the inverse image of $D$.  Notice that $\pi_* O_Y(-F) = O_X(-D)$ since $\pi$ is a small resolution.  Let me explain this point.
Choose $V \subseteq X$ an open set.  Then 
$$
\Gamma(V, \pi_* O_Y(-F)) = \Gamma(\pi^{-1} V, O_Y(-F)) = \Gamma((\pi^{-1} V) \cap U, O_Y(-F))
$$
since $\pi^{-1}V \setminus U$ is codimension 2 (see for example THIS ANSWER by Sándor Kovács).
But $\Gamma((\pi^{-1} V) \cap U, O_Y(-F)) = \Gamma(V \cap U, \pi_* O_Y(-F))$.  It follows that $\pi_* O_Y(-F)$ is determined away from the singularity, and is therefore S2 / reflexive.
Since $Y$ is smooth, $F$ is a Cartier divisor and we know that $-F$ is $\pi$-ample by construction.  Set $A = V(t)$ to be an ample Cartier divisor on $X$ (the choice of $A$ doesn't matter so much here).
(Define $\mathcal{M}$): It follows that $\mathcal{M} = O_Y(-F + n \pi^*A)$ is ample on $Y$ for $n \gg 0$.  
Then we know
$$
\pi_* O_Y(-F + n \pi^* A) = (\pi_* O_Y(-F) ) \otimes O_X(nA) = O_X(-D + nA).
$$
which is reflexive.  Note that you can also consider $-mF + mn \pi^* A$ to make  $\mathcal{M}^m$ as ample as you'd like.  The same sort of computation still holds.
Ok, so this gives a negative answer for (2).  
(Question 1.): Let's now consider (1).  Note $\pi_* \mathcal{M}$ is globally generated (since $n \gg 0$).  But on the other hand,
$$H^0(X, \pi_* \mathcal{M}) = H^0(U, \mathcal{M}) = H^0(Y, \mathcal{M})$$ again since $X \setminus U$ and $Y \setminus U$ are codimension 2.
Thus the global section of $H^0(Y, \mathcal{M})$ already globally generate $\pi_* \mathcal{M}$, so I don't think (1) works either.
Obviously if $X$ is Factorial then you won't run into this problem as you already pointed out.
