Is reflexivity an open condition? Is the condition that a module is reflexive an open condition?  
That is, if $X$ is a smooth projective complex variety, $T$ a quasi-projective variety, and $F$ a finitely presented module on $X \times T$ that is $T$-flat, then we can form the locus $T' \subset T$ of points $t$ such that the restriction of $F$ to $X \times t$ is reflexive.
Is $T' \subset T$ open?
If not, is it locally closed?
Recall that a coherent sheaf $F$ is said to be reflexive if the natural map $F \to (F^{\vee})^{\vee}$ to the double dual is an isomorphism.
 A: This locus is indeed open. I will explain why using Kollar's "Hulls and Husks" (arXiv:0805.0576). More generally, this article studies in great detail when taking the double dual commutes with base change.
First, we may restrict to the open locus of $T$ where $F_t$ is torsion-free (because reflexive sheaves are torsion-free).
  Then, we choose an ample line bundle $H$ on $X$, and we compute the Hilbert polynomials relatively to $H$. The Hilbert polynomial $P(F_t)$ of $F_t$ is constant by flatness. From the exact sequence $0\to F_t\to F^{\vee\vee}_t\to F^{\vee\vee}_t/F_t\to 0$, we see that $F_t$ is reflexive exactly when $P(F^{\vee\vee}_t)=P(F_t)$, i.e. exactly when $P(F^{\vee\vee}_t)$ takes its minimal value. But the polynomial $P(F^{\vee\vee}_t)$ is constructible and upper semicontinuous by Proposition 28 (3) of Hulls ans Husks. This proves that this locus is open.
A: [EDIT: as Sasha points out, this does not answer the question. Please see it as a comment explaining why $X$ should be proper!]  
The answer is no in general if $X$ is not proper: take $X=\mathrm{Spec}\,\mathbb{C}[x]$,  $T=\mathrm{Spec}\,\mathbb{C}[t]$, and $F=$ the structure sheaf of $Z=\mathrm{Spec}\,(\mathbb{C}[t,x]/(1-tx))$. Then $T'$ is just the origin.  
Variant: if instead you take $T=\mathrm{Spec}\,\mathbb{C}[t,u]$ and $Z=\mathrm{Spec}\,(\mathbb{C}[t,u,x]/(u,1-tx))$ (i.e. the same $Z$ as before, but embedded in 3-space), then $T'$ is the union of the origin and the complement of the $t$-axis, hence not locally closed (but still constructible).  
Of course the point here is that $Z$ "goes to infinity" at the origin. I don't have a counterexample where $X$ is proper, but the main problem then is "taking the dual in the fibers", as in Sasha's comment above.
A: I'll assume you meant to say that $F$ was locally finitely presented or coherent in your second sentence. The locus where $F$ is reflexive is the complement of the union
of the supports of the kernel and cokernel of $F\to (F^\vee)^\vee$. This will be open.
Postscript As Sasha points out the argument is incomplete because it is not clear
that duals commute with restriction to the fibres. 
PPS Now Laurent has a counter example, so I guess that finishes it.
