Maybe this is well-known, but suppose you have an invertible sheaf $\mathcal{L}$ on a scheme $X$. If $X$ is a projective space (or even projective bundle over an integral scheme, by a similar argument…I think), then if $H^0(X, \mathcal{L})\neq 0$ and $H^0(X, \mathcal{L}^\vee)\neq 0$ we get that $\mathcal{L}\simeq \mathcal{O}_X$. This is really simple because $\mathcal{L}\simeq \mathcal{O}_X(n)$, so you get that the two hypothesis mean $n\geq 0$ and $n\leq 0$, so $n=0$.
My guess is this must be true more generally. Are there conditions on $X$ such that any invertible sheaf with the property that having non-trivial global sections of both it and its dual implies it is trivial?
Best Answer
In general, this fails. Take $X = \mathrm{Spec}A$ for $A$ a Dedekind domain. Then line bundles correspond to equivalence classes of Weil divisors. A line bundle corresponds to some formal sum $\sum n_i \mathfrak{p}_i$. But you can always find something in $A$ whose divisor is bigger than this (just take it highly divisible at those primes). So any line bundle will satisfy your condition, but if $A$ is not a UFD it will not necessarily be trivial: the Picard group is the class group of $A$. (See 8.2 of http://people.fas.harvard.edu/~amathew/CRing.pdf for an explanation of this isomorphism and a general discussion of line bundles on affine schemes.)
This is true for $X$ an integral proper scheme over an algebraically closed field $k$. (It's in Mumford's Abelian Varieties.). You don't need the isomorphism of the Picard group with $\mathbb{Z}$ which is only anyway true when $X$ is all of projective space.
Suppose $\mathcal{L}$ is such a line bundle. Then there is a nonzero morphism $\mathcal{O}_X \to \mathcal{L}$ and a nonzero morphism $\mathcal{L} \to \mathcal{O}_X$. This is what it means for the space of global sections of $\mathcal{L}$ and its dual to not vanish. So we get a composition $$\mathcal{L} \to \mathcal{L},$$ which is given by a global regular function that can't be zero (as it isn't zero at the generic point). This is necessarily an element of the field $k$, so an isomorphism of line bundles. Similarly the composition $\mathcal{O}_X \to \mathcal{O}_X$ is an isomorphism. (I suppose the key property here is that a nonzero endomorphism of a line bundle is an isomorphism, and this in turn is clear from $\Gamma(X, \mathcal{O}_X) = k$, which is a consequence of properness.) It follows that $\mathcal{L} \to \mathcal{O}_X$ is an isomorphism and $\mathcal{L}$ is trivial.
This doesn't really need $k $ to be algebraically closed (though then $X$ geometrically integral over $k$ would be necessary to make the argument work, unless I'm missing something).
This is false without integrality hypotheses. (Take a disconnected scheme, and a line bundle trivial on one piece but not the other.)