Suppose $(M,\omega)$ is a compact symplectic manifold and $C$ a closed curve in it. Is there a Lagrangian submanifold containing $C$?
I have a sequence of $J\_i$-holomorphic maps from a disk to $M$, and all the maps are identical on the boundary of the disk. So, if I know that the image of the boundary is contained in a Lagrangian submanifold, I can apply Gromov convergence. 



Lagrangian submanifold containing a curve
Yes, such a Lagrangian submanifold does exist. The isotropic neighborhood theorem (see for instance p. 24 of Weinstein's *[Lectures on Symplectic Manifolds](http://books.google.com/books?id=VDZZLwaczj8C&printsec=frontcover&dq=weinstein+lectures+on+symplectic+manifolds&hl=en&ei=EyngTonwHYTDgQfo_fT8BQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CDAQ6AEwAA#v=onepage&q=weinstein%2520lectures%2520on%2520symplectic%2520manifolds&f=false)*), together with the fact that all symplectic vector bundles over the circle are trivial, implies that any closed curve in a symplectic $2n$-manifold has a neighborhood symplectomorphic to the standard product $(S^1\times(-\epsilon,\epsilon))\times B^{2n-2}(r)$, via a symplectomorphism mapping the curve to $S^1\times \{0\}\times\{0\}$. Here $B^{2n-2}(r)$ denotes the standard symplectic ball of dimension $2n-2$ and some small radius $r$. Now let $T$ be a Lagrangian torus in $B^{2n-2}(r)$ which contains the origin (for instance you could use a product of small-radius-cirles containing the origin in each of the $n-1$ factors of $\mathbb{R}^2$ in $\mathbb{R}^{2n-2}$). Then the image of $S^1\times 0\times T$ under the Weinstein symplectomorphism will be a Lagrangian submanifold containing your curve.


