Weinstein Lagrangian neighborhood theorem

Choose a compatible $J$ by [F1]. Then $J(TL)$ is a Lagrangian complement to $TL$: it is Lagrangian because $J$ preserves $\omega$, and if $Ju\in TL$ then $0=\omega (u,Ju)=g_J(u,u)$ forces $u=0$. The map $J(TL)\to T^{\ast }L$, $w\mapsto {\alpha }_w$ with ${\alpha }_w(u)=\omega (u,w)$, is an isomorphism.

At the zero section, $T(T^{\ast }L)=TL\oplus T^{\ast }L$ and ${\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}((u,\alpha ),(v,\beta ))=\beta (u)-\alpha (v)$. Hence step 1.1 gives a symplectic bundle isomorphism $TM{\mathrm{\mid }}_L\to T(T^{\ast }L){\mathrm{\mid }}_L$ equal to the identity on $TL$. Use tubular neighbourhoods from [F3] to realize it as the differential of a diffeomorphism $h$ between neighbourhoods, fixed on $L$.

The forms $h^{\ast }\omega$ and ${\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}$ agree as tensors along the zero section. Their convex interpolation is symplectic after shrinking, so relative Moser in [F3] gives a correction fixed on the zero section. Composing it with $h$ yields $\mathrm{\Phi }$ and proves the claim, including noncompact closed $L$ through variable-radius neighbourhoods.