Since $\eta$ is a holomorphic 1-form valued in $u^{*}T^{*}X$, Stokes theorem gives

$$\int_{\Sigma}\langle\bar{\partial}X_{H}\wedge\eta\rangle=\int_{\partial\Sigma}\langle X_{H},\eta\rangle,$$

where $\langle,\rangle$ stands for the pairing between $TX$ and $T^{*}X$. On $\partial\Sigma$, we can write $\eta=\omega(\cdot,Y)ds$ for some vector field $Y$ valued in $u^{*}TX$, and $s$ is any local coordinate on $\partial\Sigma$. The cokernel element condition implies $\omega(v,Y)=0$ for any $v\in u^{*}TL$, so $Y$ must in fact be valued in the Lagrangian subbundle $u^{*}TL$. Thus

$$\int_{\partial\Sigma}\langle X_{H},\eta\rangle=\int_{\partial\Sigma}\omega(X_{H},Y)ds=\int_{\partial\Sigma}dH(Y)ds.$$

We suppose for contradiction, that this pairing vanishes identically for any $H$ supported in the prescribed ball.

By the holomorphicity of $\eta$, its zeros are isolated, so without loss of generality $Y$ does not vanish in the local portion of $\partial\Sigma$ where $u$ is injective and immersed. Suppose first that $Y$ is not tangent to the image of $\Sigma$. Then we find some local function $h$ on a small ball in $X$ with $dh(Y)=1$ and $h=0$ on the local portion of $\partial\Sigma$, and another cutoff function $h_{2}\geq 0$ with $dh_{2}(Y)=0$ along $\partial\Sigma$, supported in a small ball. Taking $H=hh_{2}$, then

$$\int_{\partial\Sigma}dH(Y)ds=\int_{\partial\Sigma}h_{2}ds\neq 0.$$

This contradiction shows $Y$ is tangent to the image of $\Sigma$ in the local portion of $\partial\Sigma$. We can write $Y=f\partial_{s}$ for some local function $f$. Then requiring

$$\int_{\partial\Sigma}dH(Y)ds=\int_{\partial\Sigma}f\partial_{s}Hds=-\int_{\partial\Sigma}H\partial_{s}fds$$

for any compactly supported local function $H$, implies that $f$ is constant in the local portion of $\partial\Sigma$. Thus up to multiplying by a nonzero constant, locally

$$Y=\frac{\partial u}{\partial s}ds,\quad\eta=\omega(\cdot,\frac{\partial u}{\partial s})ds.$$

(29)

We now produce holomorphic vector fields on $\Sigma$. For holomorphic strips or polygons with $k+1\geq 3$ corners, we select one input end as $p$, and call the output $q$ as usual, and represent $\Sigma$ as a strip with $k-1$ boundary punctures. This perspective provides a natural translation vector field $\frac{\partial u}{\partial s}$, which have exponential decay along the $p,q$ ends, but may not be $L^{2}$ near the other $k-1$ ends. Instead, by thinking about the $k-1$ ends as the origin in the upper half plane model, we see

$$\frac{\partial u}{\partial s}=O(|z|^{\alpha-1}),\quad\alpha=\min\{\phi_{1}/\pi,\ldots\phi_{n}/\pi\}$$

for the characterizing angles $\phi_{1},\ldots\phi_{n}$ at the Lagrangian intersection point. The $T^{(1,0)}X$ part of $2J_{X}\frac{\partial u}{\partial s}$ is $J_{X}\frac{\partial u}{\partial s}+\sqrt{-1}\frac{\partial u}{\partial s}$. Contracting this with the $T^{*(1,0)}X\otimes T^{*(1,0)}\Sigma$ part of $\eta$ yields a 1-form on $\Sigma$

$$\zeta=\eta(J_{X}\frac{\partial u}{\partial s}+\sqrt{-1}\frac{\partial u}{\partial s})$$

which is also holomorphic, with boundary value along $\Sigma$

$$\zeta=\omega(J_{X}\frac{\partial u}{\partial s}+\sqrt{-1}\frac{\partial u}{\partial s},Y)ds=\omega(J_{X}\frac{\partial u}{\partial s},Y)ds.$$

(30)

Here $\omega(\frac{\partial u}{\partial s},Y)=0$ since both vectors satisfy the $TL$ boundary condition. Notably, the boundary condition of $\zeta$ is real valued. In the upper half plane model, the Schwartz reflection principle allows us to extend $\zeta$ meromorphically over $\mathbb{CP}^{1}$.

At any of the $k-1$ ends, since $\eta\in L^{2}$, we know by holomorphicity $|\eta|=O(|z|^{\alpha})$, so $\zeta=O(|z|^{2\alpha-1})$ in the upper half plane model, hence has no pole. At the $p,q$ ends, by the decay of the holomorphic $\frac{\partial u}{\partial s}$ and $\eta$, we likewise infer that $\zeta$ has no pole in the upper half plane model. In conclusion, the extension of $\zeta$ over $\mathbb{CP}^{1}$ has no pole, so must in fact vanish. However, by (29)(30), on a local portion of $\partial\Sigma$

$$\zeta=\omega(J_{X}\frac{\partial u}{\partial s},\frac{\partial u}{\partial s})ds\neq 0.$$

This contradiction proves the Proposition in the $k\geq 1$ case.

Finally, for the teardrop curve case $k=0$, we replace the holomorphic vector field $\frac{\partial u}{\partial s}$ by the Möbius vector fields vanishing at the corner, and the rest of the arguments are entirely similar. ∎