= Solution
Write the standard complex coordinates as $z_j=q_j+ip_j$ and let $J$ denote multiplication by $i$. Use the <standard symplectic form> $\omega_0=\sum_jdq_j\wedge dp_j$, for which $\omega_0(v,Jv)=|v|^2$. In real coordinates $s,t$ on the disk, holomorphicity and the <Cauchy-Riemann equations> imply
$$
\partial_t h=J\partial_s h.
$$
Therefore the pulled-back two-form is
$$
h^*\omega_0=\omega_0(\partial_sh,\partial_th)\,ds\wedge dt=|\partial_s h|^2\,ds\wedge dt.
$$
If this integral were zero, the continuous nonnegative integrand would vanish everywhere in the interior. The displayed Cauchy-Riemann relation would then make both first derivatives zero, forcing $h$ to be constant on the connected disk and, by continuity, on its boundary. For the nonconstant map under consideration the <symplectic area> is consequently strictly positive:
$$
\int_{D^2}h^*\omega_0=\int_{D^2}|\partial_sh|^2\,ds\,dt>0.
$$
Take the primitive $\lambda=\tfrac12\sum_j(q_jdp_j-p_jdq_j)$ from Question 3. Let $c=h|_{\partial D^2}$, oriented as the boundary of the disk. Since $c$ lies in $L$, the <Stokes theorem> gives
$$
\boxed{\int_c\lambda|_L=\int_{\partial D^2}h^*\lambda=\int_{D^2}h^*\omega_0>0}.
$$
The restriction $\lambda|_L$ is closed because $L$ is a <Lagrangian submanifold>. If its <Liouville class of a Lagrangian submanifold> were zero, it would be an <exact differential form>, and its integral along every closed curve in $L$ would vanish. The boundary loop above contradicts this. \b[The Liouville class is therefore nonzero.] This proves that a <holomorphic disk detects a nonzero Liouville class> directly by positivity and Stokes; no disk-existence theorem is being assumed beyond the disk given here.
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