= Solution
The hypothesis is $|\Sigma_M|<\Sigma$, with the physical <surface density> nonnegative. At height $z>0$, the vertical derivative of an isolated thin-disk gravitational potential has a strictly positive kernel:
$$
\partial_z\Phi(\mathbf R,z)=G\int\frac{z\Sigma(\mathbf R')}{(|\mathbf R-\mathbf R'|^2+z^2)^{3/2}}\,d^2R'.
$$
The same formula with $\Sigma_M$ represents $\partial_z\Phi_M$. The <triangle inequality> and the strict <surface density> bound therefore give the <positive-kernel comparison of thin-disk fields>
$$
|\partial_z\Phi_M|\le G\int\frac{z|\Sigma_M(\mathbf R')|}{(|\mathbf R-\mathbf R'|^2+z^2)^{3/2}}\,d^2R'<\partial_z\Phi.
$$
Use $\partial_z\Phi_M=-\sqrt{4\pi G/\mu_0}\,B_z$. The integrand in part (c) is then $(\partial_z\Phi_M)^2-(\partial_z\Phi)^2<0$ in the upper vacuum region. Reflection symmetry gives the same result below; the infinitesimally thin disk has zero three-dimensional volume. Thus, with the finite-integral assumptions of the <tensor virial theorem>,
$$
\boxed{\ddot I_\perp<0.}
$$
Since the disk starts at rest, $\dot I_\perp=0$ initially and its radial second moment begins to decrease. This is <magnetic subcriticality of a razor-thin disk>: magnetic support cannot prevent initial contraction in the global virial sense. The conclusion concerns the mass-weighted radial size, and does not by itself prove that every fluid element accelerates inward or that contraction continues indefinitely.
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