Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 24H / b

Write the plane as $\{x:\nu \cdot x=c\}$ for a fixed unit vector $\nu$, and put $h(x)=\nu \cdot x-c$. Suppose $X$ misses the plane. Choose a point of $X$ and a number $A$ larger than its distance to the plane. The hypothesis $D_n\to \infty$ makes $\{x\in X:|h(x)|\le A\}$ bounded, and closedness of $X$ makes it compact. Thus $|h|$ attains its global minimum at some $p\in X$, with $|h(p)|>0$.

Locally the sign of $h$ is constant, and $h$ has a local minimum or maximum at $p$. Its surface gradient vanishes, so the tangent plane of $X$ there is parallel to the given plane and its normal is $\pm \nu$. For a surface tangent vector, the intrinsic Hessian of $h$ is the normal component of the second fundamental form, namely ${\mathrm{Hess}}_{X}h=\pm b$ at $p$. The Hessian is semidefinite because this is a local extremum. Since $X$ is a minimal surface, ${\mathrm{tr}}_{g}b=2H=0$. A semidefinite symmetric form with zero trace must be zero, so $b(p)=0$. This makes $p$ a planar point, contradicting the assumption. Therefore $\boxed{X\cap \Pi \ne \varnothing }$.

The boundaryless hypothesis ensures that the minimizing point is an interior surface point. The condition on $D_n$ ensures that the infimum is attained; disjointness from a plane alone would not do so.