# Prove that there are no complete regular minimal surfaces lying above a paraboloid

Prove that there are no complete regular minimal surfaces lying above a paraboloid contained in $U=\{(x,y,z) \in \mathbb{R}^3 : a(x^2+y^2)<z\}$. Here $a>0$.

I've had this problem on my mind for some time and I can't seem to find a way to conclude in a non-dubious way. Intuitively I feel that such a surface would get too close to the boundary of $U$ and would go out of it because of its Gaussian curvature always being negative, or in fact there is perhaps a point where the curvature has to be positive, but I can't seem to show it correctly. I thought that if I considered a point on the boundary of $U$ as I can find a ball of radius $\varepsilon$ say, where there are no points of the surface, (because the surface is complete and therefore a closed set), then I might be able to come to some contradiction but I just can't get it to work, I mean all I can see is that the surface would have to curve around that ball (and would probably still get closer to the boundary of U), but it doesn't help me to conclude. What am I missing?

• yes sorry that I'll add that to OP.
– Jack
Oct 15 '14 at 5:43

Consider the function $f(x,y,z)=z-\frac{a}{2}(x^2+y^2)$. Its sublevel sets on $S$, that is $L_t= \{p\in S: f(p)\le t \}$, are compact since along any sequence going into infinity $f$ tends to $+\infty$. Thus, $f$ attains its minimum $m$ on $S$ at some point $p_0$.

This means that $S$ lies inside the paraboloid $z\ge m+\frac{a}{2}(x^2+y^2)$ and is internally tangent to it at $p_0$. Since the paraboloid has positive curvature everywhere, $S$ has positive curvature at $p_0$, which is a contradiction.

[To make the last sentence precise, represent $S$ as a graph $w=g(u,v)$ in coordinate system $(u,v,w)$ where the $uv$-plane is the tangent plane at $p_0$.]

• You can't do that along $S$.
– user147263
Oct 16 '14 at 15:47
• And therefore there's a minimum on each of these sets and therefore a global minimum is that right?
– Jack
Oct 16 '14 at 15:56
• Yes, you just need one nonempty compact sublevel sets; the global minimum can't be anywhere else, and it's attained there by compactness.
– user147263
Oct 16 '14 at 15:57
• @Jack: Are you sure $L_t$ is going to be compact? What if our surface is obtained by rotating a 'wild' sine curve such as $\gamma: \mathbb{R}^+\to \mathbb{R}^3$ given by $$\gamma(u)=\left(\frac{a}{2}+\frac{a}{4}\sin(1/u),\,0,\,a+u\right)$$ around the z-axis (plot from Maple)? As far as I can see, we need $S$ to be closed (rather than merely complete) to get compactness of $L_t$ and for user147263's strategy to work. Oct 12 '17 at 22:53
• And geodescially complete doesn't imply closed, as noted by @orangeskid in this thread. Oct 12 '17 at 22:59