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?
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 $p_0$.]