# Is this an example of a regular surface with planar point that is strictly locally convex?

Definition of locally convex:

(Definition) (Local Convexity and Curvature). A surface $S\subset \mathbb{R}^3$ is locally convex at a point $p\in S$ if there exists a neighborhood $V \subset S$ of $p$ such that $V$ is contained in one of the closed half-spaces determined by $T_p(S)$ in $\mathbb{R}^3$. If, in addition, $V$ has only one common point with $T_p(S)$, then $S$ is called strictly locally convex at $p$.

Consider the surface $S \subset \mathbb{R}^3$ generated by revolving the function $z = x^4$ about the $z$ axis.

Then this surface can be parameterized by $f(x,y) = (x,y,(x^2+y^2)^2)$.

which has zero normal curvature at $(0,0,0)$, in every direction, in particular, the principal direction, so it must be a planar point.

Yet, we know that every point is above the $z$ axis, so it's strictly locally convex.

So strictly locally convexity at a point does not demand that the principal curvature need to be non-zero.

Is this right?

• Not really my area but if you look at a 2D curve such as $y=x^{2n}$, or $y=\frac{x^{2n}}{(2n)!}$, then it is strictly locally convex at $x=0$ and can be made "arbitrary flat" as $n$ grows. Then you can revolve this curve around its symmetry axis and get a 3D surface with the same property. Commented Apr 30, 2018 at 2:48