# Mean curvature

1. Feb 16, 2012

### hedipaldi

Hi,
I know that the mean curvature at an extremum point where the function vanishes must be nonpositive.can this say someting about the sign of the mean curvature at the farthest point on a close surface from the origin?
Thank's
Hedi

2. Feb 17, 2012

### lavinia

You need to explain what you are talking about more clearly. What function?

3. Feb 17, 2012

### hedipaldi

I mean a real smooth function of two variables whose graph is a closed surface in R3

4. Feb 17, 2012

### lavinia

so you mean the mean curvature of the graph?

Isn't the mean curvature of the standard sphere strictly positive - in fact for any surface of positive Gauss curvature?

There is a therem that says that any closed surface in 3 space must have a point of positive Gauss curvature. At this point the mean curvature is positive.

Last edited: Feb 17, 2012
5. Feb 17, 2012

### hedipaldi

But Gauss curvature is positive also if the two principal curvatures are negative.at the farthest point from the origin, the surface is enclosed within a ball whose boundary shares a common tangent plane with the surface at this point.It seems that this implies that the mean curvature of the surface at this point must have a definite sign.I don't know what sign and how to explain it.

6. Feb 17, 2012

### lavinia

right. My mistake. But for a sphere or an ellipsoid or any convex surface of positive curvature, the principal curvatures should both be positive. Yes?.

I am having trouble visualizing the case of both negative principal curvatures. May it can happen at a single point but in a region?can you give an example?

7. Feb 17, 2012

### hedipaldi

The surface must be convex at the farthest point so if this forces positive mean corvature'we are done,but i am not sure of it.

8. Feb 17, 2012

### hedipaldi

At a saddle point Gauss curvature is negative so one of the principal directions is negative.this may help for imagining a negative principal direction.

9. Feb 17, 2012

### lavinia

Ok. Now I see what your question is.

A classical argument says - surround the surface with a very large sphere centered at the origin and let it's radius shrink until it first touches the surface.At this point the surface and the sphere are tangent and the entire surface lies on the inside of the sphere. Therefore the surface must be convex at this point.

10. Feb 18, 2012

### hedipaldi

Right.Doed it implies something about the sign of the mean curvature at the farthest point?

11. Feb 18, 2012

### lavinia

I think so. In order for the surface to be tangent it must curve away from the surrounding sphere in all directions so the principal curvatures must both be positive.

Last edited: Feb 18, 2012